# K difference permutation

• Difficulty Level : Medium
• Last Updated : 20 May, 2021

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 ```

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

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++

 `// C++ program to find k absolute difference``// permutation``#include``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;``}`

## Java

 `// 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.`

## Python3

 `# 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`

## C#

 `// 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.`

## PHP

 ``

## Javascript

 ``

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 write.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.