Count of distinct values for bitwise XOR of X and Y for X, Y at most N

Given an integer N, the task is to find the number of distinct values possible for the bit-wise XOR of X and Y where 1 ≤ X, Y ≤ N.

Examples:

Input: N = 1
Output: 1
Explanation: The possible xor values are 1⊕1=0 which has 1 unique value.

Input: N = 2
Output: 2
Explanation: The possible xor values are 1⊕1 = 0, 1⊕2 = 1, 2⊕2 = 0 which has 2 unique values.

Approach: For the values of N equals 1 and 2, the answer is simple. For the remaining cases, consider N≥3. Consider p as the highest power for which 2^p ≤ N. Suppose 2^p < N, then all the numbers from 0 to 2^{p+1} -1 can be obtained. This can be achieved in the following way:

For example,  
let N = 12, then x = 3. 
Number 12 (1100 in binary ) can be formed by (2^3(1000), 4(0100)).
By the property of xor, num⊕1 is either num +1 or num -1. 
So if  1 < num ≤ 2^p < N, 1 ≤ num⊕1 ≤ N. 
Hence these pairs of numbers will always be valid.

Now, what happens if 2^p = N? All of the above cases hold true except for the case of num = 2^p. This cannot be obtained from any xor pair (X, Y) where (1 ≤ X, Y ≤ 2^p). Since the only number with bit p set is 2^p, we must keep i = 2^p. Then for X ⊕ Y=2*p, Keep Y = 0, which cannot be done since Y ≥ 1. Hence, in this case, except 2^p, xor pair for any number from 0 to 2^{p + 1} -1 can be obtained. Follow the steps below to solve the problem:

• Initialize the variable ans as 1.
• If N equals 2 then print 2 and return.
• Traverse in a while loop till ans is less than N and multiply ans by 2.
• If ans equals N then multiply ans by 2 and reduce it’s value by 1.
• After performing the above steps, print the value of ans as the answer.

Below is the implementation of the above approach.

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
int MOD = 1e9 + 7;

// Function to find the possible values
void find(long long N)
{
    long long ans = 1;

    // Special case
    if (N == 2) {
        cout << 2 << endl;
        return;
    }

    while (ans < N) {
        ans *= 2;
    }

    if (ans == N) {
        ans *= 2;
        ans--;
    }

    cout << ans % MOD;
}

// Driver Code
int main()
{
    long long N = 7;
    find(N);
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;

class GFG {

  // Function to find the possible values
  static void find(long N)
  {
    long MOD = 1000000007;
    long ans = 1;

    // Special case
    if (N == 2) {
      System.out.println("2");
      return;
    }

    while (ans < N) {
      ans *= 2;
    }

    if (ans == N) {
      ans *= 2;
      ans--;
    }
    long temp = ans % MOD;
    System.out.print(temp);
  }

  // Driver Code
  public static void main (String[] args) {
    long N = 7;
    find(N);
  }
}

// This code is contributed by hrithikgarg03188.

Python

# Python program for the above approach

# Function to find the possible values
def find(N):

    MOD = 1000000007
    ans = 1

    # Special case
    if (N == 2):
        print(2)
        return;

    while ans < N:
        ans *= 2


    if (ans == N):
        ans *= 2
        ans -= 1

    print(ans % MOD)

if __name__ == "__main__":

    N = 7
    find(N)

  # This code is contributed by hrithikgarg03188.

C#

// C# program to implement
// the above approach
using System;
class GFG
{

// Function to find the possible values
  static void find(long N)
  {
    long MOD = 1000000007;
    long ans = 1;

    // Special case
    if (N == 2) {
      Console.WriteLine("2");
      return;
    }

    while (ans < N) {
      ans *= 2;
    }

    if (ans == N) {
      ans *= 2;
      ans--;
    }
    long temp = ans % MOD;
    Console.Write(temp);
  }

// Driver Code
public static void Main()
{
    long N = 7;
    find(N);
}
}

// This code is contributed by Samim Hossain Mondal.

Javascript

<script>
       // JavaScript code for the above approach
       let MOD = 1e9 + 7;

       // Function to find the possible values
       function find(N)
       {
           let ans = 1;

           // Special case
           if (N == 2) {
               document.write(2 + '<br>')
               return;
           }
           while (ans < N) {
               ans *= 2;
           }
           if (ans == N) {
               ans *= 2;
               ans--;
           }
           document.write(ans % MOD);
       }

       // Driver Code
       let N = 7;
       find(N);

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

Output

8

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