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

• Last Updated : 09 Apr, 2021

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) = GCD(2, 1) = 1
GCD(P, 2) = GCD(1, 2) = 1
GCD(P, 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 ``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 = 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 = 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

 `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`` = ``\$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

 ``
Output:
`3 1 2 4 5`

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