**Parity:** Parity of a number refers to whether it contains an odd or even number of 1-bits. The number has “odd parity”, if it contains odd number of 1-bits and is “even parity” if it contains even number of 1-bits.

Main idea of the below solution is – Loop while n is not 0 and in loop unset one of the set bits and invert parity.

Algorithm: getParity(n) 1. Initialize parity = 0 2. Loop while n != 0 a. Invert parity parity = !parity b. Unset rightmost set bit n = n & (n-1) 3. return parityExample:Initialize: n = 13 (1101) parity = 0 n = 13 & 12 = 12 (1100) parity = 1 n = 12 & 11 = 8 (1000) parity = 0 n = 8 & 7 = 0 (0000) parity = 1

**Program:**

## C

# include <stdio.h> # define bool int /* Function to get parity of number n. It returns 1 if n has odd parity, and returns 0 if n has even parity */ bool getParity(unsigned int n) { bool parity = 0; while (n) { parity = !parity; n = n & (n - 1); } return parity; } /* Driver program to test getParity() */ int main() { unsigned int n = 7; printf("Parity of no %d = %s", n, (getParity(n)? "odd": "even")); getchar(); return 0; }

## Java

// Java program to find parity // of an integer import java.util.*; import java.lang.*; import java.io.*; import java.math.BigInteger; class GFG { /* Function to get parity of number n. It returns 1 if n has odd parity, and returns 0 if n has even parity */ static boolean getParity(int n) { boolean parity = false; while(n != 0) { parity = !parity; n = n & (n-1); } return parity; } /* Driver program to test getParity() */ public static void main (String[] args) { int n = 12; System.out.println("Parity of no " + n + " = " + (getParity(n)? "odd": "even")); } } /* This code is contributed by Amit khandelwal*/

## Python3

# Python3 code to get parity. # Function to get parity of number n. # It returns 1 if n has odd parity, # and returns 0 if n has even parity def getParity( n ): parity = 0 while n: parity = ~parity n = n & (n - 1) return parity # Driver program to test getParity() n = 7 print ("Parity of no ", n," = ", ( "odd" if getParity(n) else "even")) # This code is contributed by "Sharad_Bhardwaj".

Output:

Parity of no 7 = odd

Above solution can be optimized by using lookup table. Please refer to Bit Twiddle Hacks[1st reference] for details.

**Time Complexity: ** The time taken by above algorithm is proportional to the number of bits set. Worst case complexity is O(Log n).

**Uses: **Parity is used in error detection and cryptography.

**References:**

http://graphics.stanford.edu/~seander/bithacks.html#ParityNaive – last checked on 30 May 2009.