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

- 0 <= a <= x
- 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

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