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 an odd number of 1-bits and is “even parity” if it contains an even number of 1-bits.
The 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 parity
Example:
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<bits/stdc++.h>
# define bool int
using namespace std;
bool getParity(unsigned int n)
{
bool parity = 0;
while (n)
{
parity = !parity;
n = n & (n - 1);
}
return parity;
}
int main()
{
unsigned int n = 7;
cout<< "Parity of no " <<n<< " = " <<(getParity(n)? "odd" : "even" );
getchar ();
return 0;
}
|
C
# include <stdio.h>
# define bool int
bool getParity(unsigned int n)
{
bool parity = 0;
while (n)
{
parity = !parity;
n = n & (n - 1);
}
return parity;
}
int main()
{
unsigned int n = 7;
printf ( "Parity of no %d = %s" , n,
(getParity(n)? "odd" : "even" ));
getchar ();
return 0;
}
|
Java
import java.util.*;
import java.lang.*;
import java.io.*;
import java.math.BigInteger;
class GFG
{
static boolean getParity( int n)
{
boolean parity = false ;
while (n != 0 )
{
parity = !parity;
n = n & (n- 1 );
}
return parity;
}
public static void main (String[] args)
{
int n = 7 ;
System.out.println( "Parity of no " + n + " = " +
(getParity(n)? "odd" : "even" ));
}
}
|
Python3
def getParity( n ):
parity = 0
while n:
parity = ~parity
n = n & (n - 1 )
return parity
n = 7
print ( "Parity of no " , n, " = " ,
( "odd" if getParity(n) else "even" ))
|
C#
using System;
class GFG {
static bool getParity( int n)
{
bool parity = false ;
while (n != 0)
{
parity = !parity;
n = n & (n-1);
}
return parity;
}
public static void Main ()
{
int n = 7;
Console.Write( "Parity of no " + n
+ " = " + (getParity(n)?
"odd" : "even" ));
}
}
|
PHP
<?php
function getParity( $n )
{
$parity = 0;
while ( $n )
{
$parity = ! $parity ;
$n = $n & ( $n - 1);
}
return $parity ;
}
$n = 7;
echo "Parity of no " , $n , " = " ,
getParity( $n )? "odd" : "even" ;
?>
|
Javascript
<script>
function getParity(n)
{
var parity = false ;
while (n != 0)
{
parity = !parity;
n = n & (n - 1);
}
return parity;
}
var n = 7;
document.write( "Parity of no " + n + " = " +
(getParity(n) ? "odd" : "even" ));
</script>
|
OutputParity 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).
Auxiliary Space: O(1)
Another approach: (Using built-in-function)
C++
# include<bits/stdc++.h>
# define bool int
using namespace std;
bool getParity(unsigned int n)
{
return __builtin_parity(n);
}
int main()
{
unsigned int n = 7;
cout<< "Parity of no " <<n<< " = " <<(getParity(n)? "odd" : "even" );
getchar ();
return 0;
}
|
Java
import java.util.*;
class Main {
public static boolean getParity( int n) {
return Integer.bitCount(n) % 2 == 1 ;
}
public static void main(String[] args) {
int n = 7 ;
System.out.println( "Parity of no " + n + " = " + (getParity(n) ? "odd" : "even" ));
}
}
|
Python3
def getParity(n):
return ( bin (n).count( "1" )) % 2
n = 7
print ( "Parity of no {0} = " . format (n),end = "")
print ( "odd" if getParity(n) else "even" )
|
C#
using System;
using System.Linq;
class GFG
{
public static bool GetParity( int n)
{
return Convert.ToInt32(Convert.ToString(n, 2).Count(x => x == '1' )) % 2 == 1;
}
public static void Main()
{
int n = 7;
Console.WriteLine( "Parity of no " + n + " = " + (GetParity(n) ? "odd" : "even" ));
}
}
|
Javascript
const getParity = (n) => {
return (n.toString(2).split( "1" ).length - 1) % 2;
};
const n = 7;
console.log(`Parity of no ${n} =`, getParity(n) ? "odd" : "even" );
|
OutputParity of no 7 = odd
Time Complexity: O(1)
Auxiliary Space: O(1)
Another Approach: Mapping numbers with the bit
We can use a map or an array of the number of bits to form a nibble (a nibble consists of 4 bits, so a 16 – length array would be required). Then, we can get the nibbles of a given number.
