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Find a permutation such that number of indices for which gcd(p[i], i) > 1 is exactly K

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  • Last Updated : 22 Jun, 2022
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Given two integers N and K, the task is to find a permutation of integers from the range [1, N] such that the number of indices (1-based indexing) where gcd(p[i], i) > 1 is exactly K. Print -1 if such permutation is not possible.

Examples: 

Input: N = 4, K = 3 
Output: 1 2 3 4 
gcd(1, 1) = 1 
gcd(2, 2) = 2 
gcd(3, 3) = 3 
gcd(4, 4) = 4 
Therefore, there are exactly 3 indices where gcd(p[i], i) > 1

Input: N = 1, K = 1 
Output: -1

Approach: A couple of observations can be made here:  

  1. gcd(i, i + 1) = 1
  2. gcd(1, i) = 1
  3. gcd(i, i) = i

Since gcd of 1 with any other number is always going to be one, K can almost be N – 1. Consider the permutation where p[i] = i, The number of indices where gcd(p[i], i) > 1 will be N – 1. Notice that swapping 2 continuous elements (excluding 1) will reduce the count of such indices by exactly 2. And swapping with 1 will reduce the count by exactly 1.
Below is the implementation of the above approach: 
 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to print the required permutation
void printPermutation(int n, int k)
{
 
    // Not possible
    if (k >= n || (n % 2 == 0 && k == 0)) {
        cout << -1;
        return;
    }
    int per[n + 1], i;
 
    // Initial permutation
    for (i = 1; i <= n; i++)
        per[i] = i;
 
    // Indices for which gcd(p[i], i) > 1
    // in the initial permutation
    int cnt = n - 1;
    i = 2;
    while (i < n) {
 
        // Reduce the count by 2
        if (cnt - 1 > k) {
            swap(per[i], per[i + 1]);
            cnt -= 2;
        }
 
        // Reduce the count by 1
        else if (cnt - 1 == k) {
            swap(per[1], per[i]);
            cnt--;
        }
        else
            break;
        i += 2;
    }
 
    // Print the permutation
    for (i = 1; i <= n; i++)
        cout << per[i] << " ";
}
 
// Driver code
int main()
{
    int n = 4, k = 3;
    printPermutation(n, k);
 
    return 0;
}

Java




// Java implementation of the approach
class GFG
{
 
// Function to print the required permutation
    static void printPermutation(int n, int k)
    {
 
        // Not possible
        if (k >= n || (n % 2 == 0 && k == 0))
        {
            System.out.print(-1);
            return;
        }
        int per[] = new int[n + 1], i;
 
        // Initial permutation
        for (i = 1; i <= n; i++)
        {
            per[i] = i;
        }
 
        // Indices for which gcd(p[i], i) > 1
        // in the initial permutation
        int cnt = n - 1;
        i = 2;
        while (i < n)
        {
 
            // Reduce the count by 2
            if (cnt - 1 > k)
            {
                swap(per, i, i + 1);
                cnt -= 2;
            }
             
            // Reduce the count by 1
            else if (cnt - 1 == k)
            {
                swap(per, 1, i);
                cnt--;
            }
            else
            {
                break;
            }
            i += 2;
        }
 
        // Print the permutation
        for (i = 1; i <= n; i++)
        {
            System.out.print(per[i] + " ");
        }
    }
 
    static int[] swap(int[] arr, int i, int j)
    {
        int temp = arr[i];
        arr[i] = arr[j];
        arr[j] = temp;
        return arr;
    }
     
    // Driver code
    public static void main(String[] args)
    {
        int n = 4, k = 3;
        printPermutation(n, k);
    }
}
 
/* This code contributed by PrinciRaj1992 */

Python3




# Python3 implementation of the approach
 
# Function to print the required permutation
def printPermutation(n, k):
 
    # Not possible
    if (k >= n or (n % 2 == 0 and
                   k == 0)):
        print(-1);
        return;
    per = [0] * (n + 1);
 
    # Initial permutation
    for i in range(1, n + 1):
        per[i] = i;
 
    # Indices for which gcd(p[i], i) > 1
    # in the initial permutation
    cnt = n - 1;
    i = 2;
    while (i < n):
 
        # Reduce the count by 2
        if (cnt - 1 > k):
             
            t = per[i];
            per[i] = per[i + 1];
            per[i + 1] = t;
             
            # swap(per[i], per[i + 1]);
            cnt -= 2;
 
        # Reduce the count by 1
        elif (cnt - 1 == k):
             
            # swap(per[1], per[i]);
            t = per[1];
            per[1] = per[i];
            per[i] = t;
            cnt -= 1;
        else:
            break;
        i += 2;
 
