Coin Change

Description

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

Code

Denote $M$ as the amount, $N$ as the size of coins vector. Time Complexity: $O(MN)$, Space Complexity: $O(M)$

邏輯和 Perfect Squares 類似。是 Knapsack 類型的題目。

class Solution {
public:
    int coinChange(vector<int>& coins, int amount) {
        vector<int> dp(amount + 1, 1e5);
        dp[0] = 0;
        for(int i = 1; i <= amount; i++) {
            for(auto coin: coins) {
                if(i >= coin) {
                    dp[i] = min(dp[i], dp[i - coin] + 1);
                }
            }
        }
        return dp[amount] == 1e5 ? -1 : dp[amount];
    } 
};

Source

• Coin Change - LeetCode