Count of values of x <= n for which (n XOR x) = (n – x)

Given an integer n, the task is to find the number of possible values of 0 ≤ x ≤ n which satisfy n XOR x = n – x.

Examples:

Input: n = 5
Output: 4
Following values of x satisfy the equation
5 XOR 0 = 5 – 0 = 5
5 XOR 1 = 5 – 1 = 4
5 XOR 4 = 5 – 4 = 1
5 XOR 5 = 5 – 5 = 0

Input: n = 2
Output: 2

Naive approach: The easy approach is to check for all values from 0 to n (both inclusive) and finding whether they satisfy the equation. The below code implements this approach:

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include<bits/stdc++.h>
using namespace std;
  
// Function to return the count of 
// valid values of x
static int countX(int n)
{
    int count = 0;
  
    for (int i = 0; i <= n; i++)
    {
  
        // If n - x = n XOR x
        if (n - i == (n ^ i))
                count++;
    }
  
        // Return the required count;
        return count;
}
  
// Driver code
int main()
{
    int n = 5;
    int answer = countX(n);
    cout << answer;
}
  
// This code is contributed by 
// Shivi_Aggarwal

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Java

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// Java implementation of the approach
public class GFG {
  
    // Function to return the count of 
    // valid values of x
    static int countX(int n)
    {
        int count = 0;
  
        for (int i = 0; i <= n; i++) {
  
            // If n - x = n XOR x
            if (n - i == (n ^ i))
                count++;
        }
  
        // Return the required count;
        return count;
    }
  
    // Driver code
    public static void main(String args[])
    {
        int n = 5;
        int answer = countX(n);
        System.out.println(answer);
    }
}

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Python3

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# Python3 implementation of the approach
import math as mt
  
# Function to return the count of
# valid values of x
def countX(n):
    count = 0
  
    for i in range(n + 1):
  
        if n - i == (n ^ i):
            count += 1
  
    return count
  
# Driver Code
if __name__ == '__main__':
    n = 5
    answer = countX(n)
    print(answer)
  
# This code is contributed by 
# Mohit kumar 29

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C#

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// C# implementation of the above approach 
using System;
  
class GFG 
  
    // Function to return the count of 
    // valid values of x 
    static int countX(int n) 
    
        int count = 0; 
  
        for (int i = 0; i <= n; i++)
        
  
            // If n - x = n XOR x 
            if (n - i == (n ^ i)) 
                count++; 
        
  
        // Return the required count; 
        return count; 
    
  
    // Driver code 
    public static void Main() 
    
        int n = 5; 
        int answer = countX(n); 
        Console.WriteLine(answer); 
    
  
// This code is contributed by Ryuga

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PHP

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<?php
// PHP implementation of the approach
  
// Function to return the count of 
// valid values of x
function countX($n)
{
    $count = 0;
  
    for ($i = 0; $i <= $n; $i++)
    {
  
        // If n - x = n XOR x
        if ($n - $i == ($n ^ $i))
            $count++;
    }
  
    // Return the required count;
    return $count;
}
  
// Driver code
$n = 5;
$answer = countX($n);
echo($answer);
  
// This code is Contributed 
// by Mukul Singh.
?>

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Output:

4

Time complexity: O(N)

Efficient Approach: Convert n to its binary representation. Now, for every 1 in the binary string whether we subtract 1 or 0 from it, it will be equivalent to XOR of 1 with 0 or 1 i.e.
(1 – 1) = (1 XOR 1) = 0
(1 – 0) = (1 XOR 0) = 1
But 0 doesn’t satisfy this condition. So, we only need to consider all the ones in the binary representation of n. Now, for every 1 there are two possibilities, either 0 or 1. Thus if we have m number of 1’s in n then our solution would be 2m.

Below is the implementation of the above approach:

Java

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// Java implementation of the approach
public class GFG {
  
    // Function to return the count of
    // valid values of x
    static int countX(int n)
    {
        // Convert n into binary String
        String binary = Integer.toBinaryString(n);
  
        // To store the count of 1s
        int count = 0;
  
        for (int i = 0; i < binary.length(); i++) {
  
            // If current bit is 1
            if (binary.charAt(i) == '1')
                count++;
        }
  
        // Calculating answer
        int answer = (int)Math.pow(2, count);
        return answer;
    }
  
    // Driver code
    public static void main(String args[])
    {
        int n = 5;
        int answer = countX(n);
        System.out.println(answer);
    }
}

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Python 3

# Python3 implementation of the approach

# Function to return the count of
# valid values of x
def countX(n):

# Convert n into binary String
binary = “{0:b}”.format(n)

# To store the count of 1s
count = 0

for i in range(len(binary)):

# If current bit is 1
if (binary[i] == ‘1’):
count += 1

# Calculating answer
answer = int(pow(2, count))
return answer

# Driver code
if __name__ == “__main__”:

n = 5
answer = countX(n)
print(answer)

# This code is contributed by ita_c

C#

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// C# implementation of the approach
using System;
  
class GFG 
{
  
// Function to return the count of
// valid values of x
static int countX(int n)
{
    // Convert n into binary String
    string binary = Convert.ToString(n, 2);
  
    // To store the count of 1s
    int count = 0;
  
    for (int i = 0; i < binary.Length; i++) 
    {
  
        // If current bit is 1
        if (binary[i] == '1')
            count++;
    }
  
    // Calculating answer
    int answer = (int)Math.Pow(2, count);
    return answer;
}
  
// Driver code
public static void Main()
{
    int n = 5;
    int answer = countX(n);
    Console.WriteLine(answer);
}
}
  
// This code is contributed
// by Akanksha Rai

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Output:

4

Time complexity:  $O(\log{}n)$



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