Find an N-length permutation that contains subarrays with sum less than Bitwise XOR

Given a positive integer N, the task is to find a permutation of length N having Bitwise OR of any of its subarray greater than or equal to the length of the subarray.

Examples:

Input: N = 5 
Output: 1 3 5 2 4 
Explanation:
Consider the subarray {1, 3, 5} from the permutation {1, 3, 5, 2, $}. 
Length of the subarray = 3 
Bitwise OR of the subarray = 1 | 3 | 5 = 7 ( which is greater than 3) 
Similarly, every subarray of any length smaller than equal to N will satisfy the condition.

Input: 4 
Output: 4 3 1 2

Approach: Actually any permutation of length N satisfies the required conditions based on the following facts:

• Bitwise OR of any set of numbers is greater than or equal to the maximum number present in that set.
• Since any subarray will contain at least one element greater than or equal to its length, any permutation satisfies the given condition.

Below is the implementation of the above approach:

C++

// C++ implementation of
// the above approach
#include <iostream>
using namespace std;

// Function to print the
// required permutation
void findPermutation(int N)
{
   for(int i = 1; i <= N; i++)
     cout << i << " ";

   cout << endl;
}

// Driver code
int main()
{

    int N = 5;

    findPermutation(N);

    return 0;
}

Java

// Java implementation of
// the above approach
import java.util.*;

class GFG{

// Function to print the
// required permutation
static void findPermutation(int N)
{
    for(int i = 1; i <= N; i++)
        System.out.print(i + " ");
}

// Driver Code
public static void main(String[] args)
{
     int N = 5;

    findPermutation(N);
}
}

// This code is contributed by susmitakundugoaldanga

Python3

# Python3 implementation of
# the above approach

# Function to print the
# required permutation
def findPermutation(N):
    for i in range(1, N + 1, 1):
        print(i, end= " ")
    print("\n", end = "")

# Driver code
if __name__ == '__main__':
    N = 5
    findPermutation(N)

    # This code is contributed by ipg2016107.

C#

// C# implementation of
// the above approach
using System;

class GFG{

// Function to print the
// required permutation
static void findPermutation(int N)
{
    for(int i = 1; i <= N; i++)
        Console.Write(i + " ");
}

// Driver code
public static void Main()
{
    int N = 5;

    findPermutation(N);
}
}

// This code is contributed by SURENDRA_GANGWAR

Javascript

<script>
// javascript implementation of
// the above approach

// Function to print the
// required permutation
function findPermutation(N)
{
   for(var i = 1; i <= N; i++)
     document.write( i + " ");

   document.write("<br>");
}

var N = 5;
findPermutation(N);

//This code in contributed by SoumikMondal
</script>

Output:

1 2 3 4 5

Time Complexity : O(N)
Auxiliary Space: O(1)