Count of all possible values of X whose Bitwise XOR with N is greater than N

Given an integer N count the number of values of X such that  X⊕N > N, where  ⊕ denotes bitwise  XOR operation

Examples:

Input: N = 10
Output: 5 
Explanation: The five possible value satisfying the above condition are:
1⊕10 = 11, 4⊕10 = 14, 5⊕10 = 15, 6⊕10 = 12, 7⊕10 = 13

Input: N = 8
Output: 7
Explanation: The seven possible value satisfying the above condition are:
1⊕8 = 9, 2⊕8 = 10, 3⊕8 = 11, 4⊕8 = 12, 5⊕8 = 13, 6⊕8 = 14, 7⊕8 = 15

Naive Approach: The simple approach to this problem is to iterate from 1 to N and increment the count if  X XOR N >N  for 0 < X < N. Finally, return the count of the number of values.

Time Complexity: O(N)
Auxiliary Space: O(1)

Efficient Approach: The most efficient approach is to find the number nearest to the next power of 2 which has all its bits set and finally subtract it from the original number. Given below is the mathematical observation of the solution:

If number N can be written by using k bits then number  X must use k-1 bits at most because:-

• If a number X has a same bits as number N Xoring both the number will result in a number less than N which wont satisfy our condition X xor N > N.
• If a number X is greater than number N then Xoring both the number  will always result X is greater than N which wont satisfy our inequality.
  So the problem reduces to find the count of the number that lies in the given range [x , 2^k - 1]. We are taking 2^k - 1 because this is the maximum number that can be form when all its n bits are set.
• The require number equals  2^k - 1 -N .

Follow the steps below to solve the problem:

• Find the nearest next power of 2.
• Subtract 1 from the nearest power of 2  so that the given number has all its bit set
• Finally, subtract from the original number and return it

Below is the implementation of the above approach: 

// C++ program for find the count of
// number of value of X such that  N XOR X >N

#include <bits/stdc++.h>
using namespace std;

// Function for  calculating the total
// possible value satisfy the condition
// X⊕N > N and 0 < X < N
void maximumXor(long long N)
{
    long long num = N;
    long long int count;

    count = 0;
    while (N > 0) {

        // Count the total number of bits
        count++;
        N = N / 2;
    }
    long long next_power = ((long long)1 << count);

    // Finding the next power of 2
    // of the given number
    long long result = next_power - 1;

    // Finding the number nearest to
    // the nextpower of 2 which has all
    // its bits set
    cout << result - num;
}

// Driver Code
int main()
{
    int N = 10;
    maximumXor(N);
    return 0;
}

// Java program for find the count of
// number of value of X such that  N XOR X >N
import java.util.*;
public class GFG {

  // Function for  calculating the total
  // possible value satisfy the condition
  // X⊕N > N and 0 < X < N
  static void maximumXor(long N)
  {
    long num = N;
    long count = 0;

    count = 0;
    while (N > 0) {

      // Count the total number of bits
      count++;
      N = N / 2;
    }
    long next_power = ((long)1 << count);

    // Finding the next power of 2
    // of the given number
    long result = next_power - 1;

    // Finding the number nearest to
    // the nextpower of 2 which has all
    // its bits set
    System.out.print(result - num);
  }

  // Driver Code
  public static void main(String args[])
  {
    int N = 10;
    maximumXor(N);
  }
}

// This code is contributed by Samim Hossain Mondal.

# python3 program for find the count of
# number of value of X such that N XOR X >N

# Function for calculating the total
# possible value satisfy the condition
# X⊕N > N and 0 < X < N
def maximumXor(N):

    num = N
    count = 0

    while (N > 0):

        # Count the total number of bits
        count += 1
        N = N // 2

    next_power = (1 << count)

    # Finding the next power of 2
    # of the given number
    result = next_power - 1

    # Finding the number nearest to
    # the nextpower of 2 which has all
    # its bits set
    print(result - num)

# Driver Code
if __name__ == "__main__":

    N = 10
    maximumXor(N)

# This code is contributed by rakeshsahni

// C# program for find the count of
// number of value of X such that  N XOR X >N
using System;

public class GFG
{

  // Function for  calculating the total
  // possible value satisfy the condition
  // X⊕N > N and 0 < X < N
  static void maximumXor(long N)
  {
    long num = N;
    long count = 0;

    count = 0;
    while (N > 0) {

      // Count the total number of bits
      count++;
      N = N / 2;
    }
    long next_power = (1 << (int)count);

    // Finding the next power of 2
    // of the given number
    long result = next_power - 1;

    // Finding the number nearest to
    // the nextpower of 2 which has all
    // its bits set
    Console.Write(result - num);
  }

  // Driver Code
  public static void Main(String []args)
  {
    int N = 10;
    maximumXor(N);
  }
}

// This code is contributed by 29AjayKumar

 <script>
        // JavaScript code for the above approach

        // Function for  calculating the total
        // possible value satisfy the condition
        // X⊕N > N and 0 < X < N
        function maximumXor(N) {
            let num = N;
            let count;

            count = 0;
            while (N > 0) {

                // Count the total number of bits
                count++;
                N = Math.floor(N / 2);
            }
            let next_power = (1 << count);

            // Finding the next power of 2
            // of the given number
            let result = next_power - 1;

            // Finding the number nearest to
            // the nextpower of 2 which has all
            // its bits set
            document.write(result - num);
        }

        // Driver Code
        let N = 10;
        maximumXor(N);

     // This code is contributed by Potta Lokesh
    </script>

5

Time Complexity: O(log(N))
Auxiliary Space: O(1)