2025-09-15 18:21

Status: adult

Tags: leetcode backtrack leetcode-medium Backtracking Intro Subsets Combination Sum

Combinations

Code

class Solution(object):
    def combine(self, n, k):
        """
        :type n: int
        :type k: int
        :rtype: List[List[int]]
        """
        
        res, sol = [], []
        nums = [x for x in range(1, n + 1)]
        
        def backtrack(i):
            # base case
            if len(sol) == k:
                res.append(sol[:])
                return
 
            if i == n:
                return
            
            # go left
            sol.append(nums[i])
            backtrack(i + 1)
            sol.pop()
 
            # go right
            backtrack(i + 1)
 
        backtrack(0)
        return res

Explanation

Base Case(s): - When the len(sol) k - When the index(i) n(length of array)

Choices: - All numbers from current index to n - Going left: this means that we deciding to add a number(add the num to sol, backtrack on i + 1 then sol.pop) - Going right: this means we are choosing not to add a number(increment i)

Constraints: the length of sol must be exactly k

Backtracking Step: Pop the last number added

References

Bitflip however my solution resembles something Greg Hogg would do.