Find permutation of first N natural numbers that satisfies the given condition

Last Updated : 7 Jun, 2022

Given two integer N and K, the task is to find the permutation P of first N natural numbers such that there are exactly K elements which satisfies the condition GCD(P[i], i) > 1 for all 1 ? i ? N.
Examples: 
 

Input: N = 3, K = 1 
Output: 2 1 3 
GCD(P[1], 1) = GCD(2, 1) = 1 
GCD(P[2], 2) = GCD(1, 2) = 1 
GCD(P[3], 3) = GCD(3, 3) = 3 
There is exactly 1 element such that GCD(P[i], i) > 1
Input: N = 5, K = 2 
Output: 3 1 2 4 5 
 


 


Approach: Keep the last K elements in their place. The rest of the elements are moved such that ith element is placed in (i + 1)th position and (N - K)th element is kept in position 1 because gcd(x, x + 1) = 1.
Below is the implementation of the above approach: 
 

C++ 
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to find permutation(p) of first N
// natural numbers such that there are exactly K
// elements of permutation such that GCD(p[i], i)>1
void Permutation(int n, int k)
{
    int p[n + 1];

    // First place all the numbers
    // in their respective places
    for (int i = 1; i <= n; i++)
        p[i] = i;

    // Modify for first n-k integers
    for (int i = 1; i < n - k; i++)
        p[i + 1] = i;

    // In first index place n-k
    p[1] = n - k;

    // Print the permutation
    for (int i = 1; i <= n; i++)
        cout << p[i] << " ";
}

// Driver code
int main()
{
    int n = 5, k = 2;
    Permutation(n, k);

    return 0;
}
Java 
// Java implementation of the approach 

class GFG
{
    
    // Function to find permutation(p) of first N 
    // natural numbers such that there are exactly K 
    // elements of permutation such that GCD(p[i], i)>1 
    static void Permutation(int n, int k) 
    { 
        int[] p = new int[n + 1]; 

        // First place all the numbers 
        // in their respective places 
        for (int i = 1; i <= n; i++) 
            p[i] = i; 

        // Modify for first n-k integers 
        for (int i = 1; i < n - k; i++) 
            p[i + 1] = i; 

        // In first index place n-k 
        p[1] = n - k; 

        // Print the permutation 
        for (int i = 1; i <= n; i++) 
            System.out.print(p[i] + " "); 
    } 

    // Driver code 
    public static void main(String[] args)
    { 
        int n = 5, k = 2; 
        Permutation(n, k);
    } 
}

// This code is contributed by Naman_Garg
Python3 
# Python 3 implementation of the approach

# Function to find permutation(p) 
# of first N natural numbers such that 
# there are exactly K elements of 
# permutation such that GCD(p[i], i)>1
def Permutation(n, k):
    p = [0 for i in range(n + 1)]

    # First place all the numbers
    # in their respective places
    for i in range(1, n + 1):
        p[i] = i

    # Modify for first n-k integers
    for i in range(1, n - k):
        p[i + 1] = i

    # In first index place n-k
    p[1] = n - k

    # Print the permutation
    for i in range(1, n + 1):
        print(p[i], end = " ")

# Driver code
if __name__ == '__main__':
    n = 5
    k = 2
    Permutation(n, k)
    
# This code is contributed by
# Surendra_Gangwar
C# 
// C# implementation of the approach 
using System;

class GFG
{
    // Function to find permutation(p) of first N 
    // natural numbers such that there are exactly K 
    // elements of permutation such that GCD(p[i], i)>1 
    static void Permutation(int n, int k) 
    { 
        int[] p = new int[n + 1]; 

        // First place all the numbers 
        // in their respective places 
        for (int i = 1; i <= n; i++) 
            p[i] = i; 

        // Modify for first n-k integers 
        for (int i = 1; i < n - k; i++) 
            p[i + 1] = i; 

        // In first index place n-k 
        p[1] = n - k; 

        // Print the permutation 
        for (int i = 1; i <= n; i++) 
            Console.Write(p[i] + " "); 
    } 

    // Driver code 
    static public void Main ()
    {
        int n = 5, k = 2; 
        Permutation(n, k);
    } 
}

// This code is contributed by ajit.
PHP 
<?php
// PHP implementation of the approach 

// Function to find permutation(p) of first N 
// natural numbers such that there are exactly K 
// elements of permutation such that GCD(p[i], i)>1 
function Permutation($n, $k) 
{ 
    $p = array(); 

    // First place all the numbers 
    // in their respective places 
    for ($i = 1; $i <= $n; $i++) 
        $p[$i] = $i; 

    // Modify for first n-k integers 
    for ($i = 1; $i < $n - $k; $i++) 
        $p[$i + 1] = $i; 

    // In first index place n-k 
    $p[1] = $n - $k; 

    // Print the permutation 
    for ($i = 1; $i <= $n; $i++) 
        echo $p[$i], " "; 
} 

// Driver code 
$n = 5; $k = 2; 
Permutation($n, $k); 

// This code is contributed by AnkitRai01
?>
JavaScript 
<script>
//Javascript implementation of the approach

// Function to find permutation(p) of first N
// natural numbers such that there are exactly K
// elements of permutation such that GCD(p[i], i)>1
function Permutation(n, k)
{
    let p  = new Array(n + 1);

    // First place all the numbers
    // in their respective places
    for (let i = 1; i <= n; i++)
        p[i] = i;

    // Modify for first n-k integers
    for (let i = 1; i < n - k; i++)
        p[i + 1] = i;

    // In first index place n-k
    p[1] = n - k;

    // Print the permutation
    for (let i = 1; i <= n; i++)
        document.write(p[i] + " ");
}

// Driver code
    let n = 5, k = 2;
    Permutation(n, k);
 
// This code is contributed by Mayank Tyagi
</script>

Output: 
3 1 2 4 5

 

Time Complexity: O(n)

Auxiliary Space: O(n)

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