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## Problem

You are given coins of different denominations and a total amount of money amount. Write a function to compute the fewest number of coins that you need to make up that amount. If that amount of money cannot be made up by any combination of the coins, return -1.

You may assume that you have an infinite number of each kind of coin.

Example 1:

Input: coins = [1,2,5], amount = 11
Output: 3
Explanation: 11 = 5 + 5 + 1


Example 2:

Input: coins = [2], amount = 3
Output: -1


Example 3:

Input: coins = [1], amount = 0
Output: 0


Example 4:

Input: coins = [1], amount = 1
Output: 1


Example 5:

Input: coins = [1], amount = 2
Output: 2


## Code

• 322: 无限硬币找和; 416: 有限硬币找和
• 322: 先遍历和在遍历硬币; 416: 先遍历硬币再遍历和
class Solution {
public int coinChange(int[] coins, int amount) {
int[] dp = new int[amount + 1];
Arrays.fill(dp, Integer.MAX_VALUE);
dp[0] = 0;

for(int i = 1; i <= amount; i++){
for (int coin : coins){
if (i - coin >= 0 && dp[i - coin] != Integer.MAX_VALUE){
dp[i] = Math.min(dp[i], dp[i - coin] + 1);
}
}
}

return dp[amount] == Integer.MAX_VALUE ? -1 : dp[amount];
}
}

class Solution:
def coinChange(self, coins: List[int], amount: int) -> int:
dp = [float("inf")] * (amount + 1)
dp[0] = 0

for i in range(1, amount + 1):
for coin in coins:
if i - coin >= 0 and dp[i - coin] != float("inf"):
dp[i] = min(dp[i - coin] + 1, dp[i])

return -1 if dp[amount] == float("inf") else dp[amount]