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

**Input**: nums = [1,1,2]

**Output**:

[[1,1,2],

[1,2,1],

[2,1,1]]

Here is the working solution for this problem in Java. It uses backtracking to solve the problem.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 | import java.util.*; import java.lang.*; import java.io.*; class Solution { public List<List<Integer>> permuteUnique(int[] nums) { List<Integer> currentResult = new ArrayList<Integer>(); List<List<Integer>> results = new ArrayList<List<Integer>>(); int startPosition = 0; calculate(nums, startPosition, results); return results; } private void calculate(int[] nums, int startPosition, List<List<Integer>> results) { if(startPosition == nums.length - 1){ List<Integer> currentResult = new ArrayList<Integer>(); for(int num : nums){ currentResult.add(num); } results.add(currentResult); } else { Set<Integer> usedNumbers = new HashSet<Integer>(); for(int i = startPosition; i < nums.length; i++){ if(usedNumbers.contains(nums[i])){ continue; } else { usedNumbers.add(nums[i]); swap(nums, startPosition, i); calculate(nums, startPosition + 1, results); //backtracking swap(nums, startPosition, i); } } } } private void swap(int[] nums, int i, int j){ int temp = nums[i]; nums[i] = nums[j]; nums[j] = temp; } } |

Time complexity in O(n!).

T(n) = n * T(n-1)

T(n-1) = (n-1) * T(n-2)

T(n-2) = (n-2) * T(n-3)

As you can see time complexity becomes n * (n-2) * (n - 3) ... * 1 which is O(n!).

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