Count pairs with Bitwise XOR as EVEN number
Given an array of N integers, the task is to find the number of pairs (i, j) such that A[i] ^ A[j] is even.
Examples:
Input: A[] = { 5, 4, 7, 2, 1}
Output: 4
Since pair of A[] =
( 5, 4 ) = 1( 5, 7 ) = 2( 5, 2 ) = 7( 5, 1 ) = 4
( 4, 7 ) = 3( 4, 2 ) = 6( 4, 1 ) = 5
( 7, 2 ) = 5( 7, 1 ) = 6
( 2, 1 ) = 3
Total XOR even pair = 4
Input: A[] = { 7, 2, 8, 1, 0, 5, 11 }
Output: 9
Since pair of A[] =
( 7, 2 ) = 5( 7, 8 ) = 15( 7, 1 ) = 6( 7, 0 ) = 7( 7, 5 ) = 2( 7, 11 ) = 12
( 2, 8 ) = 10( 2, 1 ) = 3( 2, 0 ) = 2( 2, 5 ) = 7( 2, 11 ) = 9
( 8, 1 ) = 9( 8, 0 ) = 8( 8, 5 ) = 13( 8, 11 ) = 3
( 1, 0 ) = 1( 1, 5 ) = 4( 1, 11 ) = 10
( 0, 5 ) = 5( 0, 11 ) = 11
( 5, 11 ) = 14
A naive approach is to check for every pair and print the count of pairs that are even.
Below is the implementation of the above approach:
C++
#include <iostream>
using namespace std;
int findevenPair( int A[], int N)
{
int i, j;
int evenPair = 0;
for (i = 0; i < N; i++) {
for (j = i + 1; j < N; j++) {
if ((A[i] ^ A[j]) % 2 == 0)
evenPair++;
}
}
return evenPair;
}
int main()
{
int A[] = { 5, 4, 7, 2, 1 };
int N = sizeof (A) / sizeof (A[0]);
cout << findevenPair(A, N) << endl;
return 0;
}
|
C
#include <stdio.h>
int findevenPair( int A[], int N)
{
int i, j;
int evenPair = 0;
for (i = 0; i < N; i++) {
for (j = i + 1; j < N; j++) {
if ((A[i] ^ A[j]) % 2 == 0)
evenPair++;
}
}
return evenPair;
}
int main()
{
int A[] = { 5, 4, 7, 2, 1 };
int N = sizeof (A) / sizeof (A[0]);
printf ( "%d\n" ,findevenPair(A, N));
return 0;
}
|
Java
import java.io.*;
class GFG
{
static int findevenPair( int []A, int N)
{
int i, j;
int evenPair = 0 ;
for (i = 0 ; i < N; i++)
{
for (j = i + 1 ; j < N; j++)
{
if ((A[i] ^ A[j]) % 2 == 0 )
evenPair++;
}
}
return evenPair;
}
public static void main (String[] args)
{
int A[] = { 5 , 4 , 7 , 2 , 1 };
int N = A.length;
System.out.println(findevenPair(A, N));
}
}
|
Python3
def findevenPair(A, N):
evenPair = 0
for i in range ( 0 , N):
for j in range (i + 1 , N):
if ((A[i] ^ A[j]) % 2 = = 0 ):
evenPair + = 1
return evenPair;
def main():
A = [ 5 , 4 , 7 , 2 , 1 ]
N = len (A)
print (findevenPair(A, N))
if __name__ = = '__main__' :
main()
|
C#
using System;
class GFG
{
static int findevenPair( int []A, int N)
{
int i, j;
int evenPair = 0;
for (i = 0; i < N; i++)
{
for (j = i + 1; j < N; j++)
{
if ((A[i] ^ A[j]) % 2 == 0)
evenPair++;
}
}
return evenPair;
}
public static void Main ()
{
int []A = { 5, 4, 7, 2, 1 };
int N = A.Length;
Console.WriteLine(findevenPair(A, N));
}
}
|
PHP
<?php
function findevenPair(& $A , $N )
{
$evenPair = 0;
for ( $i = 0; $i < $N ; $i ++)
{
for ( $j = $i + 1; $j < $N ; $j ++)
{
if (( $A [ $i ] ^ $A [ $j ]) % 2 == 0)
$evenPair ++;
}
}
return $evenPair ;
}
$A = array (5, 4, 7, 2, 1 );
$N = sizeof( $A );
echo (findevenPair( $A , $N ));
?>
|
Javascript
<script>
function findevenPair(A, N)
{
let i, j;
let evenPair = 0;
for (i = 0; i < N; i++) {
for (j = i + 1; j < N; j++) {
if ((A[i] ^ A[j]) % 2 == 0)
evenPair++;
}
}
return evenPair;
}
let A = [ 5, 4, 7, 2, 1 ];
let N = A.length;
document.write(findevenPair(A, N));
</script>
|
Time Complexity: O(n^2)
Auxiliary Space: O(1)
An efficient solution is to Count pairs with Bitwise XOR as ODD number i.e. oddEvenpairs. Then return totalPairs – oddEvenPairs where totalPairs = (N * (N-1) / 2) and oddEvenPairs = count * (N – count).
