This is a common question asked in interviews of Microsoft, Google & Facebook. You might have stumbled upon this question in LeetCode or InterviewBit.
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.
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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;
}
}
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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!).