This approach can be summarized into the following steps:
1. Build the 16 length array of the number of bits to form a nibble – { 0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4 }
2. Recursively count the set of the bits by taking the last nibble (4 bits) from the array using the formula num & 0xf and then getting each successive nibble by discarding the last 4 bits using >> operator.
3. Check the parity: if the number of set bits is even, ie numOfSetBits % 2 == 0, then the number is of even parity. Else, it is of odd parity.
C++
#include <bits/stdc++.h>
using namespace std;
int nibble_to_bits[16]
= { 0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4 };
unsigned int countSetBits(unsigned int num)
{
int nibble = 0;
if (0 == num)
return nibble_to_bits[0];
nibble = num & 0xf;
return nibble_to_bits[nibble] + countSetBits(num >> 4);
}
bool getParity( int num) { return countSetBits(num) % 2; }
int main()
{
unsigned int n = 7;
cout << "Parity of no " << n << " = "
<< (getParity(n) ? "odd" : "even" );
return 0;
}
|
Java
import java.util.*;
class GFG{
static int [] nibble_to_bits = {
0 , 1 , 1 , 2 , 1 , 2 , 2 , 3 , 1 , 2 , 2 , 3 , 2 , 3 , 3 , 4
};
static int countSetBits( int num)
{
int nibble = 0 ;
if ( 0 == num)
return nibble_to_bits[ 0 ];
nibble = num & 0xf ;
return nibble_to_bits[nibble]
+ countSetBits(num >> 4 );
}
static boolean getParity( int num)
{
return countSetBits(num) % 2 == 1 ;
}
public static void main(String[] args)
{
int n = 7 ;
System.out.print(
"Parity of no " + n + " = "
+ (getParity(n) ? "odd" : "even" ));
}
}
|
Python3
nibble_to_bits = [ 0 , 1 , 1 , 2 , 1 , 2 , 2 , 3 , 1 , 2 , 2 , 3 , 2 , 3 , 3 , 4 ]
def countSetBits(num):
nibble = 0
if ( 0 = = num):
return nibble_to_bits[ 0 ]
nibble = num & 0xf
return nibble_to_bits[nibble] + countSetBits(num >> 4 )
def getParity(num):
return countSetBits(num) % 2
n = 7
print ( "Parity of no" , n, " = " , [ "even" , "odd" ][getParity(n)])
|
C#
using System;
class GFG {
static int [] nibble_to_bits = {
0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4
};
static int countSetBits( int num)
{
int nibble = 0;
if (0 == num)
return nibble_to_bits[0];
nibble = num & 0xf;
return nibble_to_bits[nibble]
+ countSetBits(num >> 4);
}
static bool getParity( int num)
{
return countSetBits(num) % 2 == 1;
}
public static void Main( string [] args)
{
int n = 7;
Console.WriteLine(
"Parity of no " + n + " = "
+ (getParity(n) ? "odd" : "even" ));
}
}
|
Javascript
let nibble_to_bits
= [ 0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4 ];
function countSetBits(num)
{
let nibble = 0;
if (0 == num)
return nibble_to_bits[0];
nibble = num & 0xf;
return nibble_to_bits[nibble] + countSetBits(num >> 4);
}
function getParity(num) { return countSetBits(num) % 2; }
let n = 7;
console.log( "Parity of no " + n + " = " + (getParity(n) ? "odd" : "even" ));
|
OutputParity of no 7 = odd
Time Complexity: O(1)
Auxiliary Space: O(1)
Uses: Parity is used in error detection and cryptography.
Compute the parity of a number using XOR and table look-up
References:
http://graphics.stanford.edu/~seander/bithacks.html#ParityNaive – last checked on 30 May 2009.