Find a permutation of 2N numbers such that the result of given expression is exactly 2K

Given two integers N and K, the task is to find a permutation of first 2*N natural numbers such that the following equation is satisfied. 
\sum\limits_{i=1}^N |A_{2i-1}-A_{2i}| - |\sum\limits_{i=1}^N A_{2i-1}-A_{2i}|=2K   
Note: The value of K will always be less than or equal to N.
Examples: 
 

Input : N = 1,  K = 0
Output : 1 2
The result of the above expression will be:
|1-2|-|1-2| =0

Input : N = 2,  K = 1 
Output : 2 1 3 4
The result of the above expression will be:
(|2-1|+|3-4|)-(|2-1+3-4|) = 2

Approach:
Consider the sorted permutation: 
 

1, 2, 3, 4, 5, 6.... 

The result of the expression will come out to be exactly 0. If we swap any 2 indices 2i-1 and 2i, the result will increase by exactly 2. So we need to make K such swaps. 
Below is the implementation of the above approach: 
 

// C++ program to find the required permutation
// of first 2*N natural numbers

#include <bits/stdc++.h>
using namespace std;

// Function to find the required permutation
// of first 2*N natural numbers
void printPermutation(int n, int k)
{
    // Iterate in blocks of 2
    for (int i = 1; i <= n; i++) {
        int x = 2 * i - 1;
        int y = 2 * i;

        // We need more increments, so print in reverse order
        if (i <= k)
            cout << y << " " << x << " ";

        // We have enough increments, so print in same order
        else
            cout << x << " " << y << " ";
    }
}

// Driver Code
int main()
{
    int n = 2, k = 1;

    printPermutation(n, k);

    return 0;
}

// Java program to find the 
// required permutation
// of first 2*N natural numbers
class GFG 
{ 
        
    // Function to find the required permutation
    // of first 2*N natural numbers
    static void printPermutation(int n, int k) 
    { 
        // Iterate in blocks of 2
        for (int i = 1; i <= n; i++) 
        {
            int x = 2 * i - 1;
            int y = 2 * i;
        
        // We need more increments,
        // so print in reverse order
        if (i <= k)
            System.out.print(y + " " + x + " ");

        // We have enough increments, 
        // so print in same order
        else
            System.out.print(x + " " + y + " ");
        }
    } 
        
    // Driver code 
    public static void main(String []args) 
    { 
        int n = 2, k = 1;
        printPermutation(n, k);
    } 
}

// This code is contributed by Ita_c.
 

# Python3 program to find the required 
# permutation of first 2*N natural numbers 

# Function to find the required permutation 
# of first 2*N natural numbers 
def printPermutation(n, k) :
    
    # Iterate in blocks of 2 
    for i in range(1, n + 1) :
        x = 2 * i - 1; 
        y = 2 * i; 

        # We need more increments, 
        # so print in reverse order 
        if (i <= k) :
            print(y, x, end = " "); 

        # We have enough increments, 
        # so print in same order 
        else :
            print(x, y, end = " "); 

# Driver Code 
if __name__ == "__main__" : 
    n = 2; k = 1; 

    printPermutation(n, k); 
    
# This code is contributed by Ryuga

using System; 

// C# program to find the 
// required permutation
// of first 2*N natural numbers
    
class GFG 
{ 
        
    // Function to find the required permutation
    // of first 2*N natural numbers
    static void printPermutation(int n, int k) 
    { 
        // Iterate in blocks of 2
        for (int i = 1; i <= n; i++) 
        {
            int x = 2 * i - 1;
            int y = 2 * i;
        
        // We need more increments,
        // so print in reverse order
        if (i <= k)
            Console.Write(y + " " + x + " ");

        // We have enough increments, 
        // so print in same order
        else
            Console.Write(x + " " + y + " ");
        }
    } 
        
    // Driver code 
    public static void Main() 
    { 
        int n = 2, k = 1;
        printPermutation(n, k);
    } 
}

// This code is contributed by
// shashank_sharma

<?php
// PHP program to find the required 
// permutation of first 2*N natural numbers

// Function to find the required permutation
// of first 2*N natural numbers
function printPermutation($n, $k)
{
    // Iterate in blocks of 2
    for ($i = 1; $i <= $n; $i++)
    {
        $x = 2 * $i - 1;
        $y = 2 * $i;

        // We need more increments, so print
        // in reverse order
        if ($i <= $k)
            echo $y . " " . $x . " ";

        // We have enough increments, 
        // so print in same order
        else
            echo $x . " " . $y . " ";
    }
}

// Driver Code
$n = 2;
$k = 1;

printPermutation($n, $k);

// This code is contributed by chandan_jnu
?>

<script>

// javascript program to find the
// required permutation
// of first 2*N natural numbers
    
 
        
    // Function to find the required permutation
    // of first 2*N natural numbers
    function printPermutation( n,  k) 
    { 
        // Iterate in blocks of 2
        for (var i = 1; i <= n; i++) 
        {
            var x = 2 * i - 1;
            var y = 2 * i;
        
        // We need more increments,
        // so print in reverse order
        if (i <= k)
            document.write(y + " " + x + " ");

        // We have enough increments,
        // so print in same order
        else
            document.write(x + " " + y + " ");
        }
    } 
        
    // Driver code
 
        var n = 2, k = 1;
        printPermutation(n, k);
        
        // This code is contributed by bunnyram19.
        
</script>

2 1 3 4

Time Complexity: O(N), since there runs a loop from 1 to n.
Auxiliary Space: O(1), since no extra space has been taken.