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[5/21] 322. Coin Change

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작성일 2022.05.20 17:36

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[LeetCode 시즌 3] 2022년 5월 21일 문제입니다.

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

322. Coin Change

Medium

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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] <= 231 - 1
0 <= amount <= 104

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링크

https://leetcode.com/problems/coin-change/ 4439 회 연결

익명
작성일 2022.05.21 06:47

dynamic programming - bottom up
- compute all minimum counts up to 'amount'
- TC: O(amount * N of coins)
- SC: O(amount) for memoization table

class Solution:
    def coinChange(self, coins: List[int], amount: int) -> int:
        dp = [float("inf")] * (amount + 1)
        dp[0] = 0
        
        for a in range(1, amount + 1):
            for c in coins:
                if c <= a:
                    dp[a] = min(dp[a], dp[a - c] + 1)
                    
        return dp[-1] if dp[-1] != float("inf") else -1

익명
작성일 2022.05.24 17:17

C++
Runtime: 143 ms, faster than 39.70% of C++ online submissions for Coin Change.
Memory Usage: 14.3 MB, less than 73.09% of C++ online submissions for Coin Change.

class Solution {
public:
    int coinChange(vector<int>& coins, int amount) {
        vector<int> dp(amount + 1, 1e9);
        int n = coins.size();
        
        dp[0] = 0;
        for (int i = 1; i <= amount; ++i) {
            for (int j = 0; j < n; ++j) {
                if (coins[j] <= i) {
                    dp[i] = min(dp[i], dp[i - coins[j]] + 1);
                }
            }
        }
        return dp[amount] > amount ? -1 : dp[amount];
    }
};