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

- Difficulty Level : Medium
- Last Updated : 15 Jul, 2022

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.

This article is contributed by **Shivam Pradhan (anuj_charm)**. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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