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+---
+id: coin-change-2
+title: Coin Change 2
+sidebar_label: 0518-Coin-Change-2
+tags:
+- Dynamic Programming
+- Array
+description: "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."
+---
+
+## Problem
+
+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`.
+
+### Examples
+
+**Example 1:**
+
+**Input:** `amount = 5, coins = [1, 2, 5]`  
+**Output:** `4`  
+**Explanation:** There are four ways to make up the amount:
+1. 5 = 5
+2. 5 = 2 + 2 + 1
+3. 5 = 2 + 1 + 1 + 1
+4. 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`
+
+### Constraints
+
+- `1 <= coins.length <= 300`
+- `1 <= coins[i] <= 5000`
+- All the values of `coins` are unique.
+- `0 <= amount <= 5000`
+
+---
+
+## Approach
+
+To solve this problem, we can use dynamic programming. We will create a 1D array `dp` where `dp[i]` represents the number of ways to make the amount `i` using the given coins.
+
+### Steps:
+
+1. Initialize a 1D array `dp` of size `amount + 1` with all elements set to `0`.
+2. Set `dp[0] = 1` because there is one way to make the amount `0`, which is to use no coins.
+3. For each coin in the `coins` array:
+   - Update the `dp` array from the current coin's value up to the `amount`.
+   - For each amount `i` from the coin's value to `amount`, add the number of ways to make the amount `i - coin` to `dp[i]`.
+4. The value `dp[amount]` will be the result.
+
+### Solution
+
+#### Java
+
+```java
+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 = coin; i <= amount; i++) {
+                dp[i] += dp[i - coin];
+            }
+        }
+
+        return dp[amount];
+    }
+}
+```
+#### C++ 
+```cpp
+#include <vector>
+using namespace std;
+
+class Solution {
+public:
+    int change(int amount, vector<int>& coins) {
+        vector<int> dp(amount + 1, 0);
+        dp[0] = 1;
+
+        for (int coin : coins) {
+            for (int i = coin; i <= amount; i++) {
+                dp[i] += dp[i - coin];
+            }
+        }
+
+        return dp[amount];
+    }
+};
+```
+#### Python
+
+```python
+class Solution:
+    def change(self, amount: int, coins: List[int]) -> int:
+        dp = [0] * (amount + 1)
+        dp[0] = 1
+
+        for coin in coins:
+            for i in range(coin, amount + 1):
+                dp[i] += dp[i - coin]
+
+        return dp[amount]
+```
+### Complexity Analysis
+**Time Complexity:** O(n * amount)
+>Reason: The nested loops iterate over the coins array and the amount, leading to a time complexity of O(n * amount), where n is the number of coins.
+
+**Space Complexity:** O(amount)
+>Reason: The space complexity is O(amount) due to the 1D dp array used to store the number of combinations for each amount.
+
+### References
+**LeetCode Problem:** Coin Change 2
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