package com.kotaku.camerax2leetcode

import org.junit.Assert.*
import org.junit.Test
import java.util.*

/**
 * Example local unit test, which will execute on the development machine (host).
 *
 * See [testing documentation](http://d.android.com/tools/testing).
 */
class ExampleUnitTest {
    @Test
    fun leetcode1431() {
        /**
        There are n kids with candies. You are given an integer array candies,
        where each candies[i] represents the number of candies the ith kid has,
        and an integer extraCandies, denoting the number of extra candies that you have.
        Return a boolean array result of length n, where result[i] is true if,
        after giving the ith kid all the extraCandies, they will have the greatest number of candies among all the kids,
        or false otherwise.
        Note that multiple kids can have the greatest number of candies.

        Input: candies = [2,3,5,1,3], extraCandies = 3
        Output: [true,true,true,false,true]
        Explanation: If you give all extraCandies to:
        - Kid 1, they will have 2 + 3 = 5 candies, which is the greatest among the kids.
        - Kid 2, they will have 3 + 3 = 6 candies, which is the greatest among the kids.
        - Kid 3, they will have 5 + 3 = 8 candies, which is the greatest among the kids.
        - Kid 4, they will have 1 + 3 = 4 candies, which is not the greatest among the kids.
        - Kid 5, they will have 3 + 3 = 6 candies, which is the greatest among the kids.
         */
        fun kidsWithCandies(candies: IntArray, extraCandies: Int): List<Boolean> {
            val ret = MutableList(candies.size) { false }
            val max = candies.max()
            println("candies.size is ${candies.size}, extraCandies is $extraCandies, max is $max")

            for (i in candies.indices) {
                if (candies[i] + extraCandies >= max) {
                    ret[i] = true
                }
            }

            return ret
        }

        val ret = kidsWithCandies(intArrayOf(2, 3, 5, 1, 3), 3)
        ret.forEachIndexed { i, v ->
            println("ret[$i] = $v")
        }
    }

    @Test
    fun leetcode605() {
        /**
        You have a long flowerbed in which some of the plots are planted, and some are not.
        However, flowers cannot be planted in adjacent plots.
        Given an integer array flowerbed containing 0's and 1's, where 0 means empty and 1 means not empty,
        and an integer n, return true if n new flowers can be planted in the flowerbed without violating the no-adjacent-flowers rule and false otherwise.

        Example 1:
        Input: flowerbed = [1,0,0,0,1], n = 1
        Output: true

        Example 2:
        Input: flowerbed = [1,0,0,0,1], n = 2
        Output: false
         */
        fun canPlaceFlowers(flowerbed: IntArray, n: Int): Boolean {
            if (flowerbed.size == 1) {
                return if (n > 1) false
                else if (n == 1) return flowerbed[0] == 0
                else return true
            }
            var can = 0
            for (i in flowerbed.indices) {
                if (flowerbed[i] == 1) continue
                val leftEmpty = if (i == 0) true else flowerbed[i - 1] == 0

                val rightEmpty = if (i == flowerbed.size - 1) true else flowerbed[i + 1] == 0

                if (leftEmpty && rightEmpty) {
                    flowerbed[i] = 1
                    can++
                }
            }
            return can >= n
        }

        val ret = canPlaceFlowers(intArrayOf(1, 0, 0, 0, 1), 2)
        println("ret is $ret")
    }

    @Test
    fun leetcode345() {
        /**
        Given a string s, reverse only all the vowels in the string and return it.
        The vowels are 'a', 'e', 'i', 'o', and 'u', and they can appear in both lower and upper cases, more than once.

        Example 1:
        Input: s = "hello"
        Output: "holle"

        Example 2:
        Input: s = "leetcode"
        Output: "leotcede"
         */

        fun reverseVowels(s: String): String {
            val vowels = listOf('a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U')
            val chars = s.toCharArray()
            var left = 0
            var right = chars.size - 1

            while (left < right) {
                if (chars[left] in vowels && chars[right] in vowels) {
                    val temp = chars[left]
                    chars[left] = chars[right]
                    chars[right] = temp
                    left++
                    right--
                } else if (chars[left] in vowels) {
                    right--
                } else {
                    left++
                }
            }

            return String(chars)
        }

        val ret = reverseVowels("leetcode")
        println("ret is $ret")

    }

    @Test
    fun leetcode151() {
        /**
        Given an input string s, reverse the order of the words.
        A word is defined as a sequence of non-space characters. The words in s will be separated by at least one space.
        Return a string of the words in reverse order concatenated by a single space.
        Note that s may contain leading or trailing spaces or multiple spaces between two words.
        The returned string should only have a single space separating the words. Do not include any extra spaces.

        Example 1:
        Input: s = "the sky is blue"
        Output: "blue is sky the"

        Example 2:
        Input: s = "  hello world  "
        Output: "world hello"
        Explanation: Your reversed string should not contain leading or trailing spaces.

        Example 3:
        Input: s = "a good   example"
        Output: "example good a"
        Explanation: You need to reduce multiple spaces between two words to a single space in the reversed string.
         **/
        fun reverseWords(s: String): String {
            //return s.trim().split(" ").reversed().filter { it.trim().isNotEmpty() }.joinToString(separator = " ")

            val wordsArray = s.trim().split(" ").reversed()
            wordsArray.forEach {
                println("words is $it ;")
            }

            return wordsArray.filter { it.trim().isNotEmpty() }.joinToString(separator = " ")
        }

        val ret = reverseWords("a good   example")
        println("ret is $ret")
    }

    @Test
    fun leetcode334() {
        /**
        Given an integer array nums, return true if there exists a triple of indices (i, j, k)
        such that i < j < k and nums[i] < nums[j] < nums[k]. If no such indices exists, return false.

        Example 1:
        Input: nums = [1,2,3,4,5]
        Output: true
        Explanation: Any triplet where i < j < k is valid.

        Example 2:
        Input: nums = [5,4,3,2,1]
        Output: false
        Explanation: No triplet exists.

        Example 3:
        Input: nums = [2,1,5,0,4,6]
        Output: true
        Explanation: The triplet (3, 4, 5) is valid because nums[3] == 0 < nums[4] == 4 < nums[5] == 6.
         */
        fun increasingTriplet(nums: IntArray): Boolean {
            var small = Int.MAX_VALUE
            var big = Int.MAX_VALUE

            for (num in nums) {
                if (num <= small) {
                    small = num
                } else if (num <= big) {
                    big = num
                } else {
                    return true
                }

                println("small is $small ; big is $big ")
            }

            return false
        }

        val ret = increasingTriplet(intArrayOf(2, 1, 5, 0, 4, 6))
        println("ret is $ret")
    }

    @Test
    fun leetcode443() {
        /**
        Given an array of characters chars, compress it using the following algorithm:
        Begin with an empty string s. For each group of consecutive repeating characters in chars:

        If the group's length is 1, append the character to s.
        Otherwise, append the character followed by the group's length.
        The compressed string s should not be returned separately, but instead, be stored in the input character array chars.
        Note that group lengths that are 10 or longer will be split into multiple characters in chars.

        After you are done modifying the input array, return the new length of the array.
        You must write an algorithm that uses only constant extra space.

        Example 1:
        Input: chars = ["a","a","b","b","c","c","c"]
        Output: Return 6, and the first 6 characters of the input array should be: ["a","2","b","2","c","3"]
        Explanation: The groups are "aa", "bb", and "ccc". This compresses to "a2b2c3".
         */
        fun compress(chars: CharArray): Int {
            var writeIndex = 0
            var readIndex = 0

            while (readIndex < chars.size) {
                val currentChar = chars[readIndex]
                var count = 0

                while (readIndex < chars.size && chars[readIndex] == currentChar) {
                    readIndex++
                    count++
                }

                chars[writeIndex] = currentChar
                writeIndex++

                if (count > 1) {
                    val countChars = count.toString().toCharArray()
                    for (i in countChars.indices) {
                        chars[writeIndex] = countChars[i]
                        writeIndex++
                    }
                }
            }

            printCharArray(chars)

            return writeIndex
        }

        val ret =
            compress(charArrayOf('a', 'b', 'b', 'b', 'b', 'b', 'b', 'b', 'b', 'b', 'b', 'b', 'b'))
        println("ret is $ret")

    }

    @Test
    fun leetcode283() {
        /**
        Given an integer array nums, move all 0's to the end of it while maintaining the relative order of the non-zero elements.
        Note that you must do this in-place without making a copy of the array.

        Example 1:
        Input: nums = [0,1,0,3,12]
        Output: [1,3,12,0,0]

        Example 2:
        Input: nums = [0]
        Output: [0]
         */
        fun moveZeroes(nums: IntArray): IntArray {
            if (nums.size == 1) {
                return nums
            }

            var left = 0
            var right = 0

            while (left < nums.size && nums[left] != 0) {
                left++
                right++
            }

            while (right < nums.size) {
                if (nums[right] == 0) {
                    right++
                } else {
                    nums[left++] = nums[right]
                    nums[right++] = 0
                }
            }

            return nums
        }

        val ret = moveZeroes(intArrayOf(0, 1, 0, 3, 12))
        printIntArray(ret)

    }

    @Test
    fun leetcode392() {
        /***
        Given two strings s and t, return true if s is a subsequence of t, or false otherwise.
        A subsequence of a string is a new string that is formed from the original string by deleting some (can be none) of
        the characters without disturbing the relative positions of the remaining characters.
        (i.e., "ace" is a subsequence of "abcde" while "aec" is not).
        Example 1:
        Input: s = "abc", t = "ahbgdc"
        Output: true

        Example 2:
        Input: s = "axc", t = "ahbgdc"
        Output: false
         */

        fun isSubsequence(s: String, t: String): Boolean {
            var i = 0
            var j = 0

            while (i < s.length && j < t.length) {
                if (s[i] == t[j]) {
                    i++
                }
                j++
            }

            return i == s.length
        }

        val ret = isSubsequence("abc", "ahbgdc")
        println("ret is $ret")
    }

    @Test
    fun leetcode11() {
        /**
        You are given an integer array height of length n.
        There are n vertical lines drawn such that the two endpoints of the ith line are (i, 0) and (i, height[i]).
        Find two lines that together with the x-axis form a container, such that the container contains the most water.
        Return the maximum amount of water a container can store.

        Example 1:
        Input: height = [1,8,6,2,5,4,8,3,7]
        Output: 49
        Explanation: The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7].
        In this case, the max area of water (blue section) the container can contain is 49.
         */

        fun maxArea(height: IntArray): Int {
            var left = 0
            var right = height.size - 1
            var maxArea = 0

            while (left <= right) {
                val area = height[left].coerceAtMost(height[right]) * (right - left)
                if (height[right] > height[left]) {
                    left++
                } else {
                    right--
                }
                maxArea = maxArea.coerceAtLeast(area)
            }

            return maxArea
        }

        val ret = maxArea(intArrayOf(1, 8, 6, 2, 5, 4, 8, 3, 7))
        println("ret is $ret")
    }

    @Test
    fun leetcode1679() {
        /**
        You are given an integer array nums and an integer k.
        In one operation, you can pick two numbers from the array whose sum equals k and remove them from the array.
        Return the maximum number of operations you can perform on the array.

        Example 1:
        Input: nums = [1,2,3,4], k = 5
        Output: 2
        Explanation: Starting with nums = [1,2,3,4]:
        - Remove numbers 1 and 4, then nums = [2,3]
        - Remove numbers 2 and 3, then nums = []
        There are no more pairs that sum up to 5, hence a total of 2 operations.

        Example 2:
        Input: nums = [3,1,3,4,3], k = 6
        Output: 1
        Explanation: Starting with nums = [3,1,3,4,3]:
        - Remove the first two 3's, then nums = [1,4,3]
        There are no more pairs that sum up to 6, hence a total of 1 operation.
         */
        fun maxOperations(nums: IntArray, k: Int): Int {
            nums.sort()
            var left = 0
            var right = nums.size - 1
            var ret = 0

            while (left < right) {
                if (nums[left] + nums[right] == k) {
                    ret++
                    left++
                    right--
                } else if (nums[left] + nums[right] < k) {
                    left++
                } else if (nums[left] + nums[right] > k) {
                    right--
                }
            }

            return ret
        }

        val ret = maxOperations(intArrayOf(1, 2, 3, 4), 5)
        println("ret is $ret")
    }

    @Test
    fun leetcode643() {
        /***
        You are given an integer array nums consisting of n elements, and an integer k.
        Find a contiguous subarray whose length is equal to k that has the maximum average value and return this value.
        Any answer with a calculation error less than 10-5 will be accepted.

        Example 1:
        Input: nums = [1,12,-5,-6,50,3], k = 4
        Output: 12.75000
        Explanation: Maximum average is (12 - 5 - 6 + 50) / 4 = 51 / 4 = 12.75

        Example 2:
        Input: nums = [5], k = 1
        Output: 5.00000
         */
        fun findMaxAverage(nums: IntArray, k: Int): Double {
            var sum = 0
            for (i in 0 until k) {
                sum += nums[i]
            }
            var maxSum = sum

            for (i in k until nums.size) {
                sum += nums[i] - nums[i - k]
                maxSum = maxOf(maxSum, sum)
            }

            return maxSum.toDouble() / k.toDouble()
        }

        val ret = findMaxAverage(intArrayOf(1, 12, -5, -6, 50, 3), 4)
        println("ret is $ret")
    }

    @Test
    fun leetcode1456() {
        /**
        Given a string s and an integer k, return the maximum number of vowel letters in any substring of s with length k.
        Vowel letters in English are 'a', 'e', 'i', 'o', and 'u'.

        Example 1:
        Input: s = "abciiidef", k = 3
        Output: 3
        Explanation: The substring "iii" contains 3 vowel letters.

        Example 2:
        Input: s = "aeiou", k = 2
        Output: 2
        Explanation: Any substring of length 2 contains 2 vowels.

        Example 3:
        Input: s = "leetcode", k = 3
        Output: 2
        Explanation: "lee", "eet" and "ode" contain 2 vowels.*/
        fun maxVowels(s: String, k: Int): Int {
            var sum = 0
            val charList = charArrayOf('a', 'e', 'i', 'o', 'u')
            for (i in 0 until k) {
                if (charList.contains(s[i])) sum += 1
            }
            var maxSum = sum

            for (i in k until s.length) {
                if (charList.contains(s[i])) sum += 1
                if (charList.contains(s[i - k])) sum -= 1
                maxSum = maxOf(maxSum, sum)
            }

            return maxSum
        }

        val ret = maxVowels("abciiidef", 3)
        println("ret is $ret")
    }

    @Test
    fun leetcode1004() {
        /**
        Given a binary array nums and an integer k, return the maximum number of consecutive 1's in the array
        if you can flip at most k 0's.

        Example 1:
        Input: nums = [1,1,1,0,0,0,1,1,1,1,0], k = 2
        Output: 6
        Explanation: [1,1,1,0,0,1,1,1,1,1,1]
        Bolded numbers were flipped from 0 to 1. The longest subarray is underlined.

        Example 2:
        Input: nums = [0,0,1,1,0,0,1,1,1,0,1,1,0,0,0,1,1,1,1], k = 3
        Output: 10
        Explanation: [0,0,1,1,1,1,1,1,1,1,1,1,0,0,0,1,1,1,1]
        Bolded numbers were flipped from 0 to 1. The longest subarray is underlined.
         * */
        fun longestOnes(nums: IntArray, k: Int): Int {
            var maxCount = 0
            var left = 0
            var zerosCount = 0

            for (right in nums.indices) {
                if (nums[right] == 0) {
                    zerosCount++
                }

                while (zerosCount > k) {
                    if (nums[left] == 0) {
                        zerosCount--
                    }
                    left++
                }

                maxCount = maxCount.coerceAtLeast(right - left + 1)
                println("right is $right, left is $left, zerosCount is $zerosCount, maxCount is $maxCount")
            }

            return maxCount
        }

        val ret = longestOnes(intArrayOf(1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0), 2)
        println("ret is $ret")
    }

    @Test
    fun leetcode1493() {
        /***
        Given a binary array nums, you should delete one element from it.
        Return the size of the longest non-empty subarray containing only 1's in the resulting array.
        Return 0 if there is no such subarray.

        Example 1:
        Input: nums = [1,1,0,1]
        Output: 3
        Explanation: After deleting the number in position 2, [1,1,1] contains 3 numbers with value of 1's.

