LeetCode 솔루션

[5/21] 322. Coin Change

mingki 쪽지보내기 메일보내기 자기소개 아이디로 검색 전체게시물 작성
작성일 2022.05.20 17:36

8,292 조회
2 댓글
0 추천
0 비추천
목록 게시글 쓰기

[LeetCode 시즌 3] 2022년 5월 21일 문제입니다.

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

322. Coin Change

Medium

11188270Add to ListShare

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

#LeetCode, #Amazon, #Microsoft, #Google, #Bloomberg, #Mathworks, #Facebook, #Walmart Global Tech, #Yahoo, #Goldman Sachs, #Intuit, #Array, #Dynamic Programming(DP), #Breadth-First Search(BFS)

링크

https://leetcode.com/problems/coin-change/ 4797 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];
    }
};