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

  • Difficulty Level : Medium
  • Last Updated : 29 Apr, 2021

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 : 

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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).

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)

This article is contributed by DANISH KALEEM. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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