        Example 2:
        Input: nums = [0,1,1,1,0,1,1,0,1]
        Output: 5
        Explanation: After deleting the number in position 4, [0,1,1,1,1,1,0,1] longest subarray with value of 1's is [1,1,1,1,1].

        Example 3:
        Input: nums = [1,1,1]
        Output: 2
        Explanation: You must delete one element.
         */
        fun longestSubarray(nums: IntArray): Int {
            var maxLength = 0
            var countZeros = 0
            var left = 0

            for (right in nums.indices) {
                if (nums[right] == 0) {
                    countZeros++
                }

                while (countZeros > 1) {
                    if (nums[left] == 0) {
                        countZeros--
                    }
                    left++
                }

                maxLength = maxOf(maxLength, right - left)
            }

            return maxLength
        }

        val ret = longestSubarray(intArrayOf(0, 1, 1, 1, 0, 1, 1, 0, 1))
        println("ret is $ret")
    }

    @Test
    fun leetcode1732() {
        /**
        There is a biker going on a road trip. The road trip consists of n + 1 points at different altitudes.
        The biker starts his trip on point 0 with altitude equal 0.
        You are given an integer array gain of length n where gain[i] is the net gain in altitude
        between points i​​​​​​ and i + 1 for all (0 <= i < n). Return the highest altitude of a point.

        Example 1:
        Input: gain = [-5,1,5,0,-7]
        Output: 1
        Explanation: The altitudes are [0,-5,-4,1,1,-6]. The highest is 1.

        Example 2:
        Input: gain = [-4,-3,-2,-1,4,3,2]
        Output: 0
        Explanation: The altitudes are [0,-4,-7,-9,-10,-6,-3,-1]. The highest is 0.
         */
        fun largestAltitude(gain: IntArray): Int {
            var prix = 0
            var max = 0
            gain.forEach {
                val current = prix + it
                max = max.coerceAtLeast(current)
                prix = current
            }
            return max
        }

        var ret = largestAltitude(intArrayOf(-5, 1, 5, 0, -7))
        println("ret is $ret")

        ret = largestAltitude(intArrayOf(-4, -3, -2, -1, 4, 3, 2))
        println("ret is $ret")
    }

    @Test
    fun leetcode2215() {
        /**
        Given two 0-indexed integer arrays nums1 and nums2, return a list answer of size 2 where:
        answer[0] is a list of all distinct integers in nums1 which are not present in nums2.
        answer[1] is a list of all distinct integers in nums2 which are not present in nums1.
        Note that the integers in the lists may be returned in any order.

        Example 1:
        Input: nums1 = [1,2,3], nums2 = [2,4,6]
        Output: [[1,3],[4,6]]

        Example 2:
        Input: nums1 = [1,2,3,3], nums2 = [1,1,2,2]
        Output: [[3],[]]
         */
        fun arrayDiff(array1: IntArray, array2: IntArray): List<Int> {
            val set1 = array1.toSet()
            val set2 = array2.toSet()

            val difference = set1.subtract(set2)
            return difference.toList()
        }

        fun findDifference(nums1: IntArray, nums2: IntArray): List<List<Int>> {
            return listOf(arrayDiff(nums1, nums2), arrayDiff(nums2, nums1))
        }

        val ret = findDifference(intArrayOf(1, 2, 3, 3), intArrayOf(1, 1, 2, 2))
        ret.forEachIndexed { i1, array ->
            array.forEachIndexed { i2, v ->
                println("Int[$i1][$i2] = $v")
            }
        }
    }

    @Test
    fun leetcode1207() {
        /**
        Given an array of integers arr, return true if the number of occurrences of each value in the array is unique or false otherwise.
        Example 1:
        Input: arr = [1,2,2,1,1,3]
        Output: true
        Explanation: The value 1 has 3 occurrences, 2 has 2 and 3 has 1. No two values have the same number of occurrences.

        Example 2:
        Input: arr = [1,2]
        Output: false

        Example 3:
        Input: arr = [-3,0,1,-3,1,1,1,-3,10,0]
        Output: true
         */
        fun uniqueOccurrences(arr: IntArray): Boolean {
            var ret = true
            val set1 = arr.toSet()
            val minLength = (1..set1.size).sum()
            if (arr.size >= minLength) {
                val mutableSet = mutableSetOf<Int>()
                set1.forEach {
                    println("key = $it")
                    val arrCount = arr.count { v ->
                        v == it
                    }
                    if (mutableSet.contains(arrCount)) {
                        ret = false
                        return@forEach
                    } else {
                        mutableSet.add(arrCount)
                    }
                }
            } else {
                ret = false
            }
            return ret
        }

        var ret = uniqueOccurrences(intArrayOf(-3, 0, 1, -3, 1, 1, 1, -3, 10, 0))
        println("Boolean = $ret")

        ret = uniqueOccurrences(intArrayOf(1, 2, 2, 1, 1, 3))
        println("Boolean = $ret")

        ret = uniqueOccurrences(
            intArrayOf(
                26,
                2,
                16,
                16,
                5,
                5,
                26,
                2,
                5,
                20,
                20,
                5,
                2,
                20,
                2,
                2,
                20,
                2,
                16,
                20,
                16,
                17,
                16,
                2,
                16,
                20,
                26,
                16
            )
        )
        println("Boolean = $ret")
    }

    @Test
    fun leetcode1657() {
        /***
        Two strings are considered close if you can attain one from the other using the following operations:

        Operation 1: Swap any two existing characters.
        For example, abcde -> aecdb
        Operation 2: Transform every occurrence of one existing character into another existing character, and do the same with the other character.
        For example, aacabb -> bbcbaa (all a's turn into b's, and all b's turn into a's)
        You can use the operations on either string as many times as necessary.

        Given two strings, word1 and word2, return true if word1 and word2 are close, and false otherwise.

        Example 1:
        Input: word1 = "abc", word2 = "bca"
        Output: true
        Explanation: You can attain word2 from word1 in 2 operations.
        Apply Operation 1: "abc" -> "acb"
        Apply Operation 1: "acb" -> "bca"

        Example 2:
        Input: word1 = "a", word2 = "aa"
        Output: false
        Explanation: It is impossible to attain word2 from word1, or vice versa, in any number of operations.

        Example 3:
        Input: word1 = "cabbba", word2 = "abbccc"
        Output: true
        Explanation: You can attain word2 from word1 in 3 operations.
        Apply Operation 1: "cabbba" -> "caabbb"
        Apply Operation 2: "caabbb" -> "baaccc"
        Apply Operation 2: "baaccc" -> "abbccc"
         */
        fun closeStrings(word1: String, word2: String): Boolean {
            if (word1.length != word2.length) {
                return false
            }

            val freq1 = IntArray(26)
            val freq2 = IntArray(26)

            val set1 = mutableSetOf<Char>()
            val set2 = mutableSetOf<Char>()

            for (i in word1.indices) {
                freq1[word1[i] - 'a']++
                freq2[word2[i] - 'a']++
                set1.add(word1[i])
                set2.add(word2[i])
            }

            freq1.sort()
            freq2.sort()

            return freq1.contentEquals(freq2) && set1 == set2
        }

        val ret = closeStrings("cabbba", "abbccc")
        println("Boolean = $ret")
    }

    @Test
    fun leetcode2352() {
        /**
        Given a 0-indexed n x n integer matrix grid, return the number of pairs (ri, cj) such that row ri and column cj are equal.
        A row and column pair is considered equal if they contain the same elements in the same order (i.e., an equal array).
        Example 1:
        Input: grid = [[3,2,1],[1,7,6],[2,7,7]]
        Output: 1
        Explanation: There is 1 equal row and column pair:
        - (Row 2, Column 1): [2,7,7]

        Example 2:
        Input: grid = [[3,1,2,2],[1,4,4,5],[2,4,2,2],[2,4,2,2]]
        Output: 3
        Explanation: There are 3 equal row and column pairs:
        - (Row 0, Column 0): [3,1,2,2]
        - (Row 2, Column 2): [2,4,2,2]
        - (Row 3, Column 2): [2,4,2,2]
         */
        fun equalPairs(grid: Array<IntArray>): Int {
            var ret = 0
            val rowStrings = mutableListOf<String>()
            val columnStrings = mutableListOf<String>()
            grid.forEachIndexed { i, array ->
                rowStrings.add(i, array.joinToString(separator = "-") + "-")
            }

            for (i in grid.indices) {
                var colum = ""
                grid.forEach {
                    colum = colum + it[i] + "-"
                }
                columnStrings.add(i, colum)
            }

            printStringArray(rowStrings)
            printStringArray(columnStrings)

            rowStrings.forEach { row ->
                columnStrings.forEach { column ->
                    if(row == column) ret++
                }
            }

            return ret
        }

        // [[3,1,2,2],[1,4,4,5],[2,4,2,2],[2,4,2,2]]
        var ret = equalPairs(
            arrayOf(
                intArrayOf(3, 1, 2, 2),
                intArrayOf(1, 4, 4, 5),
                intArrayOf(2, 4, 2, 2),
                intArrayOf(2, 4, 2, 2)
            )
        )
        println("ret = $ret")

        ret = equalPairs(
            arrayOf(
                intArrayOf(11,1),
                intArrayOf(1, 11)
            )
        )
        println("ret = $ret")
    }

    @Test
    fun leetcode2390() {
        fun removeStars(s: String): String {
            /**
            You are given a string s, which contains stars *.
            In one operation, you can:
            Choose a star in s.
            Remove the closest non-star character to its left, as well as remove the star itself.
            Return the string after all stars have been removed.
            Note:
            The input will be generated such that the operation is always possible.
            It can be shown that the resulting string will always be unique.

            Example 1:
            Input: s = "leet**cod*e"
            Output: "lecoe"

            Example 2:
            Input: s = "erase*****"
            Output: ""
            Explanation: The entire string is removed, so we return an empty string.
             */
            val stack = Stack<Char>()
            // 将字符串中的字符依次入栈
            for (char in s) {
                if(char != '*') stack.push(char)
                else stack.pop()
            }
            return stack.joinToString("")
        }

        val ret = removeStars("erase*****")
        println("ret is $ret")
    }

    @Test
    fun leetcode735() {
        /**
        We are given an array asteroids of integers representing asteroids in a row.
        For each asteroid, the absolute value represents its size, and the sign represents its direction
        (positive meaning right, negative meaning left). Each asteroid moves at the same speed.
        Find out the state of the asteroids after all collisions. If two asteroids meet, the smaller one will explode.
        If both are the same size, both will explode. Two asteroids moving in the same direction will never meet.

        Example 1:
        Input: asteroids = [5,10,-5]
        Output: [5,10]
        Explanation: The 10 and -5 collide resulting in 10. The 5 and 10 never collide.

        Example 2:
        Input: asteroids = [8,-8]
        Output: []
        Explanation: The 8 and -8 collide exploding each other.

        Example 3:
        Input: asteroids = [10,2,-5]
        Output: [10]
        Explanation: The 2 and -5 collide resulting in -5. The 10 and -5 collide resulting in 10.
        */
        fun asteroidCollision(asteroids: IntArray): IntArray {
            val stack = Stack<Int>()

            for (asteroid in asteroids) {
                if (asteroid > 0) {
                    stack.push(asteroid)
                } else {
                    while (!stack.isEmpty() && stack.peek() > 0 && stack.peek() < Math.abs(asteroid)) {
                        stack.pop()
                    }

                    if (stack.isEmpty() || stack.peek() < 0) {
                        stack.push(asteroid)
                    } else if (stack.peek() == Math.abs(asteroid)) {
                        stack.pop()
                    }
                }
            }

            return stack.toIntArray()
        }

        printIntArray(asteroidCollision(intArrayOf(5,10,-5)))
        printIntArray(asteroidCollision(intArrayOf(8,-8)))
        printIntArray(asteroidCollision(intArrayOf(10,2,-5)))
    }

    @Test
    fun leetcode394() {
        /**
        Given an encoded string, return its decoded string.
        The encoding rule is: k[encoded_string], where the encoded_string inside the square brackets is being
        repeated exactly k times. Note that k is guaranteed to be a positive integer.
        You may assume that the input string is always valid; there are no extra white spaces,
        square brackets are well-formed, etc. Furthermore, you may assume that the original data does not contain
        any digits and that digits are only for those repeat numbers, k. For example, there will not be input like 3a or 2[4].
        The test cases are generated so that the length of the output will never exceed 105.

        Example 1:
        Input: s = "3[a]2[bc]"
        Output: "aaabcbc"

        Example 2:
        Input: s = "3[a2[c]]"
        Output: "accaccacc"

        Example 3:
        Input: s = "2[abc]3[cd]ef"
        Output: "abcabccdcdcdef"
         */
        fun decodeString(s: String): String {
            val stack = Stack<String>()
            var currentNumber = 0
            var currentString = ""

            for (char in s) {
                when {
                    char.isDigit() -> {
                        currentNumber = currentNumber * 10 + (char - '0')
                    }
                    char == '[' -> {
                        stack.push(currentString)
                        stack.push(currentNumber.toString())
                        currentNumber = 0
                        currentString = ""
                    }
                    char == ']' -> {
                        val number = stack.pop().toInt()
                        val previousString = stack.pop()

                        currentString = previousString + currentString.repeat(number)
                    }
                    else -> {
                        currentString += char
                    }
                }
            }

            return currentString
        }

        println("ret is ${decodeString("3[a]2[bc]")}") //"aaabcbc"
        println("ret is ${decodeString("3[a2[c]]")}") //"accaccacc"
        println("ret is ${decodeString("2[abc]3[cd]ef")}") //"abcabccdcdcdef"
        println("ret is ${decodeString("100[leetcode]")}") //"abcabccdcdcdef"

    }

    @Test
    fun leetcode328() {
        /**
        Given the head of a singly linked list, group all the nodes with odd indices together
        followed by the nodes with even indices, and return the reordered list.
        The first node is considered odd, and the second node is even, and so on.
        Note that the relative order inside both the even and odd groups should remain as it was in the input.

        Example 1:
        Input: head = [1,2,3,4,5]
        Output: [1,3,5,2,4]

        Example 2:
        Input: head = [2,1,3,5,6,4,7]
        Output: [2,3,6,7,1,5,4]
         */
        fun oddEvenList(head: ListNode?): ListNode? {
            if (head == null) {
                return null
            }

            var odd = head
            var even = head.next
            val evenHead = even

            while (even?.next != null) {
                odd?.next = even.next
                odd = odd?.next
                even.next = odd?.next
                even = even.next
            }

            odd?.next = evenHead

            return head
        }
    }

    @Test
    fun leetcode2130() {
        /**
        In a linked list of size n, where n is even, the ith node (0-indexed) of the linked list is known as
        the twin of the (n-1-i)th node, if 0 <= i <= (n / 2) - 1.
        For example, if n = 4, then node 0 is the twin of node 3, and node 1 is the twin of node 2.
        These are the only nodes with twins for n = 4.
        The twin sum is defined as the sum of a node and its twin.
        Given the head of a linked list with even length, return the maximum twin sum of the linked list.

        Example 1:
        Input: head = [5,4,2,1]
        Output: 6

        Example 2:
        Input: head = [4,2,2,3]
        Output: 7

        Example 3:
        Input: head = [1,100000]
        Output: 100001
         */
        fun pairSum(head: ListNode?): Int {
            val deque = ArrayDeque<Int>()
            var max = 0
            var curr = head

            while(curr != null) {
                deque.add(curr.`val`)
                curr = curr?.next
            }

            while(deque.isNotEmpty()) {
                val sum = deque.removeFirst() + deque.removeLast()
                if (sum > max) max = sum
            }

            return max
        }
    }

    @Test
    fun leetcode104() {
        /**
        Given the root of a binary tree, return its maximum depth.
        A binary tree's maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.

