46. Permutations

Medium LeetCode

Given an array nums of distinct integers, return all the possible permutations. You can return the answer in any order.

Example 1

Input: nums = [1,2,3]

Output: [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]

Example 2

Input: nums = [0,1]

Output: [[0,1],[1,0]]

Example 3

Input: nums = [1]

Output: [[1]]

Constraints

• 1 <= nums.length <= 6
• -10 <= nums[i] <= 10
• All the integers of nums are unique.

How to solve the problem

• Backtracking (DFS)

class Solution:
    def permute(self, nums: List[int]) -> List[List[int]]:
        n = len(nums)
        results = []
        current_perm = [0] * n # Stores the current permutation
        used = [False] * n # Tracks whether nums[i] is already used in the current permutation
        def dfs(depth: int) -> None:
            if depth == n:
                results.append(current_perm.copy()) # Add a complete permutation to results
                return
            for idx, is_used in enumerate(used):
                if not is_used:
                    current_perm[depth] = nums[idx] # Choose an unused number for this position
                    used[idx] = True # Mark as used
                    dfs(depth + 1) # Recurse to fill the next position
                    used[idx] = False # Backtrack: mark as unused for other branches
                    # No need to undo current_perm[depth], will be overwritten in the next iteration

        dfs(0)
        return results

Complexity

• Time complexity: O(K*C(n, k))
• Space complexity: O(K*C(n, k))