package com.github.solairerove.algs4.leprosorium.arrays

/**
 * You are given two integer arrays nums1 and nums2,
 * sorted in non-decreasing order, and two integers m and n,
 * representing the number of elements in nums1 and nums2 respectively.
 *
 * Merge nums1 and nums2 into a single array sorted in non-decreasing order.
 *
 * The final sorted array should not be returned by the function,
 * but instead be stored inside the array nums1.
 * To accommodate this, nums1 has a length of m + n,
 * where the first m elements denote the elements that should be merged,
 * and the last n elements are set to 0 and should be ignored. nums2 has a length of n.
 *
 * Input: nums1 = [1,2,3,0,0,0], m = 3, nums2 = [2,5,6], n = 3
 * Output: [1,2,2,3,5,6]
 * Explanation: The arrays we are merging are [1,2,3] and [2,5,6].
 * The result of the merge is [1,2,2,3,5,6] with the underlined elements coming from nums1.
 *
 * Input: nums1 = [1], m = 1, nums2 = [], n = 0
 * Output: [1]
 * Explanation: The arrays we are merging are [1] and [].
 * The result of the merge is [1].
 *
 * Input: nums1 = [0], m = 0, nums2 = [1], n = 1
 * Output: [1]
 * Explanation: The arrays we are merging are [] and [1].
 * The result of the merge is [1].
 * Note that because m = 0, there are no elements in nums1.
 * The 0 is only there to ensure the merge result can fit in nums1.
 */

// O(m + n) time | O(1) space
fun merge(nums1: IntArray, m: Int, nums2: IntArray, n: Int) {
    var high1 = m - 1
    var high2 = n - 1

    for (i in nums1.size - 1 downTo 0) {
        when {
            high2 < 0 -> return
            high1 >= 0 && nums1[high1] > nums2[high2] -> nums1[i] = nums1[high1--]
            else -> nums1[i] = nums2[high2--]
        }
    }
}

package com.github.solairerove.algs4.leprosorium.arrays

/**
 * You are given two integer arrays nums1 and nums2,
 * sorted in non-decreasing order, and two integers m and n,
 * representing the number of elements in nums1 and nums2 respectively.
 *
 * Merge nums1 and nums2 into a single array sorted in non-decreasing order.
 *
 * The final sorted array should not be returned by the function,
 * but instead be stored inside the array nums1.
 * To accommodate this, nums1 has a length of m + n,
 * where the first m elements denote the elements that should be merged,
 * and the last n elements are set to 0 and should be ignored. nums2 has a length of n.
 *
 * Input: nums1 = [1,2,3,0,0,0], m = 3, nums2 = [2,5,6], n = 3
 * Output: [1,2,2,3,5,6]
 * Explanation: The arrays we are merging are [1,2,3] and [2,5,6].
 * The result of the merge is [1,2,2,3,5,6] with the underlined elements coming from nums1.
 *
 * Input: nums1 = [1], m = 1, nums2 = [], n = 0
 * Output: [1]
 * Explanation: The arrays we are merging are [1] and [].
 * The result of the merge is [1].
 *
 * Input: nums1 = [0], m = 0, nums2 = [1], n = 1
 * Output: [1]
 * Explanation: The arrays we are merging are [] and [1].
 * The result of the merge is [1].
 * Note that because m = 0, there are no elements in nums1.
 * The 0 is only there to ensure the merge result can fit in nums1.
 */

// O(m + n) time | O(1) space
fun merge(nums1: IntArray, m: Int, nums2: IntArray, n: Int) {
    var high1 = m - 1
    var high2 = n - 1

    for (i in nums1.size - 1 downTo 0) {
        when {
            high2 < 0 -> return
            high1 >= 0 && nums1[high1] > nums2[high2] -> nums1[i] = nums1[high1--]
            else -> nums1[i] = nums2[high2--]
        }
    }
}