As, pairs that will give Even Bitwise XOR are :
Even, Even
Odd, Odd
So, find the count of pairs with both odd and even elements and subtract from total no. of pairs.
Below is the implementation of the above approach:
C++
#include <iostream>
using namespace std;
int findEvenPair( int A[], int N)
{
int count = 0;
for ( int i = 0; i < N; i++) {
if (A[i] % 2 != 0)
count++;
}
int totalPairs = (N * (N - 1) / 2);
int oddEvenPairs = count * (N - count);
return totalPairs - oddEvenPairs;
}
int main()
{
int a[] = { 5, 4, 7, 2, 1 };
int n = sizeof (a) / sizeof (a[0]);
cout << findEvenPair(a, n) << endl;
return 0;
}
|
C
#include <stdio.h>
int findEvenPair( int A[], int N)
{
int count = 0;
for ( int i = 0; i < N; i++) {
if (A[i] % 2 != 0)
count++;
}
int totalPairs = (N * (N - 1) / 2);
int oddEvenPairs = count * (N - count);
return totalPairs - oddEvenPairs;
}
int main()
{
int a[] = { 5, 4, 7, 2, 1 };
int n = sizeof (a) / sizeof (a[0]);
printf ( "%d\n" ,findEvenPair(a, n));
return 0;
}
|
Java
import java.io.*;
class GFG {
static int findEvenPair( int A[], int N)
{
int count = 0 ;
for ( int i = 0 ; i < N; i++) {
if (A[i] % 2 != 0 )
count++;
}
int totalPairs = (N * (N - 1 ) / 2 );
int oddEvenPairs = count * (N - count);
return totalPairs - oddEvenPairs;
}
public static void main (String[] args) {
int a[] = { 5 , 4 , 7 , 2 , 1 };
int n = a.length;
System.out.println(findEvenPair(a, n));
}
}
|
Python3
def findEvenPair(A, N):
count = 0
for i in range ( 0 ,N):
if (A[i] % 2 ! = 0 ):
count + = 1
totalPairs = (N * (N - 1 ) / 2 )
oddEvenPairs = count * (N - count)
return ( int )(totalPairs - oddEvenPairs)
def main():
a = [ 5 , 4 , 7 , 2 , 1 ]
n = len (a)
print (findEvenPair(a, n))
if __name__ = = '__main__' :
main()
|
C#
using System;
public class GFG {
static int findEvenPair( int []A, int N)
{
int count = 0;
for ( int i = 0; i < N; i++) {
if (A[i] % 2 != 0)
count++;
}
int totalPairs = (N * (N - 1) / 2);
int oddEvenPairs = count * (N - count);
return totalPairs - oddEvenPairs;
}
public static void Main() {
int []a = { 5, 4, 7, 2, 1 };
int n = a.Length;
Console.Write(findEvenPair(a, n));
}
}
|
PHP
<?php
function findEvenPair( $A , $N )
{
$count = 0;
for ( $i = 0; $i < $N ; $i ++)
{
if ( $A [ $i ] % 2 != 0)
$count ++;
}
$totalPairs = ( $N * ( $N - 1) / 2);
$oddEvenPairs = $count * ( $N - $count );
return $totalPairs - $oddEvenPairs ;
}
$a = array (5, 4, 7, 2, 1);
$n = sizeof( $a );
echo findEvenPair( $a , $n ) . "\n" ;
?>
|
Javascript
<script>
function findEvenPair(A, N)
{
let count = 0;
for (let i = 0; i < N; i++) {
if (A[i] % 2 != 0)
count++;
}
let totalPairs = parseInt(N * (N - 1) / 2);
let oddEvenPairs = count * (N - count);
return totalPairs - oddEvenPairs;
}
let a = [ 5, 4, 7, 2, 1 ];
let n = a.length;
document.write(findEvenPair(a, n));
</script>
|
Time Complexity: O(n)
Auxiliary Space: O(1)
Last Updated :
19 Sep, 2022
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