[LeetCode 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.

 

BackPack VI and Combination Sum IV:  Find all possible combinations that sum to a target value 

Coin Change: Find the combination that sums to a target value and uses the fewest number of elements

 

State: dp[i]: the fewest number of coins needed that sum to i

Function: dp[i] = Math.min(dp[i], dp[i - coins[j]] + 1), if i >= coins[j] && dp[i - coins[j]] < Integer.MAX_VALUE

i >= coins[j]: only consider picking a coin if its value is not greater than the target value i;

dp[i - coins[j]] < Integer.MAX_VALUE: if we did pick coins[j], then we must be able to find a combination that sums 

to i - coins[j];

Initialization: dp[0] = 0, dp[i] = Integer.MAX_VALUE, for i >= 1

Answer: dp[amount] or -1

 

public class Solution {
    public int coinChange(int[] coins, int amount) {
        if(amount <= 0){
            return 0;
        }
        if(coins == null || coins.length == 0){
            return -1;
        }
        int[] dp = new int[amount + 1];
        dp[0] = 0;
        for(int i = 1; i <= amount; i++){
            dp[i] = Integer.MAX_VALUE;
        }
        for(int i = 1; i <= amount; i++){
            for(int j = 0; j < coins.length; j++){
                if(i >= coins[j] && dp[i - coins[j]] < Integer.MAX_VALUE){
                    dp[i] = Math.min(dp[i], dp[i - coins[j]] + 1);
                }
            }
        }
        if(dp[amount] < Integer.MAX_VALUE){
            return dp[amount];
        }
        return -1;
    }
}

 

 

Related Problems

BackPack VI

Combination Sum IV