C# || How To Generate Unique Permutations From Array With Duplicate Values Using C#
The following is a module with functions which demonstrates how to generate unique permutations from an array with duplicate values using C#.
1. Permute Unique – Problem Statement
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]]
2. Permute Unique – Solution
The following is a solution which demonstrates how to generate unique permutations from an array with duplicate values.
|
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 |
// ============================================================================ // Author: Kenneth Perkins // Date: Oct 27, 2021 // Taken From: http://programmingnotes.org/ // File: Solution.cs // Description: Demonstrates how to generate unique permutations // ============================================================================ public class Solution { private List<IList<int>> result = new List<IList<int>>(); public IList<IList<int>> PermuteUnique(int[] nums) { // Sort Array Array.Sort(nums); Generate(nums, new List<int>(), new bool[nums.Length]); return result; } private void Generate(int[] nums, List<int> combination, bool[] visited) { if (combination.Count == nums.Length) { result.Add(new List<int>(combination)); return; } for (int index = 0; index < nums.Length; ++index) { // Check to see if this number has been visited if (visited[index]) { continue; } else if (index > 0 && nums[index] == nums[index - 1] && !visited[index - 1]) { continue; } // Set that this index has been visited visited[index] = true; // Add this number to the combination combination.Add(nums[index]); // Keep generating permutations Generate(nums, combination, visited); // Unset that this index has been visited visited[index] = false; // Remove last item as its already been explored combination.RemoveAt(combination.Count - 1); } } }// http://programmingnotes.org/ |
QUICK NOTES:
The highlighted lines are sections of interest to look out for.
The code is heavily commented, so no further insight is necessary. If you have any questions, feel free to leave a comment below.
Once compiled, you should get this as your output for the example cases:
[[1,1,2],[1,2,1],[2,1,1]]
[[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]
Hi, I'm a software engineer. I write articles about programming and design in my spare time.
Leave a Reply