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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