        Example 1:
        Input: root = [3,9,20,null,null,15,7]
        Output: 3

        Example 2:
        Input: root = [1,null,2]
        Output: 2
        */
        val tree3 = TreeNode(3)
        val tree9 = TreeNode(9)
        val tree20 = TreeNode(20)
        val tree15 = TreeNode(15)
        val tree7 = TreeNode(7)
        tree3.left = tree9
        tree3.right = tree20
        tree20.left = tree15
        tree20.right = tree7

        fun maxDepth(root: TreeNode?): Int {
            if (root == null) {
                return 0
            }

            val leftDepth = maxDepth(root.left)
            val rightDepth = maxDepth(root.right)

            return 1 + maxOf(leftDepth, rightDepth)
        }

        println("ret is ${maxDepth(tree3)}") //3
        assertEquals(maxDepth(tree3), 3)
    }

    @Test
    fun leetcode872() {
        /**
        Consider all the leaves of a binary tree, from left to right order, the values of those leaves form a leaf value sequence.
        For example, in the given tree above, the leaf value sequence is (6, 7, 4, 9, 8).
        Two binary trees are considered leaf-similar if their leaf value sequence is the same.
        Return true if and only if the two given trees with head nodes root1 and root2 are leaf-similar.

        Example 1:
        Input: root1 = [3,5,1,6,2,9,8,null,null,7,4], root2 = [3,5,1,6,7,4,2,null,null,null,null,null,null,9,8]
        Output: true

        Example 2:
        Input: root1 = [1,2,3], root2 = [1,3,2]
        Output: false
         */
        fun getLeaves(root: TreeNode?, leaves: MutableList<Int>) {
            if (root == null) {
                return
            }

            if (root.left == null && root.right == null) {
                leaves.add(root.`val`)
            }

            getLeaves(root.left, leaves)
            getLeaves(root.right, leaves)
        }

        fun leafSimilar(root1: TreeNode?, root2: TreeNode?): Boolean {
            val leaves1 = mutableListOf<Int>()
            val leaves2 = mutableListOf<Int>()

            getLeaves(root1, leaves1)
            getLeaves(root2, leaves2)

            return leaves1 == leaves2
        }

        println("ret is ${leafSimilar(buildTree(arrayOf(3,5,1,6,2,9,8,null,null,7,4)), 
            buildTree(arrayOf(3,5,1,6,7,4,2,null,null,null,null,null,null,9,8)))}") //true

        assertEquals(leafSimilar(buildTree(arrayOf(3,5,1,6,2,9,8,null,null,7,4)),
            buildTree(arrayOf(3,5,1,6,7,4,2,null,null,null,null,null,null,9,8))), true)

        println("ret is ${leafSimilar(buildTree(arrayOf(1,2,3)),
            buildTree(arrayOf(1,3,2)))}") //false
        assertEquals(leafSimilar(buildTree(arrayOf(1,2,3)),
            buildTree(arrayOf(1,3,2))),false)
    }

    @Test
    fun leetcode1448() {
        /**
        Given a binary tree root, a node X in the tree is named good if in the path from root to X there are no nodes with
        a value greater than X.
        Return the number of good nodes in the binary tree.

        Example 1:
        Input: root = [3,1,4,3,null,1,5]
        Output: 4
        Explanation: Nodes in blue are good.
        Root Node (3) is always a good node.
        Node 4 -> (3,4) is the maximum value in the path starting from the root.
        Node 5 -> (3,4,5) is the maximum value in the path
        Node 3 -> (3,1,3) is the maximum value in the path.

        Example 2:
        Input: root = [3,3,null,4,2]
        Output: 3
        Explanation: Node 2 -> (3, 3, 2) is not good, because "3" is higher than it.

        Example 3:
        Input: root = [1]
        Output: 1
        Explanation: Root is considered as good.
         */
        fun countGoodNodes(node: TreeNode?, maxValue: Int): Int {
            if (node == null) {
                return 0
            }

            var count = 0

            if (node.`val` >= maxValue) {
                count++
            }

            val updatedMaxValue = maxOf(maxValue, node.`val`)

            count += countGoodNodes(node.left, updatedMaxValue)
            count += countGoodNodes(node.right, updatedMaxValue)

            return count
        }

        fun goodNodes(root: TreeNode?): Int {
            return countGoodNodes(root, Int.MIN_VALUE)
        }

        println("ret is ${goodNodes(buildTree(arrayOf(3,1,4,3,null,1,5)))}") // 4
        assertEquals(goodNodes(buildTree(arrayOf(3,1,4,3,null,1,5))),4)
        println("ret is ${goodNodes(buildTree(arrayOf(3,3,null,4,2)))}") // 3
        assertEquals(goodNodes(buildTree(arrayOf(3,3,null,4,2))),3)
        println("ret is ${goodNodes(buildTree(arrayOf(1)))}") // 1
        assertEquals(goodNodes(buildTree(arrayOf(1))),1)
    }

    @Test
    fun leetcode437() {
        /**
        Given the root of a binary tree and an integer targetSum, return the number of paths where the sum of the values along the path equals targetSum.
        The path does not need to start or end at the root or a leaf, but it must go downwards (i.e., traveling only from parent nodes to child nodes).
        Example 1:
        Input: root = [10,5,-3,3,2,null,11,3,-2,null,1], targetSum = 8
        Output: 3
        Explanation: The paths that sum to 8 are shown.

        Example 2:
        Input: root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
        Output: 3
         */
        fun countPathSum(node: TreeNode?, targetSum: Int): Int {
            if (node == null) {
                return 0
            }

            var count = 0

            // 检查以当前节点为起点的路径和是否等于目标和
            if (node.`val` == targetSum) {
                count++
            }

            // 递归搜索左子树和右子树
            count += countPathSum(node.left, targetSum - node.`val`) + countPathSum(node.right, targetSum - node.`val`)

            return count
        }

        fun pathSum(root: TreeNode?, targetSum: Int): Int {
            if (root == null) {
                return 0
            }

            // 从根节点开始的路径和数量 + 左子树中的路径和数量 + 右子树中的路径和数量
            val pathCount = countPathSum(root, targetSum) + pathSum(root.left, targetSum) + pathSum(root.right, targetSum)
            return pathCount
        }

        println("ret is ${pathSum(buildTree(arrayOf(10,5,-3,3,2,null,11,3,-2,null,1)), 8)}") // 3
        assertEquals(pathSum(buildTree(arrayOf(10,5,-3,3,2,null,11,3,-2,null,1)), 8),3)
        println("ret is ${pathSum(buildTree(arrayOf(5,4,8,11,null,13,4,7,2,null,null,5,1)), 22)}") // 3
        assertEquals(pathSum(buildTree(arrayOf(5,4,8,11,null,13,4,7,2,null,null,5,1)), 22),3)
        // bellow test failed, tree item value is greater than int.max. See leetcode437_solution2
        //println("ret is ${pathSum(buildTree(arrayOf(1000000000,1000000000,null,294967296,null,1000000000,null,1000000000,null,1000000000)), 0)}") // 0
        //assertEquals(pathSum(buildTree(arrayOf(1000000000,1000000000,null,294967296,null,1000000000,null,1000000000,null,1000000000)), 0),0)
    }

    @Test
    fun leetcode437_solution2() {
        fun pathSumFromRoot(root: TreeNode?, targetSum: Long): Int {
            if (root == null) return 0

            val match = if (targetSum == root.`val`.toLong()) 1 else 0
            return pathSumFromRoot(root.left, targetSum - root.`val`) + pathSumFromRoot(root.right, targetSum - root.`val`) + match
        }

        fun pathSum(root: TreeNode?, targetSum: Int): Int {
            if(root == null) return 0
            return pathSumFromRoot(root, targetSum.toLong()) + pathSum(root.left, targetSum) + pathSum(root.right, targetSum)
        }

        println("ret is ${pathSum(buildTree(arrayOf(10,5,-3,3,2,null,11,3,-2,null,1)), 8)}") // 3
        assertEquals(pathSum(buildTree(arrayOf(10,5,-3,3,2,null,11,3,-2,null,1)), 8), 3)
        println("ret is ${pathSum(buildTree(arrayOf(5,4,8,11,null,13,4,7,2,null,null,5,1)), 22)}") // 3
        assertEquals(pathSum(buildTree(arrayOf(5,4,8,11,null,13,4,7,2,null,null,5,1)), 22), 3)
        println("ret is ${pathSum(buildTree(arrayOf(1000000000,1000000000,null,294967296,null,1000000000,null,1000000000,null,1000000000)), 0)}") // 0
        assertEquals(pathSum(buildTree(arrayOf(1000000000,1000000000,null,294967296,null,1000000000,null,1000000000,null,1000000000)), 0),0)

    }

    @Test
    fun leetcode1372() {
        /**
        You are given the root of a binary tree.
        A ZigZag path for a binary tree is defined as follow:
        Choose any node in the binary tree and a direction (right or left).
        If the current direction is right, move to the right child of the current node; otherwise, move to the left child.
        Change the direction from right to left or from left to right.
        Repeat the second and third steps until you can't move in the tree.
        Zigzag length is defined as the number of nodes visited - 1. (A single node has a length of 0).

        Return the longest ZigZag path contained in that tree.

        Example 1:
        Input: root = [1,null,1,1,1,null,null,1,1,null,1,null,null,null,1]
        Output: 3

        Example 2:
        Input: root = [1,1,1,null,1,null,null,1,1,null,1]
        Output: 4

        Example 3:
        Input: root = [1]
        Output: 0
         */
        fun dfs(root: TreeNode?, l: Int, r: Int): Int {
            if (root == null) return maxOf(l,r)-1
            return maxOf(dfs(root.left, r+1, 0), dfs(root.right, 0, l+1))
        }

        fun longestZigZag(root: TreeNode?): Int = dfs(root, 0, 0)
        val tree1 = buildTree(arrayOf(1,null,1,1,1,null,null,1,1,null,1,null,null,null,1))

        println("ret is ${longestZigZag(tree1)}") // 3
        assertEquals(longestZigZag(tree1), 3)
        println("ret is ${longestZigZag(buildTree(arrayOf(1,1,1,null,1,null,null,1,1,null,1)))}") // 4
        assertEquals(longestZigZag(buildTree(arrayOf(1,1,1,null,1,null,null,1,1,null,1))), 4)
    }

    @Test
    fun leetcode236() {
        /**
        Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.
        According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node
        in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”
        Example 1:
        Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
        Output: 3
        Explanation: The LCA of nodes 5 and 1 is 3.

        Example 2:
        Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
        Output: 5
        Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

        Example 3:
        Input: root = [1,2], p = 1, q = 2
        Output: 1
         */
        fun lowestCommonAncestor(root: TreeNode?, p: TreeNode?, q: TreeNode?): TreeNode? {
            // 如果根节点为空或者等于目标节点之一，则返回根节点
            if (root == null || root == p || root == q) {
                return root
            }

            // 在左子树中寻找目标节点
            val left = lowestCommonAncestor(root.left, p, q)
            // 在右子树中寻找目标节点
            val right = lowestCommonAncestor(root.right, p, q)

            // 如果左子树和右子树都找到了目标节点，则当前节点为最低共同祖先
            if (left != null && right != null) {
                return root
            }

            // 如果只有左子树找到了目标节点，则返回左子树的结果
            return left ?: right
        }
    }

    @Test
    fun leetcode199() {
        fun rightSideView(root: TreeNode?): List<Int> {
            val rightmostValues = mutableListOf<Int>()
            if(root == null) return rightmostValues
            val queue: Queue<TreeNode> = LinkedList()
            queue.offer(root)

            while (queue.isNotEmpty()) {
                val levelSize = queue.size
                var rightmostNode: TreeNode? = null

                repeat(levelSize) {
                    val node = queue.poll()
                    rightmostNode = node

                    node.left?.let { queue.offer(it) }
                    node.right?.let { queue.offer(it) }
                }

                rightmostNode?.let { rightmostValues.add(it.`val`) }
            }

            return rightmostValues
        }

        var tree = buildTree(arrayOf(1,2,3,null,5,null,4))
        var ret = rightSideView(tree)
        assertEquals(listOf(1, 3, 4).toList(), ret)

        tree = buildTree(arrayOf(1,2))
        ret = rightSideView(tree)
        assertEquals(listOf(1, 2).toList(), ret)

        tree = buildTree(arrayOf(1,2,3,4))
        ret = rightSideView(tree)
        assertEquals(listOf(1,3,4).toList(), ret)

    }

    @Test
    fun leetcode1161() {
        fun maxLevelSum(root: TreeNode?): Int {
            if (root == null) return 0
            var maxIndex = 0
            var maxValue = Int.MIN_VALUE
            var levelNodes = mutableListOf<TreeNode>()
            levelNodes.add(root)

            var index = 1
            while (levelNodes.isNotEmpty()) {
                val levelQueue = LinkedList(levelNodes)

                val sum = levelNodes.map {it.`val`}.sum()

                if (sum > maxValue) {
                    maxValue = sum
                    maxIndex = index
                }
                index += 1

                levelNodes = mutableListOf()

                while (levelQueue.isNotEmpty()) {
                    val cur = levelQueue.poll()

                    cur.left?.let { levelNodes.add(it) }
                    cur.right?.let { levelNodes.add(it) }
                }
            }

            return maxIndex
        }

        var tree = buildTree(arrayOf(1,7,0,7,-8,null,null))
        var ret = maxLevelSum(tree)
        assertEquals(2, ret)

        tree = buildTree(arrayOf(989,null,10250,98693,-89388,null,null,null,-32127))
        ret = maxLevelSum(tree)
        assertEquals(2, ret)

    }

    @Test
    fun leetcode700() {
        fun searchBST(root: TreeNode?, `val`: Int): TreeNode? {
            var c = root

            while(c != null) {
                if(c.`val` == `val`) return c
                else {
                    if(c.`val` > `val`) c = c.left
                    else c = c.right
                }
            }

            return c
        }

        var ret = searchBST(buildTree(arrayOf(4,2,7,1,3)), 2)
        println("ret val is ${ret?.`val`}")
        assertEquals(2, ret?.`val`)

        ret = searchBST(buildTree(arrayOf(4,2,7,1,3)), 5)
        println("ret val is ${ret?.`val`}")
        assertEquals(null, ret?.`val`)
    }

    @Test
    fun leetcode450() {
        fun findSuccessorValue(node: TreeNode?): Int {
            var current = node
            while (current?.left != null) {
                current = current.left
            }
            return current!!.`val`
        }

        fun deleteNode(root: TreeNode?, key: Int): TreeNode? {
            if (root == null) {
                return null
            }

            if (key < root.`val`) {
                root.left = deleteNode(root.left, key)
            } else if (key > root.`val`) {
                root.right = deleteNode(root.right, key)
            } else {
                if (root.left == null) {
                    return root.right
                } else if (root.right == null) {
                    return root.left
                }

                val successorValue = findSuccessorValue(root.right)
                root.`val` = successorValue
                root.right = deleteNode(root.right, successorValue)
            }

            return root
        }

        val ret = deleteNode(buildTree(arrayOf(5,3,6,2,4,null,7)), 3)
        printTree(ret)
    }

    @Test
    fun leetcode841() {
        /**
        There are n rooms labeled from 0 to n - 1 and all the rooms are locked except for room 0.
        Your goal is to visit all the rooms. However, you cannot enter a locked room without having its key.
        When you visit a room, you may find a set of distinct keys in it. Each key has a number on it, denoting which room it unlocks,
        and you can take all of them with you to unlock the other rooms.
        Given an array rooms where rooms[i] is the set of keys that you can obtain if you visited room i,
        return true if you can visit all the rooms, or false otherwise.

        Example 1:
        Input: rooms = [[1],[2],[3],[]]
        Output: true
        Explanation:
        We visit room 0 and pick up key 1.
        We then visit room 1 and pick up key 2.
        We then visit room 2 and pick up key 3.
        We then visit room 3.
        Since we were able to visit every room, we return true.

        Example 2:
        Input: rooms = [[1,3],[3,0,1],[2],[0]]
        Output: false
        Explanation: We can not enter room number 2 since the only key that unlocks it is in that room.
         */
        fun canVisitAllRooms(rooms: List<List<Int>>): Boolean {
            val visited = BooleanArray(rooms.size) // 用于记录房间的访问状态
            visited[0] = true // 将第一个房间标记为已访问
            val queue: Queue<Int> = LinkedList<Int>() // 使用队列来进行广度优先搜索
            queue.offer(0) // 将第一个房间加入队列

            while (queue.isNotEmpty()) {
                val currRoom = queue.poll()
                for (key in rooms[currRoom]) {
                    if (!visited[key]) {
                        visited[key] = true
                        queue.offer(key)
                    }
                }
            }

            // 检查是否所有房间都被访问过
            for (roomVisited in visited) {
                if (!roomVisited) {
                    return false
                }
            }

            return true
        }

        // [[1],[2],[3],[]]
        assertEquals(true, canVisitAllRooms(
            listOf(
                listOf(1),
                listOf(2),
                listOf(3),
                listOf()
            )))

        // [[1,3],[3,0,1],[2],[0]]
        assertEquals(false, canVisitAllRooms(
            listOf(
                listOf(1,3),
                listOf(3,0,1),
                listOf(2),
                listOf(0)
            )))
    }

    @Test
    fun leetcode1466() {
        /**
        There are n cities numbered from 0 to n - 1 and n - 1 roads such that there is only one way to
        travel between two different cities (this network form a tree). Last year,
        The ministry of transport decided to orient the roads in one direction because they are too narrow.

