Count numbers whose XOR with N is equal to OR with N

Last Updated : 21 Aug, 2022

Given a number N, the task is to find the count of X such that N XOR X == N OR X, where 0<=X<=N
Examples:

Input: N = 5
Output: 2
For N = 5,
5 XOR 2 == 5 OR 2
5 XOR 0 == 5 OR 0
Thus, count is 2.
Input: N = 7
Output: 1
For N = 7,
7 XOR 0 == 7 OR 0
Thus, count is 1.

Approach: The idea is to convert a given number to binary and then count the unset bits in it. 2^count gives us the number of X such that N XOR X == N OR X.
Below is the implementation of the above approach:

C++

// C++ program to find 
// the XOR equals OR count 
#include<iostream>
#include<math.h>
using namespace std; 

class gfg { 
    
    // Function to calculate count 
    // of numbers with XOR equals OR 
    public:
    int xorEqualsOrCount(int N) 
    { 
        
        // variable to store count of unset bits 
        int count = 0; 
        int bit; 
        while (N > 0) { 
            
            bit = N % 2; 
            if (bit == 0) 
                count++; 
            N = N / 2; 
        } 
        return (int)pow(2, count); 
    } };

    // Driver code 
    int main() 
    { 
        gfg g ;
        int N = 7; 
        cout<<g.xorEqualsOrCount(N);
        return 0;
    } 


// This code is contributed by Soumik

Java

// Java program to find the XOR equals OR count
import java.io.*;
import java.util.*;

class GFG {

    // Function to calculate count of numbers with XOR equals OR
    static int xorEqualsOrCount(int N)
    {
        // variable to store count of unset bits
        int count = 0;
        int bit;
        while (N > 0) {
            bit = N % 2;
            if (bit == 0)
                count++;
            N = N / 2;
        }
        return (int)Math.pow(2, count);
    }

    // Driver code
    public static void main(String args[])
    {
        int N = 7;
        System.out.println(xorEqualsOrCount(N));
    }
}

Python3

# Python3 program to find
# the XOR equals OR count

# Function to calculate count
# of numbers with XOR equals OR
def xorEqualsOrCount(N) :

    # variable to store
    # count of unset bits
    count = 0

    while(N > 0) :

        bit = N % 2

        if bit == 0 :
            count += 1

        N //= 2

    return int(pow(2, count))

# Driver code     
if __name__ == "__main__" :

    N = 7
    print(xorEqualsOrCount(N))
                
# This code is contributed by 
# ANKITRAI1

// C# program to find 
// the XOR equals OR count
using System;

class GFG {

    // Function to calculate count 
    // of numbers with XOR equals OR
    static int xorEqualsOrCount(int N)
    {
        
        // variable to store count of unset bits
        int count = 0;
        int bit;
        while (N > 0) {
            
            bit = N % 2;
            if (bit == 0)
                count++;
            N = N / 2;
        }
        return (int)Math.Pow(2, count);
    }

    // Driver code
    public static void Main()
    {
        int N = 7;
        Console.WriteLine(xorEqualsOrCount(N));
    }
}

// This code is contributed by inder_verma..

PHP

<?php 
// PHP program to find the XOR
// equals OR count 
    
// Function to calculate count 
// of numbers with XOR equals OR 
function xorEqualsOrCount($N) 
{ 
    
    // variable to store count 
    // of unset bits 
    $count = 0; 
    while ($N > 0) 
    { 
        $bit = $N % 2; 
        if ($bit == 0) 
            $count++; 
        $N = intval($N / 2); 
    } 
    return pow(2, $count); 
} 

// Driver code 
$N = 7; 
echo xorEqualsOrCount($N);

// This code is contributed
// by ChitraNayal
?>

JavaScript

<script>

// Javascript program to find 
// the XOR equals OR count 

    // Function to calculate count 
    // of numbers with XOR equals OR 
    function xorEqualsOrCount(N) 
    { 
        
        // variable to store count of unset bits 
        let count = 0; 
        let bit; 
        while (N > 0) { 
            
            bit = N % 2; 
            if (bit == 0) 
                count++; 
            N = parseInt(N / 2); 
        } 
        return Math.pow(2, count); 
    } 

    // Driver code 
        let N = 7; 
        document.write(xorEqualsOrCount(N));

</script>

Output:

Time Complexity: O(log N)

Auxiliary Space: O(1)

Count numbers less than N whose modulo with A is equal to B

rachana soma

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Count numbers whose XOR with N is equal to OR with N

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