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# Coin Change 2

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

**Thoughts:**

1. `dp[i][j]`: the number of combinations to make up amount `j`by using the first `i`types of coins`State transition`:
   1. not using the `i`th coin, only using the first `i-1`coins to make up amount `j`, then we have `dp[i-1][j]`

      ways.
   2. using the `i`th coin, since we can use unlimited same coin, we need to know how many way to make up amount

      `j - coins[i]`by using first `i`coins( including `i`th), which is `dp[i][j-coins[i]]`
   3. `Initialization`:`dp[i][0] = 1`
2. `dp[i][j]`only rely on `dp[i-1][j]`and `dp[i][j-coins[i]]`, then we can optimize the space by only using one-dimension array.

**Code: 2D DP: O(K \* N), S: O(K \* N)**

```java
class Solution {
    public int change(int amount, int[] coins) {
        int[][] dp = new int[coins.length+1][amount+1];
        dp[0][0] = 1;

        for (int i = 1; i <= coins.length; i++) {
            dp[i][0] = 1;
            for (int 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];
    }
}
```

**Code: 1D DP: O(K \* N), S: O(N)**

```java
class Solution {
    public int change(int amount, int[] coins) {
        if(amount < 0) return 0;
        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];
    }
}
```