        Roads are represented by connections where connections[i] = [ai, bi] represents a road from city ai to city bi.
        This year, there will be a big event in the capital (city 0), and many people want to travel to this city.
        Your task consists of reorienting some roads such that each city can visit the city 0. Return the minimum number of edges changed.
        It's guaranteed that each city can reach city 0 after reorder.

        Example 1:
        Input: n = 6, connections = [[0,1],[1,3],[2,3],[4,0],[4,5]]
        Output: 3
        Explanation: Change the direction of edges show in red such that each node can reach the node 0 (capital).

        Example 2:
        Input: n = 5, connections = [[1,0],[1,2],[3,2],[3,4]]
        Output: 2
        Explanation: Change the direction of edges show in red such that each node can reach the node 0 (capital).

        Example 3:
        Input: n = 3, connections = [[1,0],[2,0]]
        Output: 0
         */
        fun dfs(graph: Array<MutableList<Pair<Int, Int>>>, visited: BooleanArray, node: Int): Int {
            var count = 0 // 记录需要改变方向的连接数

            visited[node] = true

            for (next in graph[node]) {
                val neighbor = next.first
                val direction = next.second

                if (!visited[neighbor]) {
                    count += dfs(graph, visited, neighbor) + direction
                }
            }

            return count
        }

        fun minReorder(n: Int, connections: Array<IntArray>): Int {
            val graph = Array(n) { mutableListOf<Pair<Int, Int>>() } // 创建邻接表

            for (conn in connections) {
                val from = conn[0]
                val to = conn[1]
                graph[from].add(to to 1) // 正向连接
                graph[to].add(from to 0) // 反向连接
            }

            val visited = BooleanArray(n) // 记录节点是否已访问
            return dfs(graph, visited, 0) // 从节点 0 开始进行深度优先搜索
        }

        //Input: n = 6, connections = [[0,1],[1,3],[2,3],[4,0],[4,5]]
        //Output: 3
        assertEquals(3, minReorder(6,
            arrayOf(
                intArrayOf(0,1),
                intArrayOf(1,3),
                intArrayOf(2,3),
                intArrayOf(4,0),
                intArrayOf(4,5),
            )
        ))

        //Input: n = 5, connections = [[1,0],[1,2],[3,2],[3,4]]
        //Output: 2
        assertEquals(2, minReorder(5,
            arrayOf(
                intArrayOf(1,0),
                intArrayOf(1,2),
                intArrayOf(3,2),
                intArrayOf(3,4),
            )
        ))

        //Input: n = 3, connections = [[1,0],[2,0]]
        //Output: 0
        assertEquals(0, minReorder(3,
            arrayOf(
                intArrayOf(1,0),
                intArrayOf(2,0),
            )
        ))

    }

    @Test
    fun leetcode1926() {
        /**
        You are given an m x n matrix maze (0-indexed) with empty cells (represented as '.') and walls (represented as '+').
        You are also given the entrance of the maze, where entrance = [entrancerow, entrancecol] denotes the row and column of the cell you are initially standing at.

        In one step, you can move one cell up, down, left, or right. You cannot step into a cell with a wall,
        and you cannot step outside the maze. Your goal is to find the nearest exit from the entrance.
        An exit is defined as an empty cell that is at the border of the maze. The entrance does not count as an exit.

        Return the number of steps in the shortest path from the entrance to the nearest exit, or -1 if no such path exists.

        Example 1:
        Input: maze = [["+","+",".","+"],[".",".",".","+"],["+","+","+","."]], entrance = [1,2]
        Output: 1
        Explanation: There are 3 exits in this maze at [1,0], [0,2], and [2,3].
        Initially, you are at the entrance cell [1,2].
        - You can reach [1,0] by moving 2 steps left.
        - You can reach [0,2] by moving 1 step up.
        It is impossible to reach [2,3] from the entrance.
        Thus, the nearest exit is [0,2], which is 1 step away.

        Example 2:
        Input: maze = [["+","+","+"],[".",".","."],["+","+","+"]], entrance = [1,0]
        Output: 2
        Explanation: There is 1 exit in this maze at [1,2].
        [1,0] does not count as an exit since it is the entrance cell.
        Initially, you are at the entrance cell [1,0].
        - You can reach [1,2] by moving 2 steps right.
        Thus, the nearest exit is [1,2], which is 2 steps away.
        */
        fun nearestExit(maze: Array<CharArray>, entrance: IntArray): Int {
            val directions = arrayOf(intArrayOf(1, 0), intArrayOf(-1, 0), intArrayOf(0, 1), intArrayOf(0, -1))
            val rows = maze.size
            val cols = maze[0].size

            val queue = LinkedList<IntArray>()
            queue.offer(entrance)
            maze[entrance[0]][entrance[1]] = '+'

            var steps = 0
            while (queue.isNotEmpty()) {
                val size = queue.size
                for (i in 0 until size) {
                    val curr = queue.poll()
                    val row = curr[0]
                    val col = curr[1]

                    if ((row == 0 || row == rows - 1 || col == 0 || col == cols - 1) && curr != entrance) {
                        return steps
                    }

                    for (dir in directions) {
                        val newRow = row + dir[0]
                        val newCol = col + dir[1]

                        if (newRow >= 0 && newRow < rows && newCol >= 0 && newCol < cols && maze[newRow][newCol] == '.') {
                            maze[newRow][newCol] = '+'
                            queue.offer(intArrayOf(newRow, newCol))
                        }
                    }
                }
                steps++
            }

            return -1
        }

        //Input: maze = [["+","+",".","+"],[".",".",".","+"],["+","+","+","."]], entrance = [1,2]
        //Output: 1
        assertEquals(1, nearestExit(arrayOf(charArrayOf('+','+','.','+')
            ,charArrayOf('.','.','.','+')
            ,charArrayOf('+','+','+','.'))
            ,intArrayOf(1,2)))

        //Input: [["+","+","+"],[".",".","."],["+","+","+"]], entrance = [1,0]
        //Output: 2
        assertEquals(2, nearestExit(arrayOf(charArrayOf('+','+','+')
            ,charArrayOf('.','.','.')
            ,charArrayOf('+','+','+'))
            ,intArrayOf(1,2)))

    }

    @Test
    fun leetcode215() {
        /**
        Given an integer array nums and an integer k, return the kth largest element in the array.
        Note that it is the kth largest element in the sorted order, not the kth distinct element.
        You must solve it in O(n) time complexity.
        Example 1:
        Input: nums = [3,2,1,5,6,4], k = 2
        Output: 5

        Example 2:
        Input: nums = [3,2,3,1,2,4,5,5,6], k = 4
        Output: 4
         */
        fun findKthLargest(nums: IntArray, k: Int): Int? {
            val pq = PriorityQueue<Int>()

            for (num in nums) {
                pq.offer(num)

                if (pq.size > k) {
                    pq.poll()
                }
            }

            return pq.peek()
        }

        //        Input: nums = [3,2,1,5,6,4], k = 2
        //        Output: 5
        assertEquals(5, findKthLargest(intArrayOf(3,2,1,5,6,4), 2))

        //        Input: nums = [3,2,3,1,2,4,5,5,6], k = 4
        //        Output: 4
        assertEquals(4, findKthLargest(intArrayOf(3,2,3,1,2,4,5,5,6), 4))
    }

    @Test
    fun leetcode2542() {
        /**
        You are given two 0-indexed integer arrays nums1 and nums2 of equal length n and a positive integer k.
        You must choose a subsequence of indices from nums1 of length k.
        For chosen indices i0, i1, ..., ik - 1, your score is defined as:
        The sum of the selected elements from nums1 multiplied with the minimum of the selected elements from nums2.
        It can defined simply as: (nums1[i0] + nums1[i1] +...+ nums1[ik - 1]) * min(nums2[i0] , nums2[i1], ... ,nums2[ik - 1]).
        Return the maximum possible score.
        A subsequence of indices of an array is a set that can be derived from the set {0, 1, ..., n-1} by deleting some or no elements.

        Example 1:
        Input: nums1 = [1,3,3,2], nums2 = [2,1,3,4], k = 3
        Output: 12
        Explanation:
        The four possible subsequence scores are:
        - We choose the indices 0, 1, and 2 with score = (1+3+3) * min(2,1,3) = 7.
        - We choose the indices 0, 1, and 3 with score = (1+3+2) * min(2,1,4) = 6.
        - We choose the indices 0, 2, and 3 with score = (1+3+2) * min(2,3,4) = 12.
        - We choose the indices 1, 2, and 3 with score = (3+3+2) * min(1,3,4) = 8.
        Therefore, we return the max score, which is 12.

        Example 2:
        Input: nums1 = [4,2,3,1,1], nums2 = [7,5,10,9,6], k = 1
        Output: 30
        Explanation:
        Choosing index 2 is optimal: nums1[2] * nums2[2] = 3 * 10 = 30 is the maximum possible score.
         */
        fun maxScore(nums1: IntArray, nums2: IntArray, k: Int): Long {
            val list = mutableListOf<Pair<Int, Int>>()

            for (i in nums1.indices) {
                list.add(Pair(nums2[i], nums1[i]))
            }

            list.sortByDescending { it.first }

            val pq = PriorityQueue<Int>()
            var sum = 0L
            var result = 0L

            for (i in 0 until k) {
                pq.offer(list[i].second)
                sum += list[i].second
            }

            result = sum * list[k - 1].first

            for (i in k until nums1.size) {
                sum -= pq.poll()
                sum += list[i].second
                pq.offer(list[i].second)

                result = Math.max(result, sum * list[i].first)
            }

            return result
        }

        //        Input: nums1 = [1,3,3,2], nums2 = [2,1,3,4], k = 3
        //        Output: 12
        assertEquals(12, maxScore(intArrayOf(1,3,3,2), intArrayOf(2,1,3,4), 3))

        //        Input: nums1 = [4,2,3,1,1], nums2 = [7,5,10,9,6], k = 1
        //        Output: 30
        assertEquals(30, maxScore(intArrayOf(4,2,3,1,1), intArrayOf(7,5,10,9,6), 1))
    }

    @Test
    fun leetcode2462() {
        /**
        You are given a 0-indexed integer array costs where costs[i] is the cost of hiring the ith worker.

        You are also given two integers k and candidates. We want to hire exactly k workers according to the following rules:

        You will run k sessions and hire exactly one worker in each session.
        In each hiring session, choose the worker with the lowest cost from either the first candidates workers or the last candidates workers. Break the tie by the smallest index.
        For example, if costs = [3,2,7,7,1,2] and candidates = 2, then in the first hiring session, we will choose the 4th worker because they have the lowest cost [3,2,7,7,1,2].
        In the second hiring session, we will choose 1st worker because they have the same lowest cost as 4th worker but they have the smallest index [3,2,7,7,2].
        Please note that the indexing may be changed in the process.
        If there are fewer than candidates workers remaining, choose the worker with the lowest cost among them. Break the tie by the smallest index.
        A worker can only be chosen once.
        Return the total cost to hire exactly k workers.

        Example 1:
        Input: costs = [17,12,10,2,7,2,11,20,8], k = 3, candidates = 4
        Output: 11
        Explanation: We hire 3 workers in total. The total cost is initially 0.
        - In the first hiring round we choose the worker from [17,12,10,2,7,2,11,20,8]. The lowest cost is 2, and we break the tie by the smallest index, which is 3. The total cost = 0 + 2 = 2.
        - In the second hiring round we choose the worker from [17,12,10,7,2,11,20,8]. The lowest cost is 2 (index 4). The total cost = 2 + 2 = 4.
        - In the third hiring round we choose the worker from [17,12,10,7,11,20,8]. The lowest cost is 7 (index 3). The total cost = 4 + 7 = 11. Notice that the worker with index 3 was common in the first and last four workers.
        The total hiring cost is 11.

        Example 2:
        Input: costs = [1,2,4,1], k = 3, candidates = 3
        Output: 4
        Explanation: We hire 3 workers in total. The total cost is initially 0.
        - In the first hiring round we choose the worker from [1,2,4,1]. The lowest cost is 1, and we break the tie by the smallest index, which is 0. The total cost = 0 + 1 = 1. Notice that workers with index 1 and 2 are common in the first and last 3 workers.
        - In the second hiring round we choose the worker from [2,4,1]. The lowest cost is 1 (index 2). The total cost = 1 + 1 = 2.
        - In the third hiring round there are less than three candidates. We choose the worker from the remaining workers [2,4]. The lowest cost is 2 (index 0). The total cost = 2 + 2 = 4.
        The total hiring cost is 4.
         */
        fun totalCost(costs: IntArray, k: Int, candidates: Int): Long {
            var left = 0
            var right = costs.lastIndex
            val canL = PriorityQueue<Int>()
            val canR = PriorityQueue<Int>()

            var res = 0L
            var n = k
            while (n > 0) {
                while (canL.size < candidates && left <= right) canL.add(costs[left++])
                while (canR.size < candidates && left <= right) canR.add(costs[right--])

                if (canL.peek()?: 100001 <= canR.peek()?: 100001) {
                    res += canL.poll()
                } else {
                    res += canR.poll()
                }
                n--
            }

            return res
        }

        //        Input: costs = [17,12,10,2,7,2,11,20,8], k = 3, candidates = 4
        //        Output: 11
        assertEquals(11, totalCost(intArrayOf(17,12,10,2,7,2,11,20,8), 3, 4))

        //        Input: costs = [1,2,4,1], k = 3, candidates = 3
        //        Output: 4
        assertEquals(4, totalCost(intArrayOf(1,2,4,1), 3, 3))

    }

    private fun printTree(root: TreeNode?) {
        if (root == null) return

        printTreeHelper(listOf(root))
    }

    private fun printTreeHelper(nodes: List<TreeNode?>) {
        if (nodes.isEmpty() || nodes.all { it == null }) return

        val nextLevelNodes = mutableListOf<TreeNode?>()
        val line = StringBuilder()

        for (node in nodes) {
            line.append(if (node != null) "${node.`val`}" else " ")
            line.append(" ".repeat(3))

            nextLevelNodes.add(node?.left)
            nextLevelNodes.add(node?.right)
        }

        println(line)

        // Print vertical lines
        val lineLength = line.length
        val verticalLine = StringBuilder("")

        for (i in 0 until lineLength) {
            verticalLine.append(if (line[i] == '|') '|' else ' ')
        }

        println(verticalLine)

        // Recursively print the next level
        printTreeHelper(nextLevelNodes)
    }
    private fun printCharArray(array: CharArray) {
        array.forEachIndexed { i, v ->
            println("chars[$i] = $v")
        }
    }

    private fun printIntArray(array: IntArray) {
        array.forEachIndexed { i, v ->
            println("Int[$i] = $v")
        }
    }

    private fun printStringArray(array: MutableList<String>) {
        array.forEachIndexed { i, v ->
            println("String[$i] = $v")
        }
    }

    private fun buildTree(nums: Array<Int?>): TreeNode? {
        if (nums.isEmpty()) {
            return null
        }

        val root = TreeNode(nums[0]!!)
        val queue: Queue<TreeNode?> = LinkedList()
        queue.offer(root)

        var index = 1
        while (index < nums.size) {
            val node = queue.poll()
            val leftValue = nums[index++]
            if (leftValue != null) {
                node?.left = TreeNode(leftValue)
                queue.offer(node?.left)
            }

            if (index < nums.size) {
                val rightValue = nums[index++]
                if (rightValue != null) {
                    node?.right = TreeNode(rightValue)
                    queue.offer(node?.right)
                }
            }
        }

        return root
    }

    class ListNode(var `val`: Int) {
        var next: ListNode? = null
    }

    class TreeNode(var `val`: Int) {
        var left: TreeNode? = null
        var right: TreeNode? = null
    }

}
package com.kotaku.camerax2leetcode

import org.junit.Assert.*
import org.junit.Test
import java.util.*

/**
 * Example local unit test, which will execute on the development machine (host).
 *
 * See [testing documentation](http://d.android.com/tools/testing).
 */
class ExampleUnitTest {
    @Test
    fun leetcode1431() {
        /**
        There are n kids with candies. You are given an integer array candies,
        where each candies[i] represents the number of candies the ith kid has,
        and an integer extraCandies, denoting the number of extra candies that you have.
        Return a boolean array result of length n, where result[i] is true if,
        after giving the ith kid all the extraCandies, they will have the greatest number of candies among all the kids,
        or false otherwise.
        Note that multiple kids can have the greatest number of candies.

