Count smaller numbers whose XOR with n produces greater value

Given a positive integer n, count numbers x such that 0 < x <n and x^n > n where ^ is bitwise XOR operation.
Examples: 

Input  : n = 12
Output : 3
Numbers are 1, 2 and 3
1^12 > 12,  2^12 > 12 and 3^12 > 12

Input  : n = 11
Output : 4
Numbers are 4, 5, 6 and 7

A number may x produce a greater XOR value if x has a set bit at a position where n has a 0 bit. So we traverse bits of n, and one by one consider all 0 bits. For every set bit at position k (Considering k = 0 for rightmost bit, k = 1 for second rightmost bit, ..), we add 2 2^k to result. For a bit at k-th position, there are 2^k numbers with set bit 1. 
Below is the implementation of the above idea. 

C++

// C++ program to count numbers whose XOR with n
// produces a value more than n.
#include<bits/stdc++.h>
using namespace std;

int countNumbers(int n)
{
    /* If there is a number like m = 11110000,
    then it is bigger than 1110xxxx. x can either
    0 or 1. So we have pow(2, k) greater numbers
    where k is  position of rightmost 1 in m. Now
    by using XOR bit at each  position can be changed.
    To change bit at any position, it needs to XOR
    it with 1. */
    int k = 0; // Position of current bit in n

    /* Traverse bits from LSB (least significant bit)
       to MSB */
    int count = 0;  // Initialize result
    while (n > 0)
    {
        // If current bit is 0, then there are
        // 2^k numbers with current bit 1 and
        // whose XOR with n produces greater value
        if ((n&1) == 0)
            count += pow(2, k);

        // Increase position for next bit
        k += 1;

        // Reduce n to find next bit
        n >>= 1;
    }

    return count;
}

// Driver code
int main()
{
    int n = 11;
    cout << countNumbers(n) << endl;
    return 0;
}

Java

// Java program to count numbers
// whose XOR with n produces a
// value more than n.
import java.lang.*;
class GFG {

    static int countNumbers(int n)
    {

        // If there is a number like
        // m = 11110000, then it is
        // bigger than 1110xxxx. x
        // can either be 0 or 1. So
        // where k is the position of
        // rightmost 1 in m. Now by
        // using the XOR bit at each
        // position can be changed.
        // To change the bit at any
        // position, it needs to
        // XOR it with 1.
        int k = 0;
        // Position of current bit in n

        // Traverse bits from LSB (least
        // significant bit) to MSB

        int count = 0;
        // Initialize result
        while (n > 0) {
            // If the current bit is 0, then
            // there are 2^k numbers with
            // current bit 1 and whose XOR
            // with n produces greater value
            if ((n & 1) == 0)
                count += (int)(Math.pow(2, k));

            // Increase position for next bit
            k += 1;

            // Reduce n to find next bit
            n >>= 1;
        }

        return count;
    }

    // Driver code
    public static void main(String[] args)
    {
        int n = 11;
        System.out.println(countNumbers(n));
    }
}

// This code is contributed by Smitha.

Python3

# Python program to count numbers whose
# XOR with n produces a value more than n.

def countNumbers(n):

    ''' If there is a number like m = 11110000,
    then it is bigger than 1110xxxx. x can either
    0 or 1. So we have pow(2, k) greater numbers
    where k is position of rightmost 1 in m. Now
    by using XOR bit at each position can be changed.
    To change bit at any position, it needs to XOR
    it with 1. '''

    # Position of current bit in n
    k = 0

    # Traverse bits from
    # LSB to MSB
    count = 0 # Initialize result
    while (n > 0):

        # If current bit is 0, then there are
        # 2^k numbers with current bit 1 and
        # whose XOR with n produces greater value
        if ((n & 1) == 0):
            count += pow(2, k)

        # Increase position for next bit
        k += 1

        # Reduce n to find next bit
        n >>= 1

    return count

# Driver code
n = 11
print(countNumbers(n))

# This code is contributed by Anant Agarwal.

C#

// C# program to count numbers
// whose XOR with n produces a
// value more than n.
using System;

class GFG {

    static int countNumbers(int n)
    {

        // If there is a number like
        // m = 11110000, then it is
        // bigger than 1110xxxx. x
        // can either be 0 or 1. So
        // where k is the position of
        // rightmost 1 in m. Now by
        // using the XOR bit at each
        // position can be changed.
        // To change the bit at any
        // position, it needs to
        // XOR it with 1.
        int k = 0;
        // Position of current bit in n

        // Traverse bits from LSB (least
        // significant bit) to MSB

        int count = 0;
        // Initialize result
        while (n > 0) {
            // If the current bit is 0, then
            // there are 2^k numbers with
            // current bit 1 and whose XOR
            // with n produces greater value
            if ((n & 1) == 0)
                count += (int)(Math.Pow(2, k));

            // Increase position for next bit
            k += 1;

            // Reduce n to find next bit
            n >>= 1;
        }

        return count;
    }

    // Driver code
    public static void Main()
    {
        int n = 11;
        Console.WriteLine(countNumbers(n));
    }
}

// This code is contributed by Anant Agarwal.

PHP

<?php
// PHP program to count
// numbers whose XOR with n
// produces a value more than n.

function countNumbers($n)
{

    /* If there is a number
       like m = 11110000,
       then it is bigger
       than 1110xxxx. x
       can either 0 or 1.
       So we have pow(2, k)
       greater numbers where
       k is position of
       rightmost 1 in m.
       Now by using XOR bit
       at each position can
       be changed. To change
       bit at any position,
       it needs to XOR it
       with 1. */

    // Position of current bit in n
    $k = 0;

    /* Traverse bits from LSB
       (least significant bit)
       to MSB */

    // Initialize result
    $count = 0;
    while ($n > 0)
    {

        // If current bit is 0,
        // then there are 2^k
        // numbers with current
        // bit 1 and whose XOR
        // with n produces greater
        // value
        if (($n & 1) == 0)
            $count += pow(2, $k);

        // Increase position
        // for next bit
        $k += 1;

        // Reduce n to
        // find next bit
        $n >>= 1;
    }

    return $count;
}

    // Driver code
    $n = 11;
    echo countNumbers($n);

// This code is contributed by anuj_67.
?>

Javascript

<script>

// Javascript program  to count numbers whose XOR with n
// produces a value more than n.

    function countNumbers(n)
    {

        // If there is a number like
        // m = 11110000, then it is
        // bigger than 1110xxxx. x
        // can either be 0 or 1. So
        // where k is the position of
        // rightmost 1 in m. Now by
        // using the XOR bit at each
        // position can be changed.
        // To change the bit at any
        // position, it needs to
        // XOR it with 1.
        let k = 0;
        // Position of current bit in n

        // Traverse bits from LSB (least
        // significant bit) to MSB

        let count = 0;
        // Initialize result
        while (n > 0) {
            // If the current bit is 0, then
            // there are 2^k numbers with
            // current bit 1 and whose XOR
            // with n produces greater value
            if ((n & 1) == 0)
                count += (Math.pow(2, k));

            // Increase position for next bit
            k += 1;

            // Reduce n to find next bit
            n >>= 1;
        }

        return count;
    }

// Driver Code

    let n = 11;
    document.write(countNumbers(n));

</script>

Output: 

4

Time complexity: O(logn)

Auxiliary Space: O(1)