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

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; ` `} ` |

*chevron_right*

*filter_none*

## 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 ` |

*chevron_right*

*filter_none*

## 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. ` |

*chevron_right*

*filter_none*

## 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. ` |

*chevron_right*

*filter_none*

## 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 ` `?> ` |

*chevron_right*

*filter_none*

**Output:**

5

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 contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

## Recommended Posts:

- Maximum 0's between two immediate 1's in binary representation
- Binary representation of a given number
- Find value of k-th bit in binary representation
- Maximum distance between two 1's in Binary representation of N
- Largest number with binary representation is m 1's and m-1 0's
- 1 to n bit numbers with no consecutive 1s in binary representation
- 1 to n bit numbers with no consecutive 1s in binary representation.
- Binary representation of previous number
- Next greater number than N with exactly one bit different in binary representation of N
- Prime Number of Set Bits in Binary Representation | Set 1
- Occurrences of a pattern in binary representation of a number
- Decimal representation of given binary string is divisible by 20 or not
- Find the n-th number whose binary representation is a palindrome
- Decimal representation of given binary string is divisible by 10 or not
- Check if binary representation of a number is palindrome