        Input: candies = [2,3,5,1,3], extraCandies = 3
        Output: [true,true,true,false,true]
        Explanation: If you give all extraCandies to:
        - Kid 1, they will have 2 + 3 = 5 candies, which is the greatest among the kids.
        - Kid 2, they will have 3 + 3 = 6 candies, which is the greatest among the kids.
        - Kid 3, they will have 5 + 3 = 8 candies, which is the greatest among the kids.
        - Kid 4, they will have 1 + 3 = 4 candies, which is not the greatest among the kids.
        - Kid 5, they will have 3 + 3 = 6 candies, which is the greatest among the kids.
         */
        fun kidsWithCandies(candies: IntArray, extraCandies: Int): List<Boolean> {
            val ret = MutableList(candies.size) { false }
            val max = candies.max()
            println("candies.size is ${candies.size}, extraCandies is $extraCandies, max is $max")

            for (i in candies.indices) {
                if (candies[i] + extraCandies >= max) {
                    ret[i] = true
                }
            }

            return ret
        }

        val ret = kidsWithCandies(intArrayOf(2, 3, 5, 1, 3), 3)
        ret.forEachIndexed { i, v ->
            println("ret[$i] = $v")
        }
    }

    @Test
    fun leetcode605() {
        /**
        You have a long flowerbed in which some of the plots are planted, and some are not.
        However, flowers cannot be planted in adjacent plots.
        Given an integer array flowerbed containing 0's and 1's, where 0 means empty and 1 means not empty,
        and an integer n, return true if n new flowers can be planted in the flowerbed without violating the no-adjacent-flowers rule and false otherwise.

        Example 1:
        Input: flowerbed = [1,0,0,0,1], n = 1
        Output: true

        Example 2:
        Input: flowerbed = [1,0,0,0,1], n = 2
        Output: false
         */
        fun canPlaceFlowers(flowerbed: IntArray, n: Int): Boolean {
            if (flowerbed.size == 1) {
                return if (n > 1) false
                else if (n == 1) return flowerbed[0] == 0
                else return true
            }
            var can = 0
            for (i in flowerbed.indices) {
                if (flowerbed[i] == 1) continue
                val leftEmpty = if (i == 0) true else flowerbed[i - 1] == 0

                val rightEmpty = if (i == flowerbed.size - 1) true else flowerbed[i + 1] == 0

                if (leftEmpty && rightEmpty) {
                    flowerbed[i] = 1
                    can++
                }
            }
            return can >= n
        }

        val ret = canPlaceFlowers(intArrayOf(1, 0, 0, 0, 1), 2)
        println("ret is $ret")
    }

    @Test
    fun leetcode345() {
        /**
        Given a string s, reverse only all the vowels in the string and return it.
        The vowels are 'a', 'e', 'i', 'o', and 'u', and they can appear in both lower and upper cases, more than once.

        Example 1:
        Input: s = "hello"
        Output: "holle"

        Example 2:
        Input: s = "leetcode"
        Output: "leotcede"
         */

        fun reverseVowels(s: String): String {
            val vowels = listOf('a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U')
            val chars = s.toCharArray()
            var left = 0
            var right = chars.size - 1

            while (left < right) {
                if (chars[left] in vowels && chars[right] in vowels) {
                    val temp = chars[left]
                    chars[left] = chars[right]
                    chars[right] = temp
                    left++
                    right--
                } else if (chars[left] in vowels) {
                    right--
                } else {
                    left++
                }
            }

            return String(chars)
        }

        val ret = reverseVowels("leetcode")
        println("ret is $ret")

    }

    @Test
    fun leetcode151() {
        /**
        Given an input string s, reverse the order of the words.
        A word is defined as a sequence of non-space characters. The words in s will be separated by at least one space.
        Return a string of the words in reverse order concatenated by a single space.
        Note that s may contain leading or trailing spaces or multiple spaces between two words.
        The returned string should only have a single space separating the words. Do not include any extra spaces.

        Example 1:
        Input: s = "the sky is blue"
        Output: "blue is sky the"

        Example 2:
        Input: s = "  hello world  "
        Output: "world hello"
        Explanation: Your reversed string should not contain leading or trailing spaces.

        Example 3:
        Input: s = "a good   example"
        Output: "example good a"
        Explanation: You need to reduce multiple spaces between two words to a single space in the reversed string.
         **/
        fun reverseWords(s: String): String {
            //return s.trim().split(" ").reversed().filter { it.trim().isNotEmpty() }.joinToString(separator = " ")

            val wordsArray = s.trim().split(" ").reversed()
            wordsArray.forEach {
                println("words is $it ;")
            }

            return wordsArray.filter { it.trim().isNotEmpty() }.joinToString(separator = " ")
        }

        val ret = reverseWords("a good   example")
        println("ret is $ret")
    }

    @Test
    fun leetcode334() {
        /**
        Given an integer array nums, return true if there exists a triple of indices (i, j, k)
        such that i < j < k and nums[i] < nums[j] < nums[k]. If no such indices exists, return false.

        Example 1:
        Input: nums = [1,2,3,4,5]
        Output: true
        Explanation: Any triplet where i < j < k is valid.

        Example 2:
        Input: nums = [5,4,3,2,1]
        Output: false
        Explanation: No triplet exists.

        Example 3:
        Input: nums = [2,1,5,0,4,6]
        Output: true
        Explanation: The triplet (3, 4, 5) is valid because nums[3] == 0 < nums[4] == 4 < nums[5] == 6.
         */
        fun increasingTriplet(nums: IntArray): Boolean {
            var small = Int.MAX_VALUE
            var big = Int.MAX_VALUE

            for (num in nums) {
                if (num <= small) {
                    small = num
                } else if (num <= big) {
                    big = num
                } else {
                    return true
                }

                println("small is $small ; big is $big ")
            }

            return false
        }

        val ret = increasingTriplet(intArrayOf(2, 1, 5, 0, 4, 6))
        println("ret is $ret")
    }

    @Test
    fun leetcode443() {
        /**
        Given an array of characters chars, compress it using the following algorithm:
        Begin with an empty string s. For each group of consecutive repeating characters in chars:

        If the group's length is 1, append the character to s.
        Otherwise, append the character followed by the group's length.
        The compressed string s should not be returned separately, but instead, be stored in the input character array chars.
        Note that group lengths that are 10 or longer will be split into multiple characters in chars.

        After you are done modifying the input array, return the new length of the array.
        You must write an algorithm that uses only constant extra space.

        Example 1:
        Input: chars = ["a","a","b","b","c","c","c"]
        Output: Return 6, and the first 6 characters of the input array should be: ["a","2","b","2","c","3"]
        Explanation: The groups are "aa", "bb", and "ccc". This compresses to "a2b2c3".
         */
        fun compress(chars: CharArray): Int {
            var writeIndex = 0
            var readIndex = 0

            while (readIndex < chars.size) {
                val currentChar = chars[readIndex]
                var count = 0

                while (readIndex < chars.size && chars[readIndex] == currentChar) {
                    readIndex++
                    count++
                }

                chars[writeIndex] = currentChar
                writeIndex++

                if (count > 1) {
                    val countChars = count.toString().toCharArray()
                    for (i in countChars.indices) {
                        chars[writeIndex] = countChars[i]
                        writeIndex++
                    }
                }
            }

            printCharArray(chars)

            return writeIndex
        }

        val ret =
            compress(charArrayOf('a', 'b', 'b', 'b', 'b', 'b', 'b', 'b', 'b', 'b', 'b', 'b', 'b'))
        println("ret is $ret")

    }

    @Test
    fun leetcode283() {
        /**
        Given an integer array nums, move all 0's to the end of it while maintaining the relative order of the non-zero elements.
        Note that you must do this in-place without making a copy of the array.

        Example 1:
        Input: nums = [0,1,0,3,12]
        Output: [1,3,12,0,0]

        Example 2:
        Input: nums = [0]
        Output: [0]
         */
        fun moveZeroes(nums: IntArray): IntArray {
            if (nums.size == 1) {
                return nums
            }

            var left = 0
            var right = 0

            while (left < nums.size && nums[left] != 0) {
                left++
                right++
            }

            while (right < nums.size) {
                if (nums[right] == 0) {
                    right++
                } else {
                    nums[left++] = nums[right]
                    nums[right++] = 0
                }
            }

            return nums
        }

        val ret = moveZeroes(intArrayOf(0, 1, 0, 3, 12))
        printIntArray(ret)

    }

    @Test
    fun leetcode392() {
        /***
        Given two strings s and t, return true if s is a subsequence of t, or false otherwise.
        A subsequence of a string is a new string that is formed from the original string by deleting some (can be none) of
        the characters without disturbing the relative positions of the remaining characters.
        (i.e., "ace" is a subsequence of "abcde" while "aec" is not).
        Example 1:
        Input: s = "abc", t = "ahbgdc"
        Output: true

        Example 2:
        Input: s = "axc", t = "ahbgdc"
        Output: false
         */

        fun isSubsequence(s: String, t: String): Boolean {
            var i = 0
            var j = 0

            while (i < s.length && j < t.length) {
                if (s[i] == t[j]) {
                    i++
                }
                j++
            }

            return i == s.length
        }

        val ret = isSubsequence("abc", "ahbgdc")
        println("ret is $ret")
    }

    @Test
    fun leetcode11() {
        /**
        You are given an integer array height of length n.
        There are n vertical lines drawn such that the two endpoints of the ith line are (i, 0) and (i, height[i]).
        Find two lines that together with the x-axis form a container, such that the container contains the most water.
        Return the maximum amount of water a container can store.

        Example 1:
        Input: height = [1,8,6,2,5,4,8,3,7]
        Output: 49
        Explanation: The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7].
        In this case, the max area of water (blue section) the container can contain is 49.
         */

        fun maxArea(height: IntArray): Int {
            var left = 0
            var right = height.size - 1
            var maxArea = 0

            while (left <= right) {
                val area = height[left].coerceAtMost(height[right]) * (right - left)
                if (height[right] > height[left]) {
                    left++
                } else {
                    right--
                }
                maxArea = maxArea.coerceAtLeast(area)
            }

            return maxArea
        }

        val ret = maxArea(intArrayOf(1, 8, 6, 2, 5, 4, 8, 3, 7))
        println("ret is $ret")
    }

    @Test
    fun leetcode1679() {
        /**
        You are given an integer array nums and an integer k.
        In one operation, you can pick two numbers from the array whose sum equals k and remove them from the array.
        Return the maximum number of operations you can perform on the array.

        Example 1:
        Input: nums = [1,2,3,4], k = 5
        Output: 2
        Explanation: Starting with nums = [1,2,3,4]:
        - Remove numbers 1 and 4, then nums = [2,3]
        - Remove numbers 2 and 3, then nums = []
        There are no more pairs that sum up to 5, hence a total of 2 operations.

        Example 2:
        Input: nums = [3,1,3,4,3], k = 6
        Output: 1
        Explanation: Starting with nums = [3,1,3,4,3]:
        - Remove the first two 3's, then nums = [1,4,3]
        There are no more pairs that sum up to 6, hence a total of 1 operation.
         */
        fun maxOperations(nums: IntArray, k: Int): Int {
            nums.sort()
            var left = 0
            var right = nums.size - 1
            var ret = 0

            while (left < right) {
                if (nums[left] + nums[right] == k) {
                    ret++
                    left++
                    right--
                } else if (nums[left] + nums[right] < k) {
                    left++
                } else if (nums[left] + nums[right] > k) {
                    right--
                }
            }

            return ret
        }

        val ret = maxOperations(intArrayOf(1, 2, 3, 4), 5)
        println("ret is $ret")
    }

    @Test
    fun leetcode643() {
        /***
        You are given an integer array nums consisting of n elements, and an integer k.
        Find a contiguous subarray whose length is equal to k that has the maximum average value and return this value.
        Any answer with a calculation error less than 10-5 will be accepted.

        Example 1:
        Input: nums = [1,12,-5,-6,50,3], k = 4
        Output: 12.75000
        Explanation: Maximum average is (12 - 5 - 6 + 50) / 4 = 51 / 4 = 12.75

        Example 2:
        Input: nums = [5], k = 1
        Output: 5.00000
         */
        fun findMaxAverage(nums: IntArray, k: Int): Double {
            var sum = 0
            for (i in 0 until k) {
                sum += nums[i]
            }
            var maxSum = sum

            for (i in k until nums.size) {
                sum += nums[i] - nums[i - k]
                maxSum = maxOf(maxSum, sum)
            }

            return maxSum.toDouble() / k.toDouble()
        }

        val ret = findMaxAverage(intArrayOf(1, 12, -5, -6, 50, 3), 4)
        println("ret is $ret")
    }

    @Test
    fun leetcode1456() {
        /**
        Given a string s and an integer k, return the maximum number of vowel letters in any substring of s with length k.
        Vowel letters in English are 'a', 'e', 'i', 'o', and 'u'.

        Example 1:
        Input: s = "abciiidef", k = 3
        Output: 3
        Explanation: The substring "iii" contains 3 vowel letters.

        Example 2:
        Input: s = "aeiou", k = 2
        Output: 2
        Explanation: Any substring of length 2 contains 2 vowels.

        Example 3:
        Input: s = "leetcode", k = 3
        Output: 2
        Explanation: "lee", "eet" and "ode" contain 2 vowels.*/
        fun maxVowels(s: String, k: Int): Int {
            var sum = 0
            val charList = charArrayOf('a', 'e', 'i', 'o', 'u')
            for (i in 0 until k) {
                if (charList.contains(s[i])) sum += 1
            }
            var maxSum = sum

            for (i in k until s.length) {
                if (charList.contains(s[i])) sum += 1
                if (charList.contains(s[i - k])) sum -= 1
                maxSum = maxOf(maxSum, sum)
            }

            return maxSum
        }

        val ret = maxVowels("abciiidef", 3)
        println("ret is $ret")
    }

    @Test
    fun leetcode1004() {
        /**
        Given a binary array nums and an integer k, return the maximum number of consecutive 1's in the array
        if you can flip at most k 0's.

        Example 1:
        Input: nums = [1,1,1,0,0,0,1,1,1,1,0], k = 2
        Output: 6
        Explanation: [1,1,1,0,0,1,1,1,1,1,1]
        Bolded numbers were flipped from 0 to 1. The longest subarray is underlined.

        Example 2:
        Input: nums = [0,0,1,1,0,0,1,1,1,0,1,1,0,0,0,1,1,1,1], k = 3
        Output: 10
        Explanation: [0,0,1,1,1,1,1,1,1,1,1,1,0,0,0,1,1,1,1]
        Bolded numbers were flipped from 0 to 1. The longest subarray is underlined.
         * */
        fun longestOnes(nums: IntArray, k: Int): Int {
            var maxCount = 0
            var left = 0
            var zerosCount = 0

            for (right in nums.indices) {
                if (nums[right] == 0) {
                    zerosCount++
                }

                while (zerosCount > k) {
                    if (nums[left] == 0) {
                        zerosCount--
                    }
                    left++
                }

                maxCount = maxCount.coerceAtLeast(right - left + 1)
                println("right is $right, left is $left, zerosCount is $zerosCount, maxCount is $maxCount")
            }

            return maxCount
        }

        val ret = longestOnes(intArrayOf(1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0), 2)
        println("ret is $ret")
    }

    @Test
    fun leetcode1493() {
        /***
        Given a binary array nums, you should delete one element from it.
        Return the size of the longest non-empty subarray containing only 1's in the resulting array.
        Return 0 if there is no such subarray.

        Example 1:
        Input: nums = [1,1,0,1]
        Output: 3
        Explanation: After deleting the number in position 2, [1,1,1] contains 3 numbers with value of 1's.

