K difference permutation

Given two integers n and k. Consider first permutation of natural n numbers, P = “1 2 3 … n”, print a permutation “Result” such that abs(Resulti – Pi) = k where Pi denotes the position of i in permutation P. The value of Pi varies from 1 to n. If there are multiple possible results, then print the lexicographically smallest one.

Input: n = 6 k = 3
Output: 4 5 6 1 2 3
Explanation:
     P = 1 2 3 4 5 6
Result = 4 5 6 1 2 3
We can notice that the difference between
individual numbers (at same positions) of 
P and result is 3 and "4 5 6 1 2 3" is 
lexicographically smallest such permutation.
Other greater permutations could be 

Input  : n = 6 k = 2
Output : Not possible
Explanation: No permutation is possible 
with difference is k 

Naive approach is to generate all the permutation from 1 to n and pick the smallest one which satisfy the condition of absolute difference k. Time complexity of this approach is Ω(n!) which will definitely time out for large value of n.

The Efficient approach is to observe the pattern at each position of index. For each position of index i, there can only exist two candidate i.e., i + k and i – k. As we need to find lexicographically smallest permutation so we will first look for i – k candidate(if possible) and then for i + k candidate.



 Illustration:
 n = 8, k = 2
 P : 1 2 3 4 5 6 7 8

 For any ith position we will check which candidate
 is possible i.e., i + k or i - k 

 1st pos = 1 + 2 = 3 (1 - 2 not possible)
 2nd pos = 2 + 2 = 4 (2 - 2 not possible)
 3rd pos = 3 - 2 = 1 (possible)
 4th pos = 4 - 2 = 2 (possible)
 5th pos = 5 + 2 = 7 (5 - 2 already placed, not possible)
 6th pos = 6 + 2 = 8 (6 - 2 already placed, not possible)
 7th pos = 7 - 2 = 5 (possible)
 8th pos = 8 - 2 = 6 (possible)

Note: If we observe above illustration then we will find that i + k and i – k are alternating after kth consecutive interval. Another observation is that the whole permutation is only when n is even such that n can be divided into two parts where each part must be divisible by k.

C++

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// C++ program to find k absolute difference
// permutation
#include<bits/stdc++.h>
using namespace std;
  
void kDifferencePermutation(int n, int k)
{
    // If k is 0 then we just print the
    // permutation from 1 to n
    if (!k)
    {
        for (int i = 0; i < n; ++i)
            cout << i + 1 << " ";
    }
  
    // Check whether permutation is feasible or not
    else if (n % (2 * k) != 0)
        cout <<"Not Possible";
  
    else
    {
        for (int i = 0; i < n; ++i)
        {
            // Put i + k + 1 candidate if position is
            // feasible, otherwise put the i - k - 1
            // candidate
            if ((i / k) % 2 == 0)
                cout << i + k + 1 << " ";
            else
                cout << i - k + 1 << " ";
        }
    }
    cout << "\n";
}
  
// Driver code
int main()
{
    int n = 6 , k = 3;
    kDifferencePermutation(n, k);
  
    n = 6 , k = 2;
    kDifferencePermutation(n, k);
  
    n = 8 , k = 2;
    kDifferencePermutation(n, k);
  
    return 0;
}

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Java

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// Java program to find k absolute
// difference permutation
import java.io.*;
  
class GFG {
  
    static void kDifferencePermutation(int n,
                                    int k)
    {
        // If k is 0 then we just print the
        // permutation from 1 to n
        if (!(k > 0))
        {
            for (int i = 0; i < n; ++i)
                System.out.print( i + 1 + " ");
        }
      
        // Check whether permutation is
        // feasible or not
        else if (n % (2 * k) != 0)
            System.out.print("Not Possible");
      
        else
        {
            for (int i = 0; i < n; ++i)
            {
                // Put i + k + 1 candidate
                // if position is feasible,
                // otherwise put the
                // i - k - 1 candidate
                if ((i / k) % 2 == 0)
                    System.out.print( i + k 
                            + 1 + " ");
                else
                    System.out.print( i - k 
                            + 1 + " ");
            }
        }
        System.out.println() ;
    }
      
    // Driver code
    static public void main (String[] args)
    {
        int n = 6 , k = 3;
        kDifferencePermutation(n, k);
      
        n = 6 ;
        k = 2;
        kDifferencePermutation(n, k);
      
        n = 8 ;
        k = 2;
        kDifferencePermutation(n, k);
    }
}
  
// This code is contributed by anuj_67.

