216. Combination Sum III

1. Description

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.

2. Example

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.

3. Constraints

• 2 <= k <= 9
• 1 <= n <= 60

4. Solutions

Backtracking

Time complexity: I don’t know 🥺
Space complexity:

class Solution {
public:
    vector<vector<int>> combinationSum3(int k, int n) {
        vector<int> candidates;
        vector<vector<int>> combinations;

        combine_sum(k, 1, n, combinations, candidates);

        return combinations;
    }

private:
    void combine_sum(
        const int count,
        int init_value,
        int target,
        vector<vector<int>> &combinations,
        vector<int> &candidates) {
        if (target == 0 && candidates.size() == count) {
            combinations.push_back(candidates);
        } else if (target > 0) {
            int need = count - candidates.size(); // Prune as early as possible

            for (int i = init_value; i <= 10 - need; ++i) {
                candidates.push_back(i);
                combine_sum(count, i + 1, target - i, combinations, candidates);
                candidates.pop_back();
            }
        }
    }
};