47. Permutations II

Medium LeetCode

Given a collection of numbers, nums, that might contain duplicates, return all possible unique permutations in any order.

Example 1

Input: nums = [1,1,2]

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

Example 2

Input: nums = [1,2,3]

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

Constraints

• 1 <= nums.length <= 8
• -10 <= nums[i] <= 10

How to solve the problem

• Backtracking (DFS)

class Solution:
    def permuteUnique(self, nums: List[int]) -> List[List[int]]:
        nums.sort()  # Sort the array to facilitate duplicate checking
        result = []
        self.backtracking(nums, [], [False] * len(nums), result)
        return result

    def backtracking(self, nums, path, used, result):
        # If the current path includes all numbers, add a copy to result
        if len(path) == len(nums):
            result.append(path[:])
            return
        for i in range(len(nums)):
            # Skip used elements, or skip duplicates unless the previous duplicate has been used in this branch
            if (i > 0 and nums[i] == nums[i - 1] and not used[i - 1]) or used[i]:
                continue
            used[i] = True  # Mark the element as used
            path.append(nums[i])  # Add element to current path
            self.backtracking(nums, path, used, result)  # Recurse
            path.pop()  # Backtrack: remove the last element
            used[i] = False  # Mark the element as unused for future paths

Complexity

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