Problem Statement:

Given an array of distinct integers nums and a target integer target, return the number of possible combinations that add up to target.

The test cases are generated so that the answer can fit in a 32-bit integer.

Example 1:

Input: nums = [1,2,3], target = 4
Output: 7
Explanation:
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.

Example 2:

Input: nums = [9], target = 3
Output: 0

Constraints:

1 <= nums.length <= 200
1 <= nums[i] <= 1000
All the elements of nums are unique.
1 <= target <= 1000

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:

The approach is to run DFS starting with zeroth index. At each step, we reduce target and re-start DFS from the starting index. This is because, here not just repetition is allowed, but also, the same set of elements with different permutation is considered different.

Here is the recursive solution.

 
class Solution {
public:
    int combinationSum4(vector<int>& nums, int target) 
    {
        if (target==0) return 1;
        if (target<0) return 0;
        int cur=0;
        for (int i=0; i<nums.size(); i++)
            cur += combinationSum4(nums, target-nums[i]);
        return cur;
    }
};

Adding memoization:

 
class Solution {
    unordered_map<int,int> memo;
public:
    int combinationSum4(vector<int>& nums, int target) 
    {
        if (target==0) return 1;
        if (target<0) return 0;
        int cur=0;
        for (int i=0; i<nums.size(); i++)
            if (memo.count(target-nums[i])) cur += memo[target-nums[i]];
            else
                cur += combinationSum4(nums, target-nums[i]);
        return memo[target] = cur;
    }
};