        Example 2:
        Input: nums = [0,1,1,1,0,1,1,0,1]
        Output: 5
        Explanation: After deleting the number in position 4, [0,1,1,1,1,1,0,1] longest subarray with value of 1's is [1,1,1,1,1].

        Example 3:
        Input: nums = [1,1,1]
        Output: 2
        Explanation: You must delete one element.
         */
        fun longestSubarray(nums: IntArray): Int {
            var maxLength = 0
            var countZeros = 0
            var left = 0

            for (right in nums.indices) {
                if (nums[right] == 0) {
                    countZeros++
                }

                while (countZeros > 1) {
                    if (nums[left] == 0) {
                        countZeros--
                    }
                    left++
                }

                maxLength = maxOf(maxLength, right - left)
            }

            return maxLength
        }

        val ret = longestSubarray(intArrayOf(0, 1, 1, 1, 0, 1, 1, 0, 1))
        println("ret is $ret")
    }

    @Test
    fun leetcode1732() {
        /**
        There is a biker going on a road trip. The road trip consists of n + 1 points at different altitudes.
        The biker starts his trip on point 0 with altitude equal 0.
        You are given an integer array gain of length n where gain[i] is the net gain in altitude
        between points i​​​​​​ and i + 1 for all (0 <= i < n). Return the highest altitude of a point.

        Example 1:
        Input: gain = [-5,1,5,0,-7]
        Output: 1
        Explanation: The altitudes are [0,-5,-4,1,1,-6]. The highest is 1.

        Example 2:
        Input: gain = [-4,-3,-2,-1,4,3,2]
        Output: 0
        Explanation: The altitudes are [0,-4,-7,-9,-10,-6,-3,-1]. The highest is 0.
         */
        fun largestAltitude(gain: IntArray): Int {
            var prix = 0
            var max = 0
            gain.forEach {
                val current = prix + it
                max = max.coerceAtLeast(current)
                prix = current
            }
            return max
        }

        var ret = largestAltitude(intArrayOf(-5, 1, 5, 0, -7))
        println("ret is $ret")

        ret = largestAltitude(intArrayOf(-4, -3, -2, -1, 4, 3, 2))
        println("ret is $ret")
    }

    @Test
    fun leetcode2215() {
        /**
        Given two 0-indexed integer arrays nums1 and nums2, return a list answer of size 2 where:
        answer[0] is a list of all distinct integers in nums1 which are not present in nums2.
        answer[1] is a list of all distinct integers in nums2 which are not present in nums1.
        Note that the integers in the lists may be returned in any order.

        Example 1:
        Input: nums1 = [1,2,3], nums2 = [2,4,6]
        Output: [[1,3],[4,6]]

        Example 2:
        Input: nums1 = [1,2,3,3], nums2 = [1,1,2,2]
        Output: [[3],[]]
         */
        fun arrayDiff(array1: IntArray, array2: IntArray): List<Int> {
            val set1 = array1.toSet()
            val set2 = array2.toSet()

            val difference = set1.subtract(set2)
            return difference.toList()
        }

        fun findDifference(nums1: IntArray, nums2: IntArray): List<List<Int>> {
            return listOf(arrayDiff(nums1, nums2), arrayDiff(nums2, nums1))
        }

        val ret = findDifference(intArrayOf(1, 2, 3, 3), intArrayOf(1, 1, 2, 2))
        ret.forEachIndexed { i1, array ->
            array.forEachIndexed { i2, v ->
                println("Int[$i1][$i2] = $v")
            }
        }
    }

    @Test
    fun leetcode1207() {
        /**
        Given an array of integers arr, return true if the number of occurrences of each value in the array is unique or false otherwise.
        Example 1:
        Input: arr = [1,2,2,1,1,3]
        Output: true
        Explanation: The value 1 has 3 occurrences, 2 has 2 and 3 has 1. No two values have the same number of occurrences.

        Example 2:
        Input: arr = [1,2]
        Output: false

        Example 3:
        Input: arr = [-3,0,1,-3,1,1,1,-3,10,0]
        Output: true
         */
        fun uniqueOccurrences(arr: IntArray): Boolean {
            var ret = true
            val set1 = arr.toSet()
            val minLength = (1..set1.size).sum()
            if (arr.size >= minLength) {
                val mutableSet = mutableSetOf<Int>()
                set1.forEach {
                    println("key = $it")
                    val arrCount = arr.count { v ->
                        v == it
                    }
                    if (mutableSet.contains(arrCount)) {
                        ret = false
                        return@forEach
                    } else {
                        mutableSet.add(arrCount)
                    }
                }
            } else {
                ret = false
            }
            return ret
        }

        var ret = uniqueOccurrences(intArrayOf(-3, 0, 1, -3, 1, 1, 1, -3, 10, 0))
        println("Boolean = $ret")

        ret = uniqueOccurrences(intArrayOf(1, 2, 2, 1, 1, 3))
        println("Boolean = $ret")

        ret = uniqueOccurrences(
            intArrayOf(
                26,
                2,
                16,
                16,
                5,
                5,
                26,
                2,
                5,
                20,
                20,
                5,
                2,
                20,
                2,
                2,
                20,
                2,
                16,
                20,
                16,
                17,
                16,
                2,
                16,
                20,
                26,
                16
            )
        )
        println("Boolean = $ret")
    }

    @Test
    fun leetcode1657() {
        /***
        Two strings are considered close if you can attain one from the other using the following operations:

        Operation 1: Swap any two existing characters.
        For example, abcde -> aecdb
        Operation 2: Transform every occurrence of one existing character into another existing character, and do the same with the other character.
        For example, aacabb -> bbcbaa (all a's turn into b's, and all b's turn into a's)
        You can use the operations on either string as many times as necessary.

        Given two strings, word1 and word2, return true if word1 and word2 are close, and false otherwise.

        Example 1:
        Input: word1 = "abc", word2 = "bca"
        Output: true
        Explanation: You can attain word2 from word1 in 2 operations.
        Apply Operation 1: "abc" -> "acb"
        Apply Operation 1: "acb" -> "bca"

        Example 2:
        Input: word1 = "a", word2 = "aa"
        Output: false
        Explanation: It is impossible to attain word2 from word1, or vice versa, in any number of operations.

        Example 3:
        Input: word1 = "cabbba", word2 = "abbccc"
        Output: true
        Explanation: You can attain word2 from word1 in 3 operations.
        Apply Operation 1: "cabbba" -> "caabbb"
        Apply Operation 2: "caabbb" -> "baaccc"
        Apply Operation 2: "baaccc" -> "abbccc"
         */
        fun closeStrings(word1: String, word2: String): Boolean {
            if (word1.length != word2.length) {
                return false
            }

            val freq1 = IntArray(26)
            val freq2 = IntArray(26)

            val set1 = mutableSetOf<Char>()
            val set2 = mutableSetOf<Char>()

            for (i in word1.indices) {
                freq1[word1[i] - 'a']++
                freq2[word2[i] - 'a']++
                set1.add(word1[i])
                set2.add(word2[i])
            }

            freq1.sort()
            freq2.sort()

            return freq1.contentEquals(freq2) && set1 == set2
        }

        val ret = closeStrings("cabbba", "abbccc")
        println("Boolean = $ret")
    }

    @Test
    fun leetcode2352() {
        /**
        Given a 0-indexed n x n integer matrix grid, return the number of pairs (ri, cj) such that row ri and column cj are equal.
        A row and column pair is considered equal if they contain the same elements in the same order (i.e., an equal array).
        Example 1:
        Input: grid = [[3,2,1],[1,7,6],[2,7,7]]
        Output: 1
        Explanation: There is 1 equal row and column pair:
        - (Row 2, Column 1): [2,7,7]

        Example 2:
        Input: grid = [[3,1,2,2],[1,4,4,5],[2,4,2,2],[2,4,2,2]]
        Output: 3
        Explanation: There are 3 equal row and column pairs:
        - (Row 0, Column 0): [3,1,2,2]
        - (Row 2, Column 2): [2,4,2,2]
        - (Row 3, Column 2): [2,4,2,2]
         */
        fun equalPairs(grid: Array<IntArray>): Int {
            var ret = 0
            val rowStrings = mutableListOf<String>()
            val columnStrings = mutableListOf<String>()
            grid.forEachIndexed { i, array ->
                rowStrings.add(i, array.joinToString(separator = "-") + "-")
            }

            for (i in grid.indices) {
                var colum = ""
                grid.forEach {
                    colum = colum + it[i] + "-"
                }
                columnStrings.add(i, colum)
            }

            printStringArray(rowStrings)
            printStringArray(columnStrings)

            rowStrings.forEach { row ->
                columnStrings.forEach { column ->
                    if(row == column) ret++
                }
            }

            return ret
        }

        // [[3,1,2,2],[1,4,4,5],[2,4,2,2],[2,4,2,2]]
        var ret = equalPairs(
            arrayOf(
                intArrayOf(3, 1, 2, 2),
                intArrayOf(1, 4, 4, 5),
                intArrayOf(2, 4, 2, 2),
                intArrayOf(2, 4, 2, 2)
            )
        )
        println("ret = $ret")

        ret = equalPairs(
            arrayOf(
                intArrayOf(11,1),
                intArrayOf(1, 11)
            )
        )
        println("ret = $ret")
    }

    @Test
    fun leetcode2390() {
        fun removeStars(s: String): String {
            /**
            You are given a string s, which contains stars *.
            In one operation, you can:
            Choose a star in s.
            Remove the closest non-star character to its left, as well as remove the star itself.
            Return the string after all stars have been removed.
            Note:
            The input will be generated such that the operation is always possible.
            It can be shown that the resulting string will always be unique.

            Example 1:
            Input: s = "leet**cod*e"
            Output: "lecoe"

            Example 2:
            Input: s = "erase*****"
            Output: ""
            Explanation: The entire string is removed, so we return an empty string.
             */
            val stack = Stack<Char>()
            // 将字符串中的字符依次入栈
            for (char in s) {
                if(char != '*') stack.push(char)
                else stack.pop()
            }
            return stack.joinToString("")
        }

        val ret = removeStars("erase*****")
        println("ret is $ret")
    }

    @Test
    fun leetcode735() {
        /**
        We are given an array asteroids of integers representing asteroids in a row.
        For each asteroid, the absolute value represents its size, and the sign represents its direction
        (positive meaning right, negative meaning left). Each asteroid moves at the same speed.
        Find out the state of the asteroids after all collisions. If two asteroids meet, the smaller one will explode.
        If both are the same size, both will explode. Two asteroids moving in the same direction will never meet.

        Example 1:
        Input: asteroids = [5,10,-5]
        Output: [5,10]
        Explanation: The 10 and -5 collide resulting in 10. The 5 and 10 never collide.

        Example 2:
        Input: asteroids = [8,-8]
        Output: []
        Explanation: The 8 and -8 collide exploding each other.

        Example 3:
        Input: asteroids = [10,2,-5]
        Output: [10]
        Explanation: The 2 and -5 collide resulting in -5. The 10 and -5 collide resulting in 10.
        */
        fun asteroidCollision(asteroids: IntArray): IntArray {
            val stack = Stack<Int>()

            for (asteroid in asteroids) {
                if (asteroid > 0) {
                    stack.push(asteroid)
                } else {
                    while (!stack.isEmpty() && stack.peek() > 0 && stack.peek() < Math.abs(asteroid)) {
                        stack.pop()
                    }

                    if (stack.isEmpty() || stack.peek() < 0) {
                        stack.push(asteroid)
                    } else if (stack.peek() == Math.abs(asteroid)) {
                        stack.pop()
                    }
                }
            }

            return stack.toIntArray()
        }

        printIntArray(asteroidCollision(intArrayOf(5,10,-5)))
        printIntArray(asteroidCollision(intArrayOf(8,-8)))
        printIntArray(asteroidCollision(intArrayOf(10,2,-5)))
    }

    @Test
    fun leetcode394() {
        /**
        Given an encoded string, return its decoded string.
        The encoding rule is: k[encoded_string], where the encoded_string inside the square brackets is being
        repeated exactly k times. Note that k is guaranteed to be a positive integer.
        You may assume that the input string is always valid; there are no extra white spaces,
        square brackets are well-formed, etc. Furthermore, you may assume that the original data does not contain
        any digits and that digits are only for those repeat numbers, k. For example, there will not be input like 3a or 2[4].
        The test cases are generated so that the length of the output will never exceed 105.

        Example 1:
        Input: s = "3[a]2[bc]"
        Output: "aaabcbc"

        Example 2:
        Input: s = "3[a2[c]]"
        Output: "accaccacc"

        Example 3:
        Input: s = "2[abc]3[cd]ef"
        Output: "abcabccdcdcdef"
         */
        fun decodeString(s: String): String {
            val stack = Stack<String>()
            var currentNumber = 0
            var currentString = ""

            for (char in s) {
                when {
                    char.isDigit() -> {
                        currentNumber = currentNumber * 10 + (char - '0')
                    }
                    char == '[' -> {
                        stack.push(currentString)
                        stack.push(currentNumber.toString())
                        currentNumber = 0
                        currentString = ""
                    }
                    char == ']' -> {
                        val number = stack.pop().toInt()
                        val previousString = stack.pop()

                        currentString = previousString + currentString.repeat(number)
                    }
                    else -> {
                        currentString += char
                    }
                }
            }

            return currentString
        }

        println("ret is ${decodeString("3[a]2[bc]")}") //"aaabcbc"
        println("ret is ${decodeString("3[a2[c]]")}") //"accaccacc"
        println("ret is ${decodeString("2[abc]3[cd]ef")}") //"abcabccdcdcdef"
        println("ret is ${decodeString("100[leetcode]")}") //"abcabccdcdcdef"

    }

    @Test
    fun leetcode328() {
        /**
        Given the head of a singly linked list, group all the nodes with odd indices together
        followed by the nodes with even indices, and return the reordered list.
        The first node is considered odd, and the second node is even, and so on.
        Note that the relative order inside both the even and odd groups should remain as it was in the input.

        Example 1:
        Input: head = [1,2,3,4,5]
        Output: [1,3,5,2,4]

        Example 2:
        Input: head = [2,1,3,5,6,4,7]
        Output: [2,3,6,7,1,5,4]
         */
        fun oddEvenList(head: ListNode?): ListNode? {
            if (head == null) {
                return null
            }

            var odd = head
            var even = head.next
            val evenHead = even

            while (even?.next != null) {
                odd?.next = even.next
                odd = odd?.next
                even.next = odd?.next
                even = even.next
            }

            odd?.next = evenHead

            return head
        }
    }

    @Test
    fun leetcode2130() {
        /**
        In a linked list of size n, where n is even, the ith node (0-indexed) of the linked list is known as
        the twin of the (n-1-i)th node, if 0 <= i <= (n / 2) - 1.
        For example, if n = 4, then node 0 is the twin of node 3, and node 1 is the twin of node 2.
        These are the only nodes with twins for n = 4.
        The twin sum is defined as the sum of a node and its twin.
        Given the head of a linked list with even length, return the maximum twin sum of the linked list.

        Example 1:
        Input: head = [5,4,2,1]
        Output: 6

        Example 2:
        Input: head = [4,2,2,3]
        Output: 7

        Example 3:
        Input: head = [1,100000]
        Output: 100001
         */
        fun pairSum(head: ListNode?): Int {
            val deque = ArrayDeque<Int>()
            var max = 0
            var curr = head

            while(curr != null) {
                deque.add(curr.`val`)
                curr = curr?.next
            }

            while(deque.isNotEmpty()) {
                val sum = deque.removeFirst() + deque.removeLast()
                if (sum > max) max = sum
            }

            return max
        }
    }

    @Test
    fun leetcode104() {
        /**
        Given the root of a binary tree, return its maximum depth.
        A binary tree's maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.

