## Algorithm

Problem Name: 518. Coin Change II

You are given an integer array `coins` representing coins of different denominations and an integer `amount` representing a total amount of money.

Return the number of combinations that make up that amount. If that amount of money cannot be made up by any combination of the coins, return `0`.

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

The answer is guaranteed to fit into a 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 = 
Output: 0
Explanation: the amount of 3 cannot be made up just with coins of 2.
```

Example 3:

```Input: amount = 10, coins = 
Output: 1
```

Constraints:

• `1 <= coins.length <= 300`
• `1 <= coins[i] <= 5000`
• All the values of `coins` are unique.
• `0 <= amount <= 5000`

## Code Examples

### #1 Code Example with Java Programming

```Code - Java Programming```

``````
class Solution {
public static int change(int amount, int[] coins) {
int[] combinations = new int[amount+1];
combinations = 1;

for (int coin : coins) {
for (int i=coin; i``````
``` Copy The Code & Input x – + cmd amount = 5, coins = [1,2,5] Output x – + cmd 4 ```
``` #2 Code Example with Javascript Programming Code - Javascript Programming function change(amount, coins) { const dp = Array.from(new Array(coins.length + 1), () => new Array(amount + 1).fill(0) ) dp = 1 for (let i = 1; i <= coins.length; i++) { dp[i] = 1 for (let j = 1; j <= amount; j++) { dp[i][j] = dp[i - 1][j] + (j >= coins[i - 1] ? dp[i][j - coins[i - 1]] : 0) } } return dp[coins.length][amount] } Copy The Code & Input x – + cmd amount = 5, coins = [1,2,5] Output x – + cmd 4 #3 Code Example with C# Programming Code - C# Programming namespace LeetCode { public class _0518_CoinChange2 { public int Change(int amount, int[] coins) { var dp = new int[amount + 1]; dp = 1; foreach (var coin in coins) for (int x = coin; x <= amount; x++) dp[x] += dp[x - coin]; return dp[amount]; } } } Copy The Code & Input x – + cmd amount = 3, coins =  Demonstration ```
``` #517 Leetcode Super Washing Machines Solution in C, C++, Java, JavaScript, Python, C# Leetcode #519 Leetcode Random Flip Matrix Solution in C, C++, Java, JavaScript, Python, C# Leetcode ```
``` ```