Count numbers whose sum with x is equal to XOR with x

Last Updated : 10 Jun, 2022

Given a integer ‘x’, find the number of values of ‘a’ satisfying the following conditions: 

  1. 0 <= a <= x
  2. a XOR x = a + x

Examples : 

Input : 5
Output : 2
Explanation: 
For x = 5, following 2 values
of 'a' satisfy the conditions:
5 XOR 0 = 5+0
5 XOR 2 = 5+2 

Input : 10
Output : 4
Explanation: 
For x = 10, following 4 values
of 'a' satisfy the conditions:
10 XOR 0 = 10+0
10 XOR 1 = 10+1
10 XOR 4 = 10+4
10 XOR 5 = 10+5

Naive Approach: 
A Simple approach is to check for all values of ‘a’ between 0 and ‘x’ (both inclusive) and calculate its XOR with x and check if the condition 2 satisfies.

Below is the implementation of the above idea:

C++ 
// C++ program to find count of values whose XOR
// with x is equal to the sum of value and x
// and values are smaller than equal to x
#include <bits/stdc++.h>
using namespace std;

int FindValues(int x)
{
    // Initialize result
    int count = 0;

    // Traversing through all values between
    // 0 and x both inclusive and counting
    // numbers that satisfy given property
    for (int i = 0; i <= x; i++)
        if ((x + i) == (x ^ i))
            count++;

    return count;
}

// Driver code
int main()
{
    int x = 10;
  
    // Function call
    cout << FindValues(x);
    return 0;
}
Java 
// Java program to find count of values whose XOR
// with x is equal to the sum of value and x
// and values are smaller than equal to x

class Fib 
{
    static int FindValues(int x)
    {
        // Initialize result
        int count = 0;

        // Traversing through all values between
        // 0 and x both inclusive and counting
        // numbers that satisfy given property
        for (int i = 0; i <= x; i++)
            if ((x + i) == (x ^ i))
                count++;

        return count;
    }

    // Driver code
    public static void main(String[] args)
    {
        int x = 10;
      
        // Function call
        System.out.println(FindValues(x));
    }
}
Python3 
# Python3 program to find count of
# values whose XOR with x is equal
# to the sum of value and x and
# values are smaller than equal to x


def FindValues(x):

    # Initialize result
    count = 0

    # Traversing through all values
    # between 0 and x both inclusive
    # and counting numbers that
    # satisfy given property
    for i in range(0, x):
        if ((x + i) == (x ^ i)):
            count = count + 1

    return count


# Driver code
x = 10

# Function call
print(FindValues(x))

# This code is contributed
# by Shivi_Aggarwal
C# 
// C# program to find count of values whose XOR
// with x is equal to the sum of value and x
// and values are smaller than equal to x
using System;

class Fib
{
    static int FindValues(int x)
    {
        // Initialize result
        int count = 0;

        // Traversing through all values between
        // 0 and x both inclusive and counting
        // numbers that satisfy given property
        for (int i = 0; i <= x; i++)
            if ((x + i) == (x ^ i))
                count++;

        return count;
    }

    // Driver code
    public static void Main()
    {
        int x = 10;
       
        // Function call
        Console.Write(FindValues(x));
    }
}

// This code is contributed by Nitin Mittal.
PHP 
<?php
// PHP program to find count
// of values whose XOR with x
// is equal to the sum of value
// and x and values are smaller
// than equal to x

// function return the 
// value of count
function FindValues($x)
{
    
    // Initialize result
    $count = 0;

    // Traversing through all values between
    // 0 and x both inclusive and counting
    // numbers that satisfy given property
    for ($i = 0; $i <= $x; $i++)
        if (($x + $i) == ($x ^ $i))
            $count++;

    return $count;
}

    // Driver code
    $x = 10;

    // Function call
    echo FindValues($x);
    
// This code is contributed by anuj_67.
?>
JavaScript 
<script>
    // Javascript program to find count of values whose XOR
    // with x is equal to the sum of value and x
    // and values are smaller than equal to x
    
    function FindValues(x)
    {
        // Initialize result
        let count = 0;
 
        // Traversing through all values between
        // 0 and x both inclusive and counting
        // numbers that satisfy given property
        for (let i = 0; i <= x; i++)
            if ((x + i) == (x ^ i))
                count++;
 
        return count;
    }
    
    let x = 10;
        
    // Function call
    document.write(FindValues(x));
    
</script>

Output
4

Time complexity: O(x).

Auxiliary Space: O(1)

Efficient Approach: 
XOR simulates binary addition without the carry over to the next digit. For the zero digits of 'a' we can either add a 1 or 0 without getting a carry which implies xor = + whereas if a digit in 'a' is 1 then the matching digit in x is forced to be 0 in order to avoid carry. For each 0 in 'a' in the matching digit in x can either being a 1 or 0 with a total combination count of 2^(num of zero). Hence, we just need to count the number of 0's in binary representation of the number and answer will by 2^(number of zeroes).