        Example 1:
        Input: root = [3,9,20,null,null,15,7]
        Output: 3

        Example 2:
        Input: root = [1,null,2]
        Output: 2
        */
        val tree3 = TreeNode(3)
        val tree9 = TreeNode(9)
        val tree20 = TreeNode(20)
        val tree15 = TreeNode(15)
        val tree7 = TreeNode(7)
        tree3.left = tree9
        tree3.right = tree20
        tree20.left = tree15
        tree20.right = tree7

        fun maxDepth(root: TreeNode?): Int {
            if (root == null) {
                return 0
            }

            val leftDepth = maxDepth(root.left)
            val rightDepth = maxDepth(root.right)

            return 1 + maxOf(leftDepth, rightDepth)
        }

        println("ret is ${maxDepth(tree3)}") //3
        assertEquals(maxDepth(tree3), 3)
    }

    @Test
    fun leetcode872() {
        /**
        Consider all the leaves of a binary tree, from left to right order, the values of those leaves form a leaf value sequence.
        For example, in the given tree above, the leaf value sequence is (6, 7, 4, 9, 8).
        Two binary trees are considered leaf-similar if their leaf value sequence is the same.
        Return true if and only if the two given trees with head nodes root1 and root2 are leaf-similar.

        Example 1:
        Input: root1 = [3,5,1,6,2,9,8,null,null,7,4], root2 = [3,5,1,6,7,4,2,null,null,null,null,null,null,9,8]
        Output: true

        Example 2:
        Input: root1 = [1,2,3], root2 = [1,3,2]
        Output: false
         */
        fun getLeaves(root: TreeNode?, leaves: MutableList<Int>) {
            if (root == null) {
                return
            }

            if (root.left == null && root.right == null) {
                leaves.add(root.`val`)
            }

            getLeaves(root.left, leaves)
            getLeaves(root.right, leaves)
        }

        fun leafSimilar(root1: TreeNode?, root2: TreeNode?): Boolean {
            val leaves1 = mutableListOf<Int>()
            val leaves2 = mutableListOf<Int>()

            getLeaves(root1, leaves1)
            getLeaves(root2, leaves2)

            return leaves1 == leaves2
        }

        println("ret is ${leafSimilar(buildTree(arrayOf(3,5,1,6,2,9,8,null,null,7,4)), 
            buildTree(arrayOf(3,5,1,6,7,4,2,null,null,null,null,null,null,9,8)))}") //true

        assertEquals(leafSimilar(buildTree(arrayOf(3,5,1,6,2,9,8,null,null,7,4)),
            buildTree(arrayOf(3,5,1,6,7,4,2,null,null,null,null,null,null,9,8))), true)

        println("ret is ${leafSimilar(buildTree(arrayOf(1,2,3)),
            buildTree(arrayOf(1,3,2)))}") //false
        assertEquals(leafSimilar(buildTree(arrayOf(1,2,3)),
            buildTree(arrayOf(1,3,2))),false)
    }

    @Test
    fun leetcode1448() {
        /**
        Given a binary tree root, a node X in the tree is named good if in the path from root to X there are no nodes with
        a value greater than X.
        Return the number of good nodes in the binary tree.

        Example 1:
        Input: root = [3,1,4,3,null,1,5]
        Output: 4
        Explanation: Nodes in blue are good.
        Root Node (3) is always a good node.
        Node 4 -> (3,4) is the maximum value in the path starting from the root.
        Node 5 -> (3,4,5) is the maximum value in the path
        Node 3 -> (3,1,3) is the maximum value in the path.

        Example 2:
        Input: root = [3,3,null,4,2]
        Output: 3
        Explanation: Node 2 -> (3, 3, 2) is not good, because "3" is higher than it.

        Example 3:
        Input: root = [1]
        Output: 1
        Explanation: Root is considered as good.
         */
        fun countGoodNodes(node: TreeNode?, maxValue: Int): Int {
            if (node == null) {
                return 0
            }

            var count = 0

            if (node.`val` >= maxValue) {
                count++
            }

            val updatedMaxValue = maxOf(maxValue, node.`val`)

            count += countGoodNodes(node.left, updatedMaxValue)
            count += countGoodNodes(node.right, updatedMaxValue)

            return count
        }

        fun goodNodes(root: TreeNode?): Int {
            return countGoodNodes(root, Int.MIN_VALUE)
        }

        println("ret is ${goodNodes(buildTree(arrayOf(3,1,4,3,null,1,5)))}") // 4
        assertEquals(goodNodes(buildTree(arrayOf(3,1,4,3,null,1,5))),4)
        println("ret is ${goodNodes(buildTree(arrayOf(3,3,null,4,2)))}") // 3
        assertEquals(goodNodes(buildTree(arrayOf(3,3,null,4,2))),3)
        println("ret is ${goodNodes(buildTree(arrayOf(1)))}") // 1
        assertEquals(goodNodes(buildTree(arrayOf(1))),1)
    }

    @Test
    fun leetcode437() {
        /**
        Given the root of a binary tree and an integer targetSum, return the number of paths where the sum of the values along the path equals targetSum.
        The path does not need to start or end at the root or a leaf, but it must go downwards (i.e., traveling only from parent nodes to child nodes).
        Example 1:
        Input: root = [10,5,-3,3,2,null,11,3,-2,null,1], targetSum = 8
        Output: 3
        Explanation: The paths that sum to 8 are shown.

        Example 2:
        Input: root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
        Output: 3
         */
        fun countPathSum(node: TreeNode?, targetSum: Int): Int {
            if (node == null) {
                return 0
            }

            var count = 0

            // 检查以当前节点为起点的路径和是否等于目标和
            if (node.`val` == targetSum) {
                count++
            }

            // 递归搜索左子树和右子树
            count += countPathSum(node.left, targetSum - node.`val`) + countPathSum(node.right, targetSum - node.`val`)

            return count
        }

        fun pathSum(root: TreeNode?, targetSum: Int): Int {
            if (root == null) {
                return 0
            }

            // 从根节点开始的路径和数量 + 左子树中的路径和数量 + 右子树中的路径和数量
            val pathCount = countPathSum(root, targetSum) + pathSum(root.left, targetSum) + pathSum(root.right, targetSum)
            return pathCount
        }

        println("ret is ${pathSum(buildTree(arrayOf(10,5,-3,3,2,null,11,3,-2,null,1)), 8)}") // 3
        assertEquals(pathSum(buildTree(arrayOf(10,5,-3,3,2,null,11,3,-2,null,1)), 8),3)
        println("ret is ${pathSum(buildTree(arrayOf(5,4,8,11,null,13,4,7,2,null,null,5,1)), 22)}") // 3
        assertEquals(pathSum(buildTree(arrayOf(5,4,8,11,null,13,4,7,2,null,null,5,1)), 22),3)
        // bellow test failed, tree item value is greater than int.max. See leetcode437_solution2
        //println("ret is ${pathSum(buildTree(arrayOf(1000000000,1000000000,null,294967296,null,1000000000,null,1000000000,null,1000000000)), 0)}") // 0
        //assertEquals(pathSum(buildTree(arrayOf(1000000000,1000000000,null,294967296,null,1000000000,null,1000000000,null,1000000000)), 0),0)
    }

    @Test
    fun leetcode437_solution2() {
        fun pathSumFromRoot(root: TreeNode?, targetSum: Long): Int {
            if (root == null) return 0

            val match = if (targetSum == root.`val`.toLong()) 1 else 0
            return pathSumFromRoot(root.left, targetSum - root.`val`) + pathSumFromRoot(root.right, targetSum - root.`val`) + match
        }

        fun pathSum(root: TreeNode?, targetSum: Int): Int {
            if(root == null) return 0
            return pathSumFromRoot(root, targetSum.toLong()) + pathSum(root.left, targetSum) + pathSum(root.right, targetSum)
        }

        println("ret is ${pathSum(buildTree(arrayOf(10,5,-3,3,2,null,11,3,-2,null,1)), 8)}") // 3
        assertEquals(pathSum(buildTree(arrayOf(10,5,-3,3,2,null,11,3,-2,null,1)), 8), 3)
        println("ret is ${pathSum(buildTree(arrayOf(5,4,8,11,null,13,4,7,2,null,null,5,1)), 22)}") // 3
        assertEquals(pathSum(buildTree(arrayOf(5,4,8,11,null,13,4,7,2,null,null,5,1)), 22), 3)
        println("ret is ${pathSum(buildTree(arrayOf(1000000000,1000000000,null,294967296,null,1000000000,null,1000000000,null,1000000000)), 0)}") // 0
        assertEquals(pathSum(buildTree(arrayOf(1000000000,1000000000,null,294967296,null,1000000000,null,1000000000,null,1000000000)), 0),0)

    }

    @Test
    fun leetcode1372() {
        /**
        You are given the root of a binary tree.
        A ZigZag path for a binary tree is defined as follow:
        Choose any node in the binary tree and a direction (right or left).
        If the current direction is right, move to the right child of the current node; otherwise, move to the left child.
        Change the direction from right to left or from left to right.
        Repeat the second and third steps until you can't move in the tree.
        Zigzag length is defined as the number of nodes visited - 1. (A single node has a length of 0).

        Return the longest ZigZag path contained in that tree.

        Example 1:
        Input: root = [1,null,1,1,1,null,null,1,1,null,1,null,null,null,1]
        Output: 3

        Example 2:
        Input: root = [1,1,1,null,1,null,null,1,1,null,1]
        Output: 4

        Example 3:
        Input: root = [1]
        Output: 0
         */
        fun dfs(root: TreeNode?, l: Int, r: Int): Int {
            if (root == null) return maxOf(l,r)-1
            return maxOf(dfs(root.left, r+1, 0), dfs(root.right, 0, l+1))
        }

        fun longestZigZag(root: TreeNode?): Int = dfs(root, 0, 0)
        val tree1 = buildTree(arrayOf(1,null,1,1,1,null,null,1,1,null,1,null,null,null,1))

        println("ret is ${longestZigZag(tree1)}") // 3
        assertEquals(longestZigZag(tree1), 3)
        println("ret is ${longestZigZag(buildTree(arrayOf(1,1,1,null,1,null,null,1,1,null,1)))}") // 4
        assertEquals(longestZigZag(buildTree(arrayOf(1,1,1,null,1,null,null,1,1,null,1))), 4)
    }

    @Test
    fun leetcode236() {
        /**
        Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.
        According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node
        in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”
        Example 1:
        Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
        Output: 3
        Explanation: The LCA of nodes 5 and 1 is 3.

        Example 2:
        Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
        Output: 5
        Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

        Example 3:
        Input: root = [1,2], p = 1, q = 2
        Output: 1
         */
        fun lowestCommonAncestor(root: TreeNode?, p: TreeNode?, q: TreeNode?): TreeNode? {
            // 如果根节点为空或者等于目标节点之一，则返回根节点
            if (root == null || root == p || root == q) {
                return root
            }

            // 在左子树中寻找目标节点
            val left = lowestCommonAncestor(root.left, p, q)
            // 在右子树中寻找目标节点
            val right = lowestCommonAncestor(root.right, p, q)

            // 如果左子树和右子树都找到了目标节点，则当前节点为最低共同祖先
            if (left != null && right != null) {
                return root
            }

            // 如果只有左子树找到了目标节点，则返回左子树的结果
            return left ?: right
        }
    }

    @Test
    fun leetcode199() {
        fun rightSideView(root: TreeNode?): List<Int> {
            val rightmostValues = mutableListOf<Int>()
            if(root == null) return rightmostValues
            val queue: Queue<TreeNode> = LinkedList()
            queue.offer(root)

            while (queue.isNotEmpty()) {
                val levelSize = queue.size
                var rightmostNode: TreeNode? = null

                repeat(levelSize) {
                    val node = queue.poll()
                    rightmostNode = node

                    node.left?.let { queue.offer(it) }
                    node.right?.let { queue.offer(it) }
                }

                rightmostNode?.let { rightmostValues.add(it.`val`) }
            }

            return rightmostValues
        }

        var tree = buildTree(arrayOf(1,2,3,null,5,null,4))
        var ret = rightSideView(tree)
        assertEquals(listOf(1, 3, 4).toList(), ret)

        tree = buildTree(arrayOf(1,2))
        ret = rightSideView(tree)
        assertEquals(listOf(1, 2).toList(), ret)

        tree = buildTree(arrayOf(1,2,3,4))
        ret = rightSideView(tree)
        assertEquals(listOf(1,3,4).toList(), ret)

    }

    @Test
    fun leetcode1161() {
        fun maxLevelSum(root: TreeNode?): Int {
            if (root == null) return 0
            var maxIndex = 0
            var maxValue = Int.MIN_VALUE
            var levelNodes = mutableListOf<TreeNode>()
            levelNodes.add(root)

            var index = 1
            while (levelNodes.isNotEmpty()) {
                val levelQueue = LinkedList(levelNodes)

                val sum = levelNodes.map {it.`val`}.sum()

                if (sum > maxValue) {
                    maxValue = sum
                    maxIndex = index
                }
                index += 1

                levelNodes = mutableListOf()

                while (levelQueue.isNotEmpty()) {
                    val cur = levelQueue.poll()

                    cur.left?.let { levelNodes.add(it) }
                    cur.right?.let { levelNodes.add(it) }
                }
            }

            return maxIndex
        }

        var tree = buildTree(arrayOf(1,7,0,7,-8,null,null))
        var ret = maxLevelSum(tree)
        assertEquals(2, ret)

        tree = buildTree(arrayOf(989,null,10250,98693,-89388,null,null,null,-32127))
        ret = maxLevelSum(tree)
        assertEquals(2, ret)

    }

    @Test
    fun leetcode700() {
        fun searchBST(root: TreeNode?, `val`: Int): TreeNode? {
            var c = root

            while(c != null) {
                if(c.`val` == `val`) return c
                else {
                    if(c.`val` > `val`) c = c.left
                    else c = c.right
                }
            }

            return c
        }

        var ret = searchBST(buildTree(arrayOf(4,2,7,1,3)), 2)
        println("ret val is ${ret?.`val`}")
        assertEquals(2, ret?.`val`)

        ret = searchBST(buildTree(arrayOf(4,2,7,1,3)), 5)
        println("ret val is ${ret?.`val`}")
        assertEquals(null, ret?.`val`)
    }

    @Test
    fun leetcode450() {
        fun findSuccessorValue(node: TreeNode?): Int {
            var current = node
            while (current?.left != null) {
                current = current.left
            }
            return current!!.`val`
        }

        fun deleteNode(root: TreeNode?, key: Int): TreeNode? {
            if (root == null) {
                return null
            }

            if (key < root.`val`) {
                root.left = deleteNode(root.left, key)
            } else if (key > root.`val`) {
                root.right = deleteNode(root.right, key)
            } else {
                if (root.left == null) {
                    return root.right
                } else if (root.right == null) {
                    return root.left
                }

                val successorValue = findSuccessorValue(root.right)
                root.`val` = successorValue
                root.right = deleteNode(root.right, successorValue)
            }

            return root
        }

        val ret = deleteNode(buildTree(arrayOf(5,3,6,2,4,null,7)), 3)
        printTree(ret)
    }

    @Test
    fun leetcode841() {
        /**
        There are n rooms labeled from 0 to n - 1 and all the rooms are locked except for room 0.
        Your goal is to visit all the rooms. However, you cannot enter a locked room without having its key.
        When you visit a room, you may find a set of distinct keys in it. Each key has a number on it, denoting which room it unlocks,
        and you can take all of them with you to unlock the other rooms.
        Given an array rooms where rooms[i] is the set of keys that you can obtain if you visited room i,
        return true if you can visit all the rooms, or false otherwise.

        Example 1:
        Input: rooms = [[1],[2],[3],[]]
        Output: true
        Explanation:
        We visit room 0 and pick up key 1.
        We then visit room 1 and pick up key 2.
        We then visit room 2 and pick up key 3.
        We then visit room 3.
        Since we were able to visit every room, we return true.

        Example 2:
        Input: rooms = [[1,3],[3,0,1],[2],[0]]
        Output: false
        Explanation: We can not enter room number 2 since the only key that unlocks it is in that room.
         */
        fun canVisitAllRooms(rooms: List<List<Int>>): Boolean {
            val visited = BooleanArray(rooms.size) // 用于记录房间的访问状态
            visited[0] = true // 将第一个房间标记为已访问
            val queue: Queue<Int> = LinkedList<Int>() // 使用队列来进行广度优先搜索
            queue.offer(0) // 将第一个房间加入队列

            while (queue.isNotEmpty()) {
                val currRoom = queue.poll()
                for (key in rooms[currRoom]) {
                    if (!visited[key]) {
                        visited[key] = true
                        queue.offer(key)
                    }
                }
            }

            // 检查是否所有房间都被访问过
            for (roomVisited in visited) {
                if (!roomVisited) {
                    return false
                }
            }

            return true
        }

        // [[1],[2],[3],[]]
        assertEquals(true, canVisitAllRooms(
            listOf(
                listOf(1),
                listOf(2),
                listOf(3),
                listOf()
            )))

        // [[1,3],[3,0,1],[2],[0]]
        assertEquals(false, canVisitAllRooms(
            listOf(
                listOf(1,3),
                listOf(3,0,1),
                listOf(2),
                listOf(0)
            )))
    }

    @Test
    fun leetcode1466() {
        /**
        There are n cities numbered from 0 to n - 1 and n - 1 roads such that there is only one way to
        travel between two different cities (this network form a tree). Last year,
        The ministry of transport decided to orient the roads in one direction because they are too narrow.

