Problem Statement:

Find all valid combinations of k numbers that sum up to n such that the following conditions are true:

Only numbers 1 through 9 are used.
Each number is used at most once.

Return a list of all possible valid combinations. The list must not contain the same combination twice, and the combinations may be returned in any order.

Example 1:

Input: k = 3, n = 7
Output: [[1,2,4]]
Explanation:
1 + 2 + 4 = 7
There are no other valid combinations.

Example 2:

Input: k = 3, n = 9
Output: [[1,2,6],[1,3,5],[2,3,4]]
Explanation:
1 + 2 + 6 = 9
1 + 3 + 5 = 9
2 + 3 + 4 = 9
There are no other valid combinations.

Example 3:

Input: k = 4, n = 1
Output: []
Explanation: There are no valid combinations.
Using 4 different numbers in the range [1,9], the smallest sum we can get is 1+2+3+4 = 10 and since 10 > 1, there are no valid combination.

Constraints:

2 <= k <= 9
1 <= n <= 60

Solution:

We can do DFS and check for current sum. If meets the condition add to result. C++ solution:

class Solution {
public:
    void combn(int curr, int target, int count,  vector<vector<int>> &res, vector<int> &v)
    {
        if (count<0 || target<0) return;
        if (count==0 && target==0)
        {
            res.push_back(v);
            return;
        }
        for (int i=curr; i<=9; i++)
        {
            v.push_back(i);
            combn(i+1, target-i, count-1, res, v);
            v.pop_back();
        }
    }
    vector<vector<int>> combinationSum3(int k, int n) {
        vector<vector<int>> res;
        vector<int>v;
        combn(1,n,k,res,v);
        return res;
    }
};

We can also do all combinations and get an AC: Python solution:

class Solution:
    def combinationSum3(self, k: int, n: int) -> List[List[int]]:
        res = []
        for cb in combinations(range(1,10),k):
            if sum(cb)==n:
                res.append(list(cb))
        return res