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Python3

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# Python 3 program to find k 
# absolute difference permutation
def kDifferencePermutation(n, k):
      
    # If k is 0 then we just print the
    # permutation from 1 to n
    if (k == 0):
        for i in range(n):
            print(i + 1, end = " ")
  
    # Check whether permutation
    # is feasible or not
    elif (n % (2 * k) != 0):
        print("Not Possible", end = "")
  
    else:
        for i in range(n):
              
            # Put i + k + 1 candidate if position is
            # feasible, otherwise put the i - k - 1
            # candidate
            if (int(i / k) % 2 == 0):
                print(i + k + 1, end = " ")
            else:
                print(i - k + 1, end = " ")
  
    print("\n", end = "")
  
# Driver code
if __name__ == '__main__':
    n = 6
    k = 3
    kDifferencePermutation(n, k)
  
    n = 6
    k = 2
    kDifferencePermutation(n, k)
  
    n = 8
    k = 2
    kDifferencePermutation(n, k)
      
# This code is contributed by
# Surendra_Gangwar

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C#

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// C# program to find k absolute
// difference permutation
using System;
  
class GFG {
  
    static void kDifferencePermutation(int n,
                                       int k)
    {
        // If k is 0 then we just print the
        // permutation from 1 to n
        if (!(k > 0))
        {
            for (int i = 0; i < n; ++i)
                Console.Write( i + 1 + " ");
        }
      
        // Check whether permutation is
        // feasible or not
        else if (n % (2 * k) != 0)
            Console.Write("Not Possible");
      
        else
        {
            for (int i = 0; i < n; ++i)
            {
                // Put i + k + 1 candidate
                // if position is feasible,
                // otherwise put the
                // i - k - 1 candidate
                if ((i / k) % 2 == 0)
                    Console.Write( i + k 
                              + 1 + " ");
                else
                    Console.Write( i - k 
                              + 1 + " ");
            }
        }
        Console.WriteLine() ;
    }
      
    // Driver code
    static public void Main ()
    {
        int n = 6 , k = 3;
        kDifferencePermutation(n, k);
      
        n = 6 ;
        k = 2;
        kDifferencePermutation(n, k);
      
        n = 8 ;
        k = 2;
        kDifferencePermutation(n, k);
    }
}
  
// This code is contributed by anuj_67.

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PHP

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<?php
// PHP program to find k absolute 
// difference permutation
  
function kDifferencePermutation( $n, $k)
{
      
    // If k is 0 then we just print the
    // permutation from 1 to n
    if (!$k)
    {
        for($i = 0; $i < $n; ++$i)
            echo $i + 1 ," ";
    }
  
    // Check whether permutation
    // is feasible or not
    else if ($n % (2 * $k) != 0)
        echo"Not Possible";
  
    else
    {
        for($i = 0; $i < $n; ++$i)
        {
              
            // Put i + k + 1 candidate
            // if position is feasible, 
            // otherwise put the i - k - 1
            // candidate
            if (($i / $k) % 2 == 0)
                echo $i + $k + 1 , " ";
            else
                echo $i - $k + 1 , " ";
        }
    }
    echo "\n";
}
  
    // Driver Code
    $n = 6 ; $k = 3;
    kDifferencePermutation($n, $k);
  
    $n = 6 ; $k = 2;
    kDifferencePermutation($n, $k);
  
    $n = 8 ;$k = 2;
    kDifferencePermutation($n, $k);
  
// This code is contributed by anuj_67.
?>

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Output:

4 5 6 1 2 3 
Not Possible
3 4 1 2 7 8 5 6 

Time complexity: O(n)
Auxiliary space: O(1)

This article is contributed by Shubham Bansal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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