Given a number, the task is to find XOR of count of 0s and count of 1s in binary representation of a given number.
Examples:
Input : 5 Output : 3 Binary representation : 101 Count of 0s = 1, Count of 1s = 2 1 XOR 2 = 3. Input : 7 Output : 3 Binary representation : 111 Count of 0s = 0 Count of 1s = 3 0 XOR 3 = 3.
The idea is simple, we traverse through all bits of a number, count 0s and 1s and finally return XOR of two counts.
C++
// C++ program to find XOR of counts 0s and 1s in // binary representation of n. #include<iostream> using namespace std;
// Returns XOR of counts 0s and 1s in // binary representation of n. int countXOR( int n)
{ int count0 = 0, count1 = 0;
while (n)
{
//calculating count of zeros and ones
(n % 2 == 0) ? count0++ :count1++;
n /= 2;
}
return (count0 ^ count1);
} // Driver Program int main()
{ int n = 31;
cout << countXOR (n);
return 0;
} |
Java
// Java program to find XOR of counts 0s // and 1s in binary representation of n. class GFG {
// Returns XOR of counts 0s and 1s
// in binary representation of n.
static int countXOR( int n)
{
int count0 = 0 , count1 = 0 ;
while (n != 0 )
{
//calculating count of zeros and ones
if (n % 2 == 0 )
count0++ ;
else
count1++;
n /= 2 ;
}
return (count0 ^ count1);
}
// Driver Program
public static void main(String[] args)
{
int n = 31 ;
System.out.println(countXOR (n));
}
} // This code is contributed by prerna saini |
Python3
# Python3 program to find XOR of counts 0s # and 1s in binary representation of n. # Returns XOR of counts 0s and 1s # in binary representation of n. def countXOR(n):
count0, count1 = 0 , 0
while (n ! = 0 ):
# calculating count of zeros and ones
if (n % 2 = = 0 ):
count0 + = 1
else :
count1 + = 1
n / / = 2
return (count0 ^ count1)
# Driver Code n = 31
print (countXOR(n))
# This code is contributed by Anant Agarwal. |
C#
// C# program to find XOR of counts 0s // and 1s in binary representation of n. using System;
class GFG {
// Returns XOR of counts 0s and 1s
// in binary representation of n.
static int countXOR( int n)
{
int count0 = 0, count1 = 0;
while (n != 0)
{
// calculating count of zeros
// and ones
if (n % 2 == 0)
count0++ ;
else
count1++;
n /= 2;
}
return (count0 ^ count1);
}
// Driver Program
public static void Main()
{
int n = 31;
Console.WriteLine(countXOR (n));
}
} // This code is contributed by Anant Agarwal. |
PHP
<?PHP // PHP program to find XOR of // counts 0s and 1s in binary // representation of n. // Returns XOR of counts 0s and 1s // in binary representation of n. function countXOR( $n )
{ $count0 = 0;
$count1 = 0;
while ( $n )
{
// calculating count of
// zeros and ones
( $n % 2 == 0) ? $count0 ++ : $count1 ++;
$n = intval ( $n / 2);
}
return ( $count0 ^ $count1 );
} // Driver Code $n = 31;
echo countXOR ( $n );
// This code is contributed // by ChitraNayal ?> |
Javascript
<script> // Javascript program to find XOR of counts 0s // and 1s in binary representation of n. // Returns XOR of counts 0s and 1s
// in binary representation of n.
function countXOR(n)
{
let count0 = 0, count1 = 0;
while (n != 0)
{
//calculating count of zeros and ones
if (n % 2 == 0)
count0++ ;
else
count1++;
n = Math.floor(n/2);
}
return (count0 ^ count1);
}
// Driver Program
let n = 31;
document.write(countXOR (n));
// This code is contributed by avanitrachhadiya2155
</script> |
Output:
5
Time Complexity: O(log(N))
Auxiliary Space: O(1)
One observation is, for a number of the form 2^x – 1, the output is always x. We can directly produce answer for this case by first checking n+1 is a power of two or not.