Question 322 Coin Change

Link:

https://leetcode.com/problems/coin-change/

My Code:

class Solution:
    def coinChange(self, coins: List[int], amount: int) -> int:
        if amount == 0:
            return 0
        else:
            ResultList = [0]
            MinCoin = min(coins)
            for i in range(1,(amount+1)):
                if i < MinCoin:
                    ResultList.append(0)
                else:
                    AvailableCoinList = []
                    for j in range(len(coins)):
                        if coins[j]<=i:
                            AvailableCoinList.append(coins[j])
                    Min = 2**31
                    if i in AvailableCoinList:
                        ResultList.append(1)
                    else:
                        for k in AvailableCoinList:
                            if ResultList[i-k] != 0:
                                Min = min(ResultList[i-k],Min)
                        if Min == 2**31:
                            ResultList.append(0)
                        else:
                            ResultList.append(Min+1)
            if ResultList[amount] == 0:
                return -1
            else:
                return ResultList[amount]

Explanation: The code follows this idea: For an amount n, we check all the coins that is smaller than it, for example k_1,…,k_j as j coins that is available. The DP[i] list is the minimum amount of coins that can achieve amount i with given coin List. so We check all the available coin i in the list, then we take the minimum of all n-i. There are several situations: first, if n is in the coin list, then the answer is 1. Next if n is not in the coin list, then if all the previous n-k_j  are zero, then it means we cannot get the number here, hence we return 0. Otherwise we take the minimum of all non- zero n- k_j and then take the minimum and plus 1.