# Combination Sum III Problem

## Description

LeetCode Problem 216.

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:

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

Example 2:

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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:

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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.
``````

Example 4:

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Input: k = 3, n = 2
Output: []
Explanation: There are no valid combinations.
``````

Example 5:

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Input: k = 9, n = 45
Output: [[1,2,3,4,5,6,7,8,9]]
Explanation:
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45
There are no other valid combinations.
``````

Constraints:

• 2 <= k <= 9
• 1 <= n <= 60

## Sample C++ Code

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class Solution {
public:
vector<vector<int>> ans;
int steps;

void dfs(vector<int>& candidates, vector<int>&comb,
int target, int cnt, int idx) {
if (cnt == steps && target == 0) {
ans.push_back(comb);
return;
}
if (cnt >= steps || target <= 0)
return;

for (int i = idx; i < candidates.size(); i ++) {
if (target >= candidates[i]) {
comb.push_back(candidates[i]);
dfs(candidates, comb, target-candidates[i], cnt+1, i+1);
comb.pop_back();
}
}
}

vector<vector<int>> combinationSum3(int k, int n) {
vector<int> candidates = {1,2,3,4,5,6,7,8,9};
steps = k;
vector<int> comb;
dfs(candidates, comb, n, 0, 0);
return ans;
}
};
``````