C# || How To Generate Next Permutation From Array Using C#

October 27, 2021 admin 1 Comment

The following is a module with functions which demonstrates how to generate the next permutation from an array using C#.

1. Next Permutation – Problem Statement

Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers.

If such an arrangement is not possible, it must rearrange it as the lowest possible order (i.e., sorted in ascending order).

The replacement must be in place and use only constant extra memory.

Example 1:

Input: nums = [1,2,3] Output: [1,3,2]

Example 2:

Input: nums = [3,2,1] Output: [1,2,3]

Example 3:

Input: nums = [1,1,5] Output: [1,5,1]

Example 4:

Input: nums = [1] Output: [1]

2. Next Permutation – Solution

The following is a solution which demonstrates how to generate the next permutation from an array





		
		
			2. Next Permutation - Solution
			
C#
			
			// ============================================================================
//    Author: Kenneth Perkins
//    Date:   Oct 27, 2021
//    Taken From: http://programmingnotes.org/
//    File:  Solution.cs
//    Description: Demonstrates how to generate the next permutation
// ============================================================================
public class Solution {
    public void NextPermutation(int[] nums) {
        // Starting from the end, look for the first index that is not in descending order
        var startIndex = nums.Length - 2;

        while (startIndex >= 0) {
            // Find first decreasing element
            if (nums[startIndex] < nums[startIndex + 1]) {
                break;
            }
            --startIndex;
        }

        if (startIndex >= 0) {
            // Starting from the end, look for the first element greater than the start index
            int endIndex = nums.Length - 1;

            while (endIndex > startIndex) {
                // Find first greater element
                if (nums[endIndex] > nums[startIndex]) {
                    break;
                }
                --endIndex;
            }

            // Swap first and last elements
            Swap(ref nums[startIndex], ref nums[endIndex]);
        }

        // Reverse the array
        Reverse(nums, startIndex + 1);
    }

    private void Reverse(int[] nums, int startIndex) {
        for (int start = startIndex, end = nums.Length - 1; start < end; ++start, --end) {
            Swap(ref nums[start], ref nums[end]);
        }
    }

    private void Swap(ref int a, ref int b) {
        var tmp = a;
        a = b;
        b = tmp;
    }
}// http://programmingnotes.org/
			
				
					
				
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						// ============================================================================
//    Author: Kenneth Perkins
//    Date:   Oct 27, 2021
//    Taken From: http://programmingnotes.org/
//    File:  Solution.cs
//    Description: Demonstrates how to generate the next permutation
// ============================================================================
public class Solution {
    public void NextPermutation(int[] nums) {
        // Starting from the end, look for the first index that is not in descending order
        var startIndex = nums.Length - 2;
 
        while (startIndex >= 0) {
            // Find first decreasing element
            if (nums[startIndex] < nums[startIndex + 1]) {
                break;
            }
            --startIndex;
        }
 
        if (startIndex >= 0) {
            // Starting from the end, look for the first element greater than the start index
            int endIndex = nums.Length - 1;
 
            while (endIndex > startIndex) {
                // Find first greater element
                if (nums[endIndex] > nums[startIndex]) {
                    break;
                }
                --endIndex;
            }
 
            // Swap first and last elements
            Swap(ref nums[startIndex], ref nums[endIndex]);
        }
 
        // Reverse the array
        Reverse(nums, startIndex + 1);
    }
 
    private void Reverse(int[] nums, int startIndex) {
        for (int start = startIndex, end = nums.Length - 1; start < end; ++start, --end) {
            Swap(ref nums[start], ref nums[end]);
        }
    }
 
    private void Swap(ref int a, ref int b) {
        var tmp = a;
        a = b;
        b = tmp;
    }
}// 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:

[1,3,2] [1,2,3] [1,5,1] [1]

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Posted in: C#, Console Tags: c-sharp, leetcode, permutations

One Response to C# || How To Generate Next Permutation From Array Using C#

Sandeep Kumar says:

July 6, 2023 at 11:38 pm

very nice

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C# || How To Generate Next Permutation From Array Using C#

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