Equal Sum and XOR

Given a positive integer n, find count of positive integers i such that 0 <= i <= n and n+i = n^i 

Examples :

Input  : n = 7
Output : 1
Explanation:
7^i = 7+i holds only for only for i = 0
7+0 = 7^0 = 7

Input: n = 12
Output: 4
12^i = 12+i hold only for i = 0, 1, 2, 3
for i=0, 12+0 = 12^0 = 12
for i=1, 12+1 = 12^1 = 13
for i=2, 12+2 = 12^2 = 14
for i=3, 12+3 = 12^3 = 15

Method 1 (Simple) :
One simple solution is to iterate over all values of i 0<= i <= n and count all satisfying values.  

C++

/* C++ program to print count of values such
   that n+i = n^i */
#include <iostream>
using namespace std;

// function to count number of values less than
// equal to n that satisfy the given condition
int countValues (int n)
{
    int countV = 0;

    // Traverse all numbers from 0 to n and
    // increment result only when given condition
    // is satisfied.
    for (int i=0; i<=n; i++ )
        if ((n+i) == (n^i) )
            countV++;

    return countV;
}

// Driver program
int main()
{
    int n = 12;
    cout << countValues(n);
    return 0;
}

Java

/* Java program to print count of values
 such that n+i = n^i */
import java.util.*;

class GFG {

    // function to count number of values
    // less than equal to n that satisfy
    // the given condition
    public static int countValues (int n)
    {
        int countV = 0;

        // Traverse all numbers from 0 to n
        // and increment result only when
        // given condition is satisfied.
        for (int i = 0; i <= n; i++ )
            if ((n + i) == (n ^ i) )
                countV++;

        return countV;
    }

    /* Driver program to test above function */
    public static void main(String[] args)
    {
        int n = 12;
        System.out.println(countValues(n));

    }
}

// This code is contributed by Arnav Kr. Mandal.

Python3

# Python3 program to print count
# of values such that n+i = n^i

# function to count number
# of values less than
# equal to n that satisfy
# the given condition
def countValues (n):
    countV = 0;

    # Traverse all numbers
    # from 0 to n and
    # increment result only
    # when given condition
    # is satisfied.
    for i in range(n + 1):
        if ((n + i) == (n ^ i)):
            countV += 1;

    return countV;

# Driver Code
n = 12;
print(countValues(n));

# This code is contributed by mits

C#

/* C# program to print count of values
such that n+i = n^i */
using System;

class GFG {

    // function to count number of values
    // less than equal to n that satisfy
    // the given condition
    public static int countValues (int n)
    {
        int countV = 0;

        // Traverse all numbers from 0 to n
        // and increment result only when
        // given condition is satisfied.
        for (int i = 0; i <= n; i++ )
            if ((n + i) == (n ^ i) )
                countV++;

        return countV;
    }

    /* Driver program to test above function */
    public static void Main()
    {
        int n = 12;
        Console.WriteLine(countValues(n));

    }
}

// This code is contributed by anuj_67.

PHP

<?php
// PHP program to print count
// of values such that n+i = n^i

// function to count number
// of values less than
// equal to n that satisfy
// the given condition
function countValues ($n)
{
    $countV = 0;

    // Traverse all numbers
    // from 0 to n and
    // increment result only
    // when given condition
    // is satisfied.
    for ($i = 0; $i <= $n; $i++ )
        if (($n + $i) == ($n^$i) )
            $countV++;

    return $countV;
}

    // Driver Code
    $n = 12;
    echo countValues($n);

// This code is contributed by m_kit
?>

Javascript

<script>

/* JavaScript program to print count of values such
that n+i = n^i */

// function to count number of values less than
// equal to n that satisfy the given condition
function countValues (n)
{
    let countV = 0;

    // Traverse all numbers from 0 to n and
    // increment result only when given condition
    // is satisfied.
    for (let i=0; i<=n; i++ )
        if ((n+i) == (n^i) )
            countV++;

    return countV;
}

// Driver program

    let n = 12;
    document.write(countValues(n));

// This code is contributed by Surbhi Tyagi.

