package leetcode

/**
 *
322. 零钱兑换
 * https://leetcode-cn.com/problems/coin-change/
 * Created by test
 * Date 2019/12/1 12:01
 * Description
 * 给定不同面额的硬币 coins 和一个总金额 amount。编写一个函数来计算可以凑成总金额所需的最少的硬币个数。
 * 如果没有任何一种硬币组合能组成总金额，返回 -1。

示例 1:

输入: coins = [1, 2, 5], amount = 11
输出: 3
解释: 11 = 5 + 5 + 1
示例 2:

输入: coins = [2], amount = 3
输出: -1
说明:
你可以认为每种硬币的数量是无限的。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/coin-change
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
 */

object CoinChange {

    class Solution {
        /**
         * 回溯法，暴力法
         * 跟跳跃游戏，加油站一类解法
         */
        fun coinChange(coins: IntArray, amount: Int): Int {
            return coin(0,coins,amount)
        }

        fun coin(coinIndx: Int,coins: IntArray,amount: Int): Int {
            if (amount == 0) {
                return 0
            } else if (coinIndx < coins.size && amount > 0){
                val maxCoin = amount / coins[coinIndx]
                var minCost = Int.MAX_VALUE
                for (i in 0 .. maxCoin) {
                    if (amount >= coins[coinIndx] * coinIndx) {
                        val res = coin(coinIndx + 1, coins,amount - coins[coinIndx] * i)
                        if (res != -1) {
                            minCost = Math.min(minCost,res + i)
                        }
                    }
                }
                return if (minCost == Int.MAX_VALUE) {
                    -1
                } else minCost
            }
            return -1
        }

        /**
         * 带备忘录的解法
         * 1. 状态转移公式：
         * f(n) = {0 ,n = 0
         *  1+ min{ f(n - ci) , i 属于 [0,k]}
         * }
         * n为金额
         * ci 为币值
         */
        fun coinChange2(coins: IntArray, amount: Int): Int {
            val memo = IntArray(amount + 1)
            memo.forEachIndexed { index, i ->
                memo[index] = -2
            }
            return coin2(0,coins,amount,memo)
        }

        fun coin2(coinIndx: Int,coins: IntArray,amount: Int,memo: IntArray): Int {
            if (amount == 0) {
                return 0
            } else {
                if (memo[amount] != -2) return memo[amount]
                if (coinIndx < coins.size && amount > 0) {
                    val maxCoin = amount / coins[coinIndx]
                    var minCost = Int.MAX_VALUE
                    for (i in 0..maxCoin) {
                        if (amount >= coins[coinIndx] * coinIndx) {
                            val res = coin2(coinIndx + 1, coins, amount - coins[coinIndx] * i, memo)
                            if (res != -1) {
                                minCost = Math.min(minCost, res + i)
                            }
                        }
                    }
                    return if (minCost == Int.MAX_VALUE) {
                        -1
                    } else {
                        memo[amount] = minCost
                        memo[amount]
                    }
                }
                return -1
            }
        }




    }

    fun coinChange3(coins: IntArray, amount: Int): Int {
        val dp = IntArray(amount + 1)
        dp.forEachIndexed { index, i ->
            dp[index] = amount + 1
    }
        dp[0] = 0
        for (i in 1 .. amount) {
            for (coin in coins) {
                if (coin <= i) {
                    dp[i] = Math.min(dp[i],dp[i - coin] + 1)
                }
            }
        }
        return if (dp[amount] > amount) -1 else dp[amount]
    }


    @JvmStatic
    fun main(args: Array<String>) {

        println(coinChange3(intArrayOf(1,2,5),11))
    }
}
package leetcode

/**
 *
322. 零钱兑换
 * https://leetcode-cn.com/problems/coin-change/
 * Created by test
 * Date 2019/12/1 12:01
 * Description
 * 给定不同面额的硬币 coins 和一个总金额 amount。编写一个函数来计算可以凑成总金额所需的最少的硬币个数。
 * 如果没有任何一种硬币组合能组成总金额，返回 -1。

示例 1:

输入: coins = [1, 2, 5], amount = 11
输出: 3
解释: 11 = 5 + 5 + 1
示例 2:

输入: coins = [2], amount = 3
输出: -1
说明:
你可以认为每种硬币的数量是无限的。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/coin-change
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
 */

object CoinChange {

    class Solution {
        /**
         * 回溯法，暴力法
         * 跟跳跃游戏，加油站一类解法
         */
        fun coinChange(coins: IntArray, amount: Int): Int {
            return coin(0,coins,amount)
        }

        fun coin(coinIndx: Int,coins: IntArray,amount: Int): Int {
            if (amount == 0) {
                return 0
            } else if (coinIndx < coins.size && amount > 0){
                val maxCoin = amount / coins[coinIndx]
                var minCost = Int.MAX_VALUE
                for (i in 0 .. maxCoin) {
                    if (amount >= coins[coinIndx] * coinIndx) {
                        val res = coin(coinIndx + 1, coins,amount - coins[coinIndx] * i)
                        if (res != -1) {
                            minCost = Math.min(minCost,res + i)
                        }
                    }
                }
                return if (minCost == Int.MAX_VALUE) {
                    -1
                } else minCost
            }
            return -1
        }

        /**
         * 带备忘录的解法
         * 1. 状态转移公式：
         * f(n) = {0 ,n = 0
         *  1+ min{ f(n - ci) , i 属于 [0,k]}
         * }
         * n为金额
         * ci 为币值
         */
        fun coinChange2(coins: IntArray, amount: Int): Int {
            val memo = IntArray(amount + 1)
            memo.forEachIndexed { index, i ->
                memo[index] = -2
            }
            return coin2(0,coins,amount,memo)
        }

        fun coin2(coinIndx: Int,coins: IntArray,amount: Int,memo: IntArray): Int {
            if (amount == 0) {
                return 0
            } else {
                if (memo[amount] != -2) return memo[amount]
                if (coinIndx < coins.size && amount > 0) {
                    val maxCoin = amount / coins[coinIndx]
                    var minCost = Int.MAX_VALUE
                    for (i in 0..maxCoin) {
                        if (amount >= coins[coinIndx] * coinIndx) {
                            val res = coin2(coinIndx + 1, coins, amount - coins[coinIndx] * i, memo)
                            if (res != -1) {
                                minCost = Math.min(minCost, res + i)
                            }
                        }
                    }
                    return if (minCost == Int.MAX_VALUE) {
                        -1
                    } else {
                        memo[amount] = minCost
                        memo[amount]
                    }
                }
                return -1
            }
        }




    }

    fun coinChange3(coins: IntArray, amount: Int): Int {
        val dp = IntArray(amount + 1)
        dp.forEachIndexed { index, i ->
            dp[index] = amount + 1
    }
        dp[0] = 0
        for (i in 1 .. amount) {
            for (coin in coins) {
                if (coin <= i) {
                    dp[i] = Math.min(dp[i],dp[i - coin] + 1)
                }
            }
        }
        return if (dp[amount] > amount) -1 else dp[amount]
    }


    @JvmStatic
    fun main(args: Array<String>) {

        println(coinChange3(intArrayOf(1,2,5),11))
    }
}