Coin Change

Given an integer array coins representing coins of different denominations and 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 coins, return −1.

Example:

Input: coins = [1, 2, 5], amount = 11;
Output: 3;
Input: coins = [2], amount = 3;
Output: -1;

Constraints:

• 1 ≤ l_coins ≤ 12
• 1 ≤ coins_i ≤ 2³¹ − 1
• 0 ≤ amount ≤ 10⁴

Solution

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

Explanation:

dp_i represents the minimum number of coins that makes up the amount i. Assume that the values from dp₀ to dp_i − 1 are already known, thus the transition equation of dp_i can be deduced as dp_i = min_j = 0^{l_coins − 1}dp_{i − coinsj} + 1 which means the value of dp_i is determined by dp_{i − coinsj}. If dp_i can transfer from dp_{i − coinsj} via coins_j, the number of coins that used should add 1.