# XOR counts of 0s and 1s in binary representation

• Difficulty Level : Medium
• Last Updated : 09 Sep, 2021

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``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

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## Javascript

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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