</script>

Output:

4

Time Complexity: O(n)

Space Complexity: O(1)

Method 2 (Efficient) : 
An efficient solution is as follows

we know that (n+i)=(n^i)+2*(n&i)
So n + i = n ^ i implies n & i = 0
Hence our problem reduces to finding values of i such that n & i = 0. How to find count of such pairs? We can use the count of unset-bits in the binary representation of n. For n & i to be zero, i must unset all set-bits of n. If the kth bit is set at a particular in n, kth bit in i must be 0 always, else kth bit of i can be 0 or 1
Hence, total such combinations are 2^(count of unset bits in n)
For example, consider n = 12 (Binary representation : 1 1 0 0). 
All possible values of i that can unset all bits of n are 0 0 0/1 0/1 where 0/1 implies either 0 or 1. Number of such values of i are 2^2 = 4. 

The following is the program following the above idea.  

C++

/* c++ program to print count of values such
  that n+i = n^i */
#include <bits/stdc++.h>
using namespace std;

// function to count number of values less than
// equal to n that satisfy the given condition
int countValues(int n)
{
    // unset_bits keeps track of count of un-set
    // bits in binary representation of n
    int unset_bits=0;
    while (n)
    {
        if ((n & 1) == 0)
            unset_bits++;
        n=n>>1;
    }

    // Return 2 ^ unset_bits
    return 1 << unset_bits;
}

// Driver code
int main()
{
    int n = 12;
    cout << countValues(n);
    return 0;
}

Java

/* Java program to print count of values
  such that n+i = n^i */
import java.util.*;

class GFG {

    // function to count number of values
    // less than equal to n that satisfy
    // the given condition
    public static int countValues(int n)
    {
        // unset_bits keeps track of count
        // of un-set bits in binary
        // representation of n
        int unset_bits=0;
        while (n > 0)
        {
            if ((n & 1) == 0)
                unset_bits++;
            n=n>>1;
        }

        // Return 2 ^ unset_bits
        return 1 << unset_bits;
    }

    /* Driver program to test above
    function */
    public static void main(String[] args)
    {
        int n = 12;
        System.out.println(countValues(n));

    }
}

// This code is contributed by Arnav Kr. Mandal.

C#

/* C# program to print count of values
  such that n+i = n^i */
using System;
public class GFG {

    // function to count number of values
    // less than equal to n that satisfy
    // the given condition
    public static int countValues(int n)
    {

        // unset_bits keeps track of count
        // of un-set bits in binary
        // representation of n
        int unset_bits=0;
        while (n > 0)
        {
            if ((n & 1) == 0)
                unset_bits++;
            n=n>>1;
        }

        // Return 2 ^ unset_bits
        return 1 << unset_bits;
    }

    /* Driver program to test above
    function */
    public static void Main(String[] args)
    {
        int n = 12;
        Console.WriteLine(countValues(n));

    }
}

// This code is contributed by umadevi9616

Python3

# Python3 program to print count of values such
# that n+i = n^i

# function to count number of values less than
# equal to n that satisfy the given condition
def countValues(n):

    # unset_bits keeps track of count of un-set
    # bits in binary representation of n
    unset_bits = 0

    while(n):
        if n & 1 == 0:
            unset_bits += 1
        n = n >> 1

    # Return 2 ^ unset_bits
    return 1 << unset_bits

# Driver code
if __name__=='__main__':
    n = 12
    print(countValues(n))

# This code is contributed by rutvik

PHP

<?php
/* PHP program to print count
of values such that n+i = n^i */


// function to count number of
// values less than equal to n
// that satisfy the given
// condition
function countValues( $n)
{

    // unset_bits keeps track
    // of count of un-set bits
    // in binary representation
    // of n
    $unset_bits = 0;
    while ($n)
    {
        if (($n & 1) == 0)
            $unset_bits++;
        $n = $n >> 1;
    }

    // Return 2 ^ unset_bits
    return 1 << $unset_bits;
}

// Driver code

    $n = 12;
    echo countValues($n);

// This code is contributed
// by Anuj_67.
?>

Javascript

<script>

// Javascript program to print count of values
// such that n+i = n^i

// Function to count number of values
// less than equal to n that satisfy
// the given condition
function countValues(n)
{

    // unset_bits keeps track of count
    // of un-set bits in binary
    // representation of n
    let unset_bits = 0;

    while (n > 0)
    {
        if ((n & 1) == 0)
            unset_bits++;

        n = n >> 1;
    }

    // Return 2 ^ unset_bits
    return 1 << unset_bits;
}

// Driver Code
let n = 12;

document.write(countValues(n));

// This code is contributed by susmitakundugoaldanga

</script>

Output :

4

Time Complexity: O(log(n))

Space Complexity: O(1)