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# Combination Sum IV

## Description

Given an integer array with all positive numbers and no duplicates, find the number of possible combinations that add up to a positive integer target.

**Example:**

```
nums = [1, 2, 3]
target = 4

The possible combination ways are:
(1, 1, 1, 1)
(1, 1, 2)
(1, 2, 1)
(1, 3)
(2, 1, 1)
(2, 2)
(3, 1)

Note that different sequences are counted as different combinations.

Therefore the output is 7.
```

**Follow up:**\
What if negative numbers are allowed in the given array?\
How does it change the problem?\
What limitation we need to add to the question to allow negative numbers?

## Solution

Let `dp[i]` be the number of combinations that add up to `i`. Then there are `n` exclusive and exhaustive cases: the combinations with the first number being `nums[j]`, for `0 <= j < n`. We can sort `nums` to prune `nums[j]` that are larger that `i`.

```cpp
class Solution {
public:
    int combinationSum4(vector<int>& nums, int target) {
        vector<int> dp(target + 1);
        sort(nums.begin(), nums.end());
        dp[0] = 1;
        for(int i = 1; i <= target; ++i){
            for(int num : nums){
                if(num > i)
                    break;
                dp[i] += dp[i - num];
            }
        }
        return dp[target];
    }
};
```

### Follow up

Limitation: there are no combinations that add up to 0. Otherwise there are either 0 or infinite number of combinations.