Below is the implementation of the above idea:

C++ 
// C++ program to count numbers whose bitwise
// XOR and sum with x are equal
#include <bits/stdc++.h>
using namespace std;

// Function to find total 0 bit in a number
long CountZeroBit(long x)
{
    unsigned int count = 0;
    while (x)
    {
       if (!(x & 1LL))
           count++;
       x >>= 1LL;
    }
    return count;
}

// Function to find Count of non-negative numbers
// less than or equal to x, whose bitwise XOR and
// SUM with x are equal.
long CountXORandSumEqual(long x)
{
    // count number of zero bit in x
    long count = CountZeroBit(x);

    // power of 2 to count
    return (1LL << count);
}

// Driver code
int main()
{
   long x = 10;
  
   // Function call
   cout << CountXORandSumEqual(x);
   return 0;
}
Java 
// Java program to count
// numbers whose bitwise
// XOR and sum with x
// are equal
import java.io.*;

class GFG {

    // Function to find total
    // 0 bit in a number
    static long CountZeroBit(long x)
    {
        long count = 0;
        while (x > 0) {
            if ((x & 1L) == 0)
                count++;
            x >>= 1L;
        }
        return count;
    }

    // Function to find Count
    // of non-negative numbers
    // less than or equal to x,
    // whose bitwise XOR and
    // SUM with x are equal.
    static long CountXORandSumEqual(long x)
    {
        // count number of
        // zero bit in x
        long count = CountZeroBit(x);

        // power of 2 to count
        return (1L << count);
    }

    // Driver code
    public static void main(String[] args)
    {

        long x = 10;

        // Function call
        System.out.println(CountXORandSumEqual(x));
    }
}

// The code is contributed by ajit
Python3 
# Python3 program to count numbers whose bitwise
# XOR and sum with x are equal

# Function to find total 0 bit in a number


def CountZeroBit(x):

    count = 0
    while (x):

        if ((x & 1) == 0):
            count += 1
        x >>= 1

    return count

# Function to find Count of non-negative numbers
# less than or equal to x, whose bitwise XOR and
# SUM with x are equal.


def CountXORandSumEqual(x):

    # count number of zero bit in x
    count = CountZeroBit(x)

    # power of 2 to count
    return (1 << count)


# Driver code
if __name__ == '__main__':
    x = 10
    
    # Function call
    print(CountXORandSumEqual(x))

# This code is contributed by 29AjayKumar
C# 
// C# program to count
// numbers whose bitwise
// XOR and sum with x
// are equal
using System;

class GFG {

    // Function to find total
    // 0 bit in a number
    static int CountZeroBit(int x)
    {
        int count = 0;
        while (x > 0) {
            if ((x & 1) == 0)
                count++;
            x >>= 1;
        }
        return count;
    }

    // Function to find Count
    // of non-negative numbers
    // less than or equal to x,
    // whose bitwise XOR and
    // SUM with x are equal.
    static int CountXORandSumEqual(int x)
    {
        // count number of
        // zero bit in x
        int count = CountZeroBit(x);

        // power of 2 to count
        return (1 << count);
    }

    // Driver code
    static public void Main()
    {
        int x = 10;
      
        // Function call
        Console.WriteLine(CountXORandSumEqual(x));
    }
}

// The code is contributed by ajit
PHP 
<?php
// PHP program to count numbers whose bitwise
// XOR and sum with x are equal

// Function to find total 0 bit in a number
function CountZeroBit($x)
{
    $count = 0;
    while ($x)
    {
    if (!($x & 1))
        $count++;
    $x >>= 1;
    }
    return $count;
}

// Function to find Count of
// non-negative numbers less
// than or equal to x, whose
// bitwise XOR and SUM with
// x are equal.
function CountXORandSumEqual($x)
{
    // count number of zero bit in x
    $count = CountZeroBit($x);

    // power of 2 to count
    return (1 << $count);
}

    // Driver code
    $x = 10;

    // Function call
    echo CountXORandSumEqual($x);

// This code is contributed by m_kit
?>
JavaScript 
<script>

// Javascript program to count
// numbers whose bitwise
// XOR and sum with x
// are equal

// Function to find total
// 0 bit in a number
function CountZeroBit(x)
{
    let count = 0;
    
    while (x > 0)
    {
        if ((x & 1) == 0)
            count++;
        x >>= 1;
    }
    return count;
}

// Function to find Count
// of non-negative numbers
// less than or equal to x,
// whose bitwise XOR and
// SUM with x are equal.
function CountXORandSumEqual(x)
{
    
    // count number of
    // zero bit in x
    let count = CountZeroBit(x);

    // power of 2 to count
    return (1 << count);
}

// Driver code
let x = 10;
   
// Function call
document.write(CountXORandSumEqual(x));

// This code is contributed by suresh07

</script>

Output
4

Time complexity: O(Log x)

Auxiliary Space: O(1)


Comment
Article Tags:
Bit Magic
DSA
Bitwise-XOR

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