322. Coin Change

You are given an integer array coins representing coins of different denominations and an integer amount representing a total amount of money.

Return 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

Constraints:

• 1 <= coins.length <= 12
• 1 <= coins[i] <= 2^31 - 1
• 0 <= amount <= 10^4

动态规划解法

动态规划递推子函数dp[i] = min(dp[i], dp[i - coin] + 1)， 这里dp从1到amount，从小到大逐个递推

class Solution:
    def coinChange(self, coins: List[int], amount: int) -> int:
        dp = [math.inf] * (amount + 1)
        dp[0] = 0
        for i in range(1, amount+1):
            for coin in coins:
                if i >= coin:
                    dp[i] = min(dp[i], dp[i - coin] + 1)
        return dp[-1] if dp[-1] != math.inf else -1