Problem

You are given coins of different denominations and a total amount of money. Write a function to compute the number of combinations that make up that amount. You may assume that you have infinite number of each kind of coin.

Note: You can assume that

0 <= amount <= 5000
1 <= coin <= 5000
the number of coins is less than 500
the answer is guaranteed to fit into signed 32-bit integer

Example 1:

Input: amount = 5, coins = [1, 2, 5]
Output: 4
Explanation: there are four ways to make up the amount:
5=5
5=2+2+1
5=2+1+1+1
5=1+1+1+1+1
 

Example 2:

Input: amount = 3, coins = [2]
Output: 0
Explanation: the amount of 3 cannot be made up just with coins of 2.
 

Example 3:

Input: amount = 10, coins = [10] 
Output: 1

Solution

DP

class Solution {
    public int change(int amount, int[] coins) {
        int[] dp = new int[amount+1];
        dp[0] = 1;
        for (int coin: coins) {
            for (int i = 1; i <= amount; i++) {
                if (i - coin >= 0) dp[i] += dp[i-coin];
            }
        }
        return dp[amount];
    }
}

DFS - TLE

class Solution {
    int count = 0;
    public int change(int amount, int[] coins) {
        if (amount == 0) return 1;
        dfs(coins, 0, 0, amount);
        return count;
    }
    private void dfs(int[] coins, int start, int sum, int amount) {
        if (start == coins.length) return;
        if (sum == amount) {
            count++;
            return;
        }
        for (int i = start; i < coins.length; i++) {
            if (coins[i] + sum > amount) continue;
            else dfs(coins, i, sum+coins[i], amount);
        }
    }
}

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