# 377. Combination Sum IV (Medium)

https://leetcode.com/problems/combination-sum-iv/

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.
```

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?

Credits:
Special thanks to @pbrother for adding this problem and creating all test cases.

## Solutions

``````class Solution {

// We must use below variable to record the number of combinations for a specific target.
// This approach really saves a great deal of time.

Map<Integer, Integer> combs = new HashMap<>();

public int combinationSum4(int[] nums, int target) {
if (nums == null || nums.length == 0) {
return 0;
}

// Here add LinkedList to track the elements that sums to target.
// Nevertheless, it's not necessary since this problem only requires the number of combinations.
}

private int recurse(int[] nums, int target, LinkedList<Integer> track) {
if (target == 0) {
//                System.out.println(track);
return 1;
}

// No need to keep the track of elements that sums up to target

int ans = 0;
for (int i = 0; i < nums.length; i++) {
if (target < nums[i]) {
continue;
}

int nextRT = target - nums[i];
if (combs.containsKey(nextRT)) {
ans += combs.get(nextRT);
} else {
int count = recurse(nums, nextRT, track);
track.removeLast();

ans += count;
combs.put(nextRT, count);
}
}

return ans;
}
}
``````