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++

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// 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;
}

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Python3

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# 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

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Output:

2 1 3 4

Time Complexity: O(N)
Auxiliary Space: O(1)



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