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

3.5

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.

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;
    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;
    }
    
    public static void main (String[] args) 
    {
        int x=10;
        System.out.println(FindValues(x));
    }
}


The time complexity of the above approach is 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).

// 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
unsigned int CountZeroBit(int x)
{
    unsigned int 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.
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
int main()
{
   int x = 10;
   cout << CountXORandSumEqual(x);
   return 0;
}

Time complexity of this solution is O(Log x)

Output:

 4

This article is contributed by DANISH KALEEM. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.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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