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# Generate permutation of [1, N] having bitwise XOR of adjacent differences as 0

• Difficulty Level : Basic
• Last Updated : 26 Sep, 2022

Given an integer N, the task is to generate a permutation from 1 to N such that the bitwise XOR of differences between adjacent elements is 0 i.e., | A− A | ^ | A− A | ^ . . . ^ | A[N −1] − A[N] | = 0, where |X – Y| represents absolute difference between X and Y.

Examples:

Input: N = 4
Output: 2 3 1 4
Explanation:  |2 -3| ^ |3 -1| ^ |1-4| = 1 ^ 2 ^ 3 = 0

Input: N = 3
Output: 1 2 3

Approach: This problem can be solved based on the following observation:

• The XOR of even number of same elements is 0. So if odd number of elements are there (which implies even number of adjacent differences) then arrange them in a  way such that the difference between any two adjacent elements is same.
• Else if N is even (which implies odd number of adjacent differences) then arrange the first four in such a way that the XOR of first three differences is 0. Then the remaining elements in the above mentioned away.

Follow the steps mentioned below to implement the above observation:

• If N is odd, arrange all the N elements in a sorted manner because the difference between any two adjacent elements will be 1 and the number of adjacent differences are even.
• If N is even:
• Keep 2, 3, 1, 4 as the first four elements because the 3 differences have XOR 0.
• Now start from 5 and print the remaining elements in sorted order, which will give the difference as 1 for all the remaining even number of differences.

Below is the implementation of the above approach.

## C++

 `// C++ implementation of above approach``#include ``using` `namespace` `std;` `// Function to print shuffle array``vector<``int``> shuffleArray(``int` `n)``{``    ``vector<``int``> res;` `    ``// Base case``    ``if` `(n < 3)``        ``cout << -1 << endl;` `    ``// When n is odd print array in``    ``// increasing order``    ``else` `if` `(n % 2 != 0) {``        ``for` `(``int` `i = 1; i <= n; i++)``            ``res.push_back(i);``    ``}` `    ``// When n is even print first 2 3 1 4``    ``// rest element in increasing order``    ``else` `{``        ``res = { 2, 3, 1, 4 };``        ``for` `(``int` `i = 5; i <= n; i++)``            ``res.push_back(i);``    ``}``    ``return` `res;``}` `// Driver code``int` `main()``{``    ``int` `N = 4;` `    ``vector<``int``> ans = shuffleArray(N);``    ``for` `(``int` `x : ans)``        ``cout << x << ``" "``;``    ``return` `0;``}`

## Java

 `// Java implementation of above approach``import` `java.util.*;``public` `class` `GFG {` `  ``// Function to print shuffle array``  ``static` `ArrayList shuffleArray(``int` `n)``  ``{``    ``ArrayList res = ``new` `ArrayList();` `    ``// Base case``    ``if` `(n < ``3``)``      ``System.out.println(-``1``);` `    ``// When n is odd print array in``    ``// increasing order``    ``else` `if` `(n % ``2` `!= ``0``) {``      ``for` `(``int` `i = ``1``; i <= n; i++)``        ``res.add(i);``    ``}` `    ``// When n is even print first 2 3 1 4``    ``// rest element in increasing order``    ``else` `{``      ``res.clear();` `      ``res.add(``2``);``      ``res.add(``3``);``      ``res.add(``1``);``      ``res.add(``4``);` `      ``for` `(``int` `i = ``5``; i <= n; i++)``        ``res.add(i);``    ``}``    ``return` `res;``  ``}` `  ``// Driver code``  ``public` `static` `void` `main(String args[])``  ``{``    ``int` `N = ``4``;` `    ``ArrayList ans = shuffleArray(N);``    ``for` `(``int` `i = ``0``; i < ans.size(); i++)``      ``System.out.print(ans.get(i) + ``" "``);``  ``}``}` `// This code is contributed by Samim Hossain Mondal.`

## Python3

 `# Python3 implementation of above approach` `# Function to print shuffle array``def` `shuffleArray(n):` `    ``res ``=` `[]` `    ``# Base case``    ``if` `(n < ``3``):``        ``print``(``-``1``)` `    ``# When n is odd print array in``    ``# increasing order``    ``elif` `(n ``%` `2` `!``=` `0``):``        ``for` `i ``in` `range``(``1``, n):``            ``res.append(i)` `    ``# When n is even print first 2 3 1 4``    ``# rest element in increasing order``    ``else``:``        ``res ``=` `[``2``, ``3``, ``1``, ``4``]``        ``for` `i ``in` `range``(``5``, n):``            ``res.append(i)` `    ``return` `res` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ``n ``=` `4``    ``res ``=` `shuffleArray(n)``    ``for` `i ``in` `res:``        ``print``(i, end``=``' '``)` `        ``# This code is contributed by richasalan57.`

## C#

 `// C# implementation of above approach``using` `System;``using` `System.Collections;` `class` `GFG``{` `// Function to print shuffle array``static` `ArrayList shuffleArray(``int` `n)``{``    ``ArrayList res = ``new` `ArrayList();` `    ``// Base case``    ``if` `(n < 3)``        ``Console.WriteLine(-1);` `    ``// When n is odd print array in``    ``// increasing order``    ``else` `if` `(n % 2 != 0) {``        ``for` `(``int` `i = 1; i <= n; i++)``            ``res.Add(i);``    ``}` `    ``// When n is even print first 2 3 1 4``    ``// rest element in increasing order``    ``else` `{``        ``res.Clear();``        ` `        ``res.Add(2);``        ``res.Add(3);``        ``res.Add(1);``        ``res.Add(4);``        ` `        ``for` `(``int` `i = 5; i <= n; i++)``            ``res.Add(i);``    ``}``    ``return` `res;``}` `// Driver code``public` `static` `void` `Main()``{``    ``int` `N = 4;` `    ``ArrayList ans = shuffleArray(N);``    ``foreach` `(``int` `x ``in` `ans)``        ``Console.Write(x + ``" "``);``}``}` `// This code is contributed by Samim Hossain Mondal.`

## Javascript

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Output

`2 3 1 4 `

Time Complexity: O(N)
Auxiliary space: O(N) because it is using extra space for vector res

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