# LeetCode 216. Combination Sum III

## Description

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

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.
```

Example 4:

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

Example 5:

```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`

## Python Solution

Depth-first search.

``````class Solution:
def combinationSum3(self, k: int, n: int) -> List[List[int]]:

nums = [i for i in range(1, 10)]

nums.sort()

combinations = []
self.dfs(nums, 0, k, n, [], combinations)

return combinations

def dfs(self, nums, index, k, target, combination, combinations):
if k == 0 and target == 0:
combinations.append(list(combination))
return

if k == 0 or target <= 0:
return

for i in range(index, len(nums)):
combination.append(nums[i])
self.dfs(nums, i + 1, k - 1, target - nums[i], combination, combinations)
combination.pop()``````
• Time Complexity: O(2^n *n).
• Space Complexity: O(N).

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