-
Notifications
You must be signed in to change notification settings - Fork 2
/
PermutationOfElementsWithDuplicates.java
75 lines (64 loc) · 2.32 KB
/
PermutationOfElementsWithDuplicates.java
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
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
package Leetcode;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
/**
* @author kalpak
*
* 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
*/
public class PermutationOfElementsWithDuplicates {
public static List<List<Integer>> permuteUnique(int[] nums) {
List<List<Integer>> result = new ArrayList<>();
List<Integer> temp = new ArrayList<>();
boolean[] isUsed = new boolean[nums.length];
Arrays.sort(nums);
backtrack(result, temp, isUsed, nums);
return result;
}
private static void backtrack(List<List<Integer>> result, List<Integer> temp, boolean[] isUsed, int[] nums) {
// Base Condition
if(temp.size() == nums.length) {
result.add(new ArrayList<>(temp));
return;
}
for(int i = 0; i < nums.length; i++) {
// Check if the number is currently in use. We use the index as opposed to the number itself.
// Also check, the number is a duplicate or not and whether it has been used to the present recursive stack or not.
// isUsed[i - 1] == false mean that the in current recursion tree, nums[i-1] has not been used but is a duplicate. We then need to avoid the number because a same entry has already been done in the previous call.
// Draw the recursion tree to understand the same
if(isUsed[i] || (i > 0 && nums[i] == nums[i-1] && isUsed[i-1] == false))
continue;
// Mark the current number as visited and add it to the temporary list
isUsed[i] = true;
temp.add(nums[i]);
// Recurse down the tree
backtrack(result, temp, isUsed, nums);
// Unmark the number and go up the recursive stack
isUsed[i] = false;
temp.remove(temp.size() - 1);
}
}
public static void main(String[] args) {
int[] nums = new int[]{1, 2, 1};
System.out.println(permuteUnique(nums));
}
}