Non-Repeating Bitwise OR Permutation

Given an integer N (N >= 3). Then the task is to output a permutation such that the bitwise OR of the previous two elements is not equal to the current element, given that the previous two element exists.

Note: If multiple permutation exists, just print any valid permutation.

Examples:

Input: N = 3
Output: 1 3 2
Explanation:

• First element: A[1] = 1. Two previous elements don't exist. Therefore, no need to check.
• Second element: A[2] = 3. Two previous elements don't exist. Therefore, no need to check.
• Third element: A[3] = 2. Two previous elements are 1 and 3. Bitwise OR of 1 and 3 is (1 | 3) = 3, which is not equal to 2.

Thus, permutation satisfies the problem constraints.

Input: N = 5
Output: 2 1 5 3 4
Explanation:

• First element: A[1] = 2. Two previous elements don't exist. Therefore, no need to check.
• Second element: A[2] = 1. Two previous elements don't exist. Therefore, no need to check.
• Third element: A[3] = 5. Two previous elements are 1 and 2. Bitwise OR of 1 and 2 is (1 | 2) = 3, which is not equal to 5.
• Fourth element: A[4] = 3. Two previous elements are 1 and 5. Bitwise OR of 1 and 5 is (1 | 5) = 5, which is not equal to 3.
• Fifth element: A[5] = 4. Two previous elements are 5 and 3. Bitwise OR of 5 and 3 is (5 | 3) = 7, which is not equal to 4.

Approach: Implement the idea below to solve the problem

The problem is observation based. It must be noted that if X, Y, Z are positive integers such that Z = (X | Y), Then Z >= max(X, Y). Therefore, If we output N numbers in decreasing order then there's no way our permutation can be wrong. Hence, it's our required answer.

Steps taken to solve the problem:

• Run a loop reversely from i = N to i = 1 and output i.

Code to implement the approach:

#include <iostream>
using namespace std;

void PrintPermutation(int N) {
    for (int i = N; i >= 1; i--) {
        cout << i << " ";
    }
}

int main() {
    // Input
    int N = 5;

    // Function call
    PrintPermutation(N);

    return 0;
}

/*package whatever //do not write package name here */

import java.io.*;

class GFG {
    public static void main(String[] args)
    {
        // Input
        int N = 5;

        // Function call
        Print_permutation(N);
    }
    public static void Print_permutation(int N)
    {
        for (int i = N; i >= 1; i--) {
            System.out.print(i + " ");
        }
    }
}

# Python program for the above approach
def print_permutation(N):
    for i in range(N, 0, -1):
        print(i, end=" ")

# Input
N = 5

# Function call
print_permutation(N)

# This code is contributed by Susobhan Akhuli

using System;

public class GFG
{
    public static void Main(string[] args)
    {
        // Input
        int N = 5;

        // Function call
        PrintPermutation(N);
    }

    public static void PrintPermutation(int N)
    {
        for (int i = N; i >= 1; i--)
        {
            Console.Write(i + " ");
        }
    }
}

<script>
// Javascript program for the above approach

// Function to print permutation
function printPermutation(N) {
    for (let i = N; i >= 1; i--) {
        document.write(i + " ");
    }
}

// Input
const N = 5;

// Function call
printPermutation(N);

// This code is contributed by Susobhan Akhuli
</script>

5 4 3 2 1 

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