322. Coin Change

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.

Example 1:
coins = [1, 2, 5], amount = 11
return 3 (11 = 5 + 5 + 1)

Example 2:
coins = [2], amount = 3
return -1.

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

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

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Dynamic programming.

dp[i] stores the minimum required coins whose sum is i.

Do not use greedy algorithm to solve this problem. It will lead you to a local optimal rather than the correct answer.

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