        Roads are represented by connections where connections[i] = [ai, bi] represents a road from city ai to city bi.
        This year, there will be a big event in the capital (city 0), and many people want to travel to this city.
        Your task consists of reorienting some roads such that each city can visit the city 0. Return the minimum number of edges changed.
        It's guaranteed that each city can reach city 0 after reorder.

        Example 1:
        Input: n = 6, connections = [[0,1],[1,3],[2,3],[4,0],[4,5]]
        Output: 3
        Explanation: Change the direction of edges show in red such that each node can reach the node 0 (capital).

        Example 2:
        Input: n = 5, connections = [[1,0],[1,2],[3,2],[3,4]]
        Output: 2
        Explanation: Change the direction of edges show in red such that each node can reach the node 0 (capital).

        Example 3:
        Input: n = 3, connections = [[1,0],[2,0]]
        Output: 0
         */
        fun dfs(graph: Array<MutableList<Pair<Int, Int>>>, visited: BooleanArray, node: Int): Int {
            var count = 0 // 记录需要改变方向的连接数

            visited[node] = true

            for (next in graph[node]) {
                val neighbor = next.first
                val direction = next.second

                if (!visited[neighbor]) {
                    count += dfs(graph, visited, neighbor) + direction
                }
            }

            return count
        }

        fun minReorder(n: Int, connections: Array<IntArray>): Int {
            val graph = Array(n) { mutableListOf<Pair<Int, Int>>() } // 创建邻接表

            for (conn in connections) {
                val from = conn[0]
                val to = conn[1]
                graph[from].add(to to 1) // 正向连接
                graph[to].add(from to 0) // 反向连接
            }

            val visited = BooleanArray(n) // 记录节点是否已访问
            return dfs(graph, visited, 0) // 从节点 0 开始进行深度优先搜索
        }

        //Input: n = 6, connections = [[0,1],[1,3],[2,3],[4,0],[4,5]]
        //Output: 3
        assertEquals(3, minReorder(6,
            arrayOf(
                intArrayOf(0,1),
                intArrayOf(1,3),
                intArrayOf(2,3),
                intArrayOf(4,0),
                intArrayOf(4,5),
            )
        ))

        //Input: n = 5, connections = [[1,0],[1,2],[3,2],[3,4]]
        //Output: 2
        assertEquals(2, minReorder(5,
            arrayOf(
                intArrayOf(1,0),
                intArrayOf(1,2),
                intArrayOf(3,2),
                intArrayOf(3,4),
            )
        ))

        //Input: n = 3, connections = [[1,0],[2,0]]
        //Output: 0
        assertEquals(0, minReorder(3,
            arrayOf(
                intArrayOf(1,0),
                intArrayOf(2,0),
            )
        ))

    }

    @Test
    fun leetcode1926() {
        /**
        You are given an m x n matrix maze (0-indexed) with empty cells (represented as '.') and walls (represented as '+').
        You are also given the entrance of the maze, where entrance = [entrancerow, entrancecol] denotes the row and column of the cell you are initially standing at.

        In one step, you can move one cell up, down, left, or right. You cannot step into a cell with a wall,
        and you cannot step outside the maze. Your goal is to find the nearest exit from the entrance.
        An exit is defined as an empty cell that is at the border of the maze. The entrance does not count as an exit.

        Return the number of steps in the shortest path from the entrance to the nearest exit, or -1 if no such path exists.

        Example 1:
        Input: maze = [["+","+",".","+"],[".",".",".","+"],["+","+","+","."]], entrance = [1,2]
        Output: 1
        Explanation: There are 3 exits in this maze at [1,0], [0,2], and [2,3].
        Initially, you are at the entrance cell [1,2].
        - You can reach [1,0] by moving 2 steps left.
        - You can reach [0,2] by moving 1 step up.
        It is impossible to reach [2,3] from the entrance.
        Thus, the nearest exit is [0,2], which is 1 step away.

        Example 2:
        Input: maze = [["+","+","+"],[".",".","."],["+","+","+"]], entrance = [1,0]
        Output: 2
        Explanation: There is 1 exit in this maze at [1,2].
        [1,0] does not count as an exit since it is the entrance cell.
        Initially, you are at the entrance cell [1,0].
        - You can reach [1,2] by moving 2 steps right.
        Thus, the nearest exit is [1,2], which is 2 steps away.
        */
        fun nearestExit(maze: Array<CharArray>, entrance: IntArray): Int {
            val directions = arrayOf(intArrayOf(1, 0), intArrayOf(-1, 0), intArrayOf(0, 1), intArrayOf(0, -1))
            val rows = maze.size
            val cols = maze[0].size

            val queue = LinkedList<IntArray>()
            queue.offer(entrance)
            maze[entrance[0]][entrance[1]] = '+'

            var steps = 0
            while (queue.isNotEmpty()) {
                val size = queue.size
                for (i in 0 until size) {
                    val curr = queue.poll()
                    val row = curr[0]
                    val col = curr[1]

                    if ((row == 0 || row == rows - 1 || col == 0 || col == cols - 1) && curr != entrance) {
                        return steps
                    }

                    for (dir in directions) {
                        val newRow = row + dir[0]
                        val newCol = col + dir[1]

                        if (newRow >= 0 && newRow < rows && newCol >= 0 && newCol < cols && maze[newRow][newCol] == '.') {
                            maze[newRow][newCol] = '+'
                            queue.offer(intArrayOf(newRow, newCol))
                        }
                    }
                }
                steps++
            }

            return -1
        }

        //Input: maze = [["+","+",".","+"],[".",".",".","+"],["+","+","+","."]], entrance = [1,2]
        //Output: 1
        assertEquals(1, nearestExit(arrayOf(charArrayOf('+','+','.','+')
            ,charArrayOf('.','.','.','+')
            ,charArrayOf('+','+','+','.'))
            ,intArrayOf(1,2)))

        //Input: [["+","+","+"],[".",".","."],["+","+","+"]], entrance = [1,0]
        //Output: 2
        assertEquals(2, nearestExit(arrayOf(charArrayOf('+','+','+')
            ,charArrayOf('.','.','.')
            ,charArrayOf('+','+','+'))
            ,intArrayOf(1,2)))

    }

    @Test
    fun leetcode215() {
        /**
        Given an integer array nums and an integer k, return the kth largest element in the array.
        Note that it is the kth largest element in the sorted order, not the kth distinct element.
        You must solve it in O(n) time complexity.
        Example 1:
        Input: nums = [3,2,1,5,6,4], k = 2
        Output: 5

        Example 2:
        Input: nums = [3,2,3,1,2,4,5,5,6], k = 4
        Output: 4
         */
        fun findKthLargest(nums: IntArray, k: Int): Int? {
            val pq = PriorityQueue<Int>()

            for (num in nums) {
                pq.offer(num)

                if (pq.size > k) {
                    pq.poll()
                }
            }

            return pq.peek()
        }

        //        Input: nums = [3,2,1,5,6,4], k = 2
        //        Output: 5
        assertEquals(5, findKthLargest(intArrayOf(3,2,1,5,6,4), 2))

        //        Input: nums = [3,2,3,1,2,4,5,5,6], k = 4
        //        Output: 4
        assertEquals(4, findKthLargest(intArrayOf(3,2,3,1,2,4,5,5,6), 4))
    }

    @Test
    fun leetcode2542() {
        /**
        You are given two 0-indexed integer arrays nums1 and nums2 of equal length n and a positive integer k.
        You must choose a subsequence of indices from nums1 of length k.
        For chosen indices i0, i1, ..., ik - 1, your score is defined as:
        The sum of the selected elements from nums1 multiplied with the minimum of the selected elements from nums2.
        It can defined simply as: (nums1[i0] + nums1[i1] +...+ nums1[ik - 1]) * min(nums2[i0] , nums2[i1], ... ,nums2[ik - 1]).
        Return the maximum possible score.
        A subsequence of indices of an array is a set that can be derived from the set {0, 1, ..., n-1} by deleting some or no elements.

        Example 1:
        Input: nums1 = [1,3,3,2], nums2 = [2,1,3,4], k = 3
        Output: 12
        Explanation:
        The four possible subsequence scores are:
        - We choose the indices 0, 1, and 2 with score = (1+3+3) * min(2,1,3) = 7.
        - We choose the indices 0, 1, and 3 with score = (1+3+2) * min(2,1,4) = 6.
        - We choose the indices 0, 2, and 3 with score = (1+3+2) * min(2,3,4) = 12.
        - We choose the indices 1, 2, and 3 with score = (3+3+2) * min(1,3,4) = 8.
        Therefore, we return the max score, which is 12.

        Example 2:
        Input: nums1 = [4,2,3,1,1], nums2 = [7,5,10,9,6], k = 1
        Output: 30
        Explanation:
        Choosing index 2 is optimal: nums1[2] * nums2[2] = 3 * 10 = 30 is the maximum possible score.
         */
        fun maxScore(nums1: IntArray, nums2: IntArray, k: Int): Long {
            val list = mutableListOf<Pair<Int, Int>>()

            for (i in nums1.indices) {
                list.add(Pair(nums2[i], nums1[i]))
            }

            list.sortByDescending { it.first }

            val pq = PriorityQueue<Int>()
            var sum = 0L
            var result = 0L

            for (i in 0 until k) {
                pq.offer(list[i].second)
                sum += list[i].second
            }

            result = sum * list[k - 1].first

            for (i in k until nums1.size) {
                sum -= pq.poll()
                sum += list[i].second
                pq.offer(list[i].second)

                result = Math.max(result, sum * list[i].first)
            }

            return result
        }

        //        Input: nums1 = [1,3,3,2], nums2 = [2,1,3,4], k = 3
        //        Output: 12
        assertEquals(12, maxScore(intArrayOf(1,3,3,2), intArrayOf(2,1,3,4), 3))

        //        Input: nums1 = [4,2,3,1,1], nums2 = [7,5,10,9,6], k = 1
        //        Output: 30
        assertEquals(30, maxScore(intArrayOf(4,2,3,1,1), intArrayOf(7,5,10,9,6), 1))
    }

    @Test
    fun leetcode2462() {
        /**
        You are given a 0-indexed integer array costs where costs[i] is the cost of hiring the ith worker.

        You are also given two integers k and candidates. We want to hire exactly k workers according to the following rules:

        You will run k sessions and hire exactly one worker in each session.
        In each hiring session, choose the worker with the lowest cost from either the first candidates workers or the last candidates workers. Break the tie by the smallest index.
        For example, if costs = [3,2,7,7,1,2] and candidates = 2, then in the first hiring session, we will choose the 4th worker because they have the lowest cost [3,2,7,7,1,2].
        In the second hiring session, we will choose 1st worker because they have the same lowest cost as 4th worker but they have the smallest index [3,2,7,7,2].
        Please note that the indexing may be changed in the process.
        If there are fewer than candidates workers remaining, choose the worker with the lowest cost among them. Break the tie by the smallest index.
        A worker can only be chosen once.
        Return the total cost to hire exactly k workers.

        Example 1:
        Input: costs = [17,12,10,2,7,2,11,20,8], k = 3, candidates = 4
        Output: 11
        Explanation: We hire 3 workers in total. The total cost is initially 0.
        - In the first hiring round we choose the worker from [17,12,10,2,7,2,11,20,8]. The lowest cost is 2, and we break the tie by the smallest index, which is 3. The total cost = 0 + 2 = 2.
        - In the second hiring round we choose the worker from [17,12,10,7,2,11,20,8]. The lowest cost is 2 (index 4). The total cost = 2 + 2 = 4.
        - In the third hiring round we choose the worker from [17,12,10,7,11,20,8]. The lowest cost is 7 (index 3). The total cost = 4 + 7 = 11. Notice that the worker with index 3 was common in the first and last four workers.
        The total hiring cost is 11.

        Example 2:
        Input: costs = [1,2,4,1], k = 3, candidates = 3
        Output: 4
        Explanation: We hire 3 workers in total. The total cost is initially 0.
        - In the first hiring round we choose the worker from [1,2,4,1]. The lowest cost is 1, and we break the tie by the smallest index, which is 0. The total cost = 0 + 1 = 1. Notice that workers with index 1 and 2 are common in the first and last 3 workers.
        - In the second hiring round we choose the worker from [2,4,1]. The lowest cost is 1 (index 2). The total cost = 1 + 1 = 2.
        - In the third hiring round there are less than three candidates. We choose the worker from the remaining workers [2,4]. The lowest cost is 2 (index 0). The total cost = 2 + 2 = 4.
        The total hiring cost is 4.
         */
        fun totalCost(costs: IntArray, k: Int, candidates: Int): Long {
            var left = 0
            var right = costs.lastIndex
            val canL = PriorityQueue<Int>()
            val canR = PriorityQueue<Int>()

            var res = 0L
            var n = k
            while (n > 0) {
                while (canL.size < candidates && left <= right) canL.add(costs[left++])
                while (canR.size < candidates && left <= right) canR.add(costs[right--])

                if (canL.peek()?: 100001 <= canR.peek()?: 100001) {
                    res += canL.poll()
                } else {
                    res += canR.poll()
                }
                n--
            }

            return res
        }

        //        Input: costs = [17,12,10,2,7,2,11,20,8], k = 3, candidates = 4
        //        Output: 11
        assertEquals(11, totalCost(intArrayOf(17,12,10,2,7,2,11,20,8), 3, 4))

        //        Input: costs = [1,2,4,1], k = 3, candidates = 3
        //        Output: 4
        assertEquals(4, totalCost(intArrayOf(1,2,4,1), 3, 3))

    }

    private fun printTree(root: TreeNode?) {
        if (root == null) return

        printTreeHelper(listOf(root))
    }

    private fun printTreeHelper(nodes: List<TreeNode?>) {
        if (nodes.isEmpty() || nodes.all { it == null }) return

        val nextLevelNodes = mutableListOf<TreeNode?>()
        val line = StringBuilder()

        for (node in nodes) {
            line.append(if (node != null) "${node.`val`}" else " ")
            line.append(" ".repeat(3))

            nextLevelNodes.add(node?.left)
            nextLevelNodes.add(node?.right)
        }

        println(line)

        // Print vertical lines
        val lineLength = line.length
        val verticalLine = StringBuilder("")

        for (i in 0 until lineLength) {
            verticalLine.append(if (line[i] == '|') '|' else ' ')
        }

        println(verticalLine)

        // Recursively print the next level
        printTreeHelper(nextLevelNodes)
    }
    private fun printCharArray(array: CharArray) {
        array.forEachIndexed { i, v ->
            println("chars[$i] = $v")
        }
    }

    private fun printIntArray(array: IntArray) {
        array.forEachIndexed { i, v ->
            println("Int[$i] = $v")
        }
    }

    private fun printStringArray(array: MutableList<String>) {
        array.forEachIndexed { i, v ->
            println("String[$i] = $v")
        }
    }

    private fun buildTree(nums: Array<Int?>): TreeNode? {
        if (nums.isEmpty()) {
            return null
        }

        val root = TreeNode(nums[0]!!)
        val queue: Queue<TreeNode?> = LinkedList()
        queue.offer(root)

        var index = 1
        while (index < nums.size) {
            val node = queue.poll()
            val leftValue = nums[index++]
            if (leftValue != null) {
                node?.left = TreeNode(leftValue)
                queue.offer(node?.left)
            }

            if (index < nums.size) {
                val rightValue = nums[index++]
                if (rightValue != null) {
                    node?.right = TreeNode(rightValue)
                    queue.offer(node?.right)
                }
            }
        }

        return root
    }

    class ListNode(var `val`: Int) {
        var next: ListNode? = null
    }

    class TreeNode(var `val`: Int) {
        var left: TreeNode? = null
        var right: TreeNode? = null
    }

}