    # Print the permutation
    for i in range(1, n + 1):
        print(per[i], end = " ");
 
# Driver code
n = 4;
k = 3;
printPermutation(n, k);
 
# This code is contributed by mits

C#




// C# implementation of the approach
using System;
 
class GFG
{
 
    // Function to print the required permutation
    static void printPermutation(int n, int k)
    {
 
        // Not possible
        if (k >= n || (n % 2 == 0 && k == 0))
        {
            Console.Write(-1);
            return;
        }
        int []per = new int[n + 1] ;
        int i ;
 
        // Initial permutation
        for (i = 1; i <= n; i++)
        {
            per[i] = i;
        }
 
        // Indices for which gcd(p[i], i) > 1
        // in the initial permutation
        int cnt = n - 1;
        i = 2;
        while (i < n)
        {
 
            // Reduce the count by 2
            if (cnt - 1 > k)
            {
                swap(per, i, i + 1);
                cnt -= 2;
            }
             
            // Reduce the count by 1
            else if (cnt - 1 == k)
            {
                swap(per, 1, i);
                cnt--;
            }
            else
            {
                break;
            }
            i += 2;
        }
 
        // Print the permutation
        for (i = 1; i <= n; i++)
        {
            Console.Write(per[i] + " ");
        }
    }
 
    static int[] swap(int[] arr, int i, int j)
    {
        int temp = arr[i];
        arr[i] = arr[j];
        arr[j] = temp;
        return arr;
    }
     
    // Driver code
    public static void Main()
    {
        int n = 4, k = 3;
        printPermutation(n, k);
    }
}
 
/* This code contributed by Ryuga */

PHP




<?php
// PHP implementation of the approach
 
// Function to print the required permutation
function printPermutation($n, $k)
{
 
    // Not possible
    if ($k >= $n || ($n % 2 == 0 &&
                     $k == 0))
    {
        echo -1;
        return;
    }
    $per[$n + 1] = array();
    $i;
 
    // Initial permutation
    for ($i = 1; $i <= $n; $i++)
        $per[$i] = $i;
 
    // Indices for which gcd(p[i], i) > 1
    // in the initial permutation
    $cnt = $n - 1;
    $i = 2;
    while ($i < $n)
    {
 
        // Reduce the count by 2
        if ($cnt - 1 > $k)
        {
             
            list($per[$i],
                 $per[$i + 1]) = array($per[$i + 1],
                                       $per[$i]);
            //swap(per[i], per[i + 1]);
            $cnt -= 2;
        }
 
        // Reduce the count by 1
        else if ($cnt - 1 == $k)
        {
             
            // swap(per[1], per[i]);
            list($per[1],
                 $per[$i]) = array($per[$i],
                                   $per[1]);
            $cnt--;
        }
        else
            break;
        $i += 2;
    }
 
    // Print the permutation
    for ($i = 1; $i <= $n; $i++)
        echo $per[$i], " ";
}
 
// Driver code
$n = 4;
$k = 3;
printPermutation($n, $k);
 
// This code is contributed by ajit.
?>

Javascript




<script>
    // Javascript implementation of the approach
     
    // Function to print the required permutation
    function printPermutation(n, k)
    {
   
        // Not possible
        if (k >= n || (n % 2 == 0 && k == 0))
        {
            document.write(-1);
            return;
        }
        let per = new Array(n + 1);
        per.fill(0);
        let i ;
   
        // Initial permutation
        for (i = 1; i <= n; i++)
        {
            per[i] = i;
        }
   
        // Indices for which gcd(p[i], i) > 1
        // in the initial permutation
        let cnt = n - 1;
        i = 2;
        while (i < n)
        {
   
            // Reduce the count by 2
            if (cnt - 1 > k)
            {
                swap(per, i, i + 1);
                cnt -= 2;
            }
               
            // Reduce the count by 1
            else if (cnt - 1 == k)
            {
                swap(per, 1, i);
                cnt--;
            }
            else
            {
                break;
            }
            i += 2;
        }
   
        // Print the permutation
        for (i = 1; i <= n; i++)
        {
            document.write(per[i] + " ");
        }
    }
   
    function swap(arr, i, j)
    {
        let temp = arr[i];
        arr[i] = arr[j];
        arr[j] = temp;
        return arr;
    }
     
    let n = 4, k = 3;
      printPermutation(n, k);
 
</script>

Output: 

1 2 3 4

 

Time Complexity: O(N)

Auxiliary Space: O(N)
 


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