Given an array A[] of size N. The task is to find the how many pair(i, j) exists such that A[i] OR A[j] is even.
Examples:
Input : N = 4
A[] = { 5, 6, 2, 8 }
Output :3
Explanation :
Since pair of A[] = ( 5, 6 ), ( 5, 2 ), ( 5, 8 ),
( 6, 2 ), ( 6, 8 ), ( 2, 8 )
5 OR 6 = 7, 5 OR 2 = 7, 5 OR 8 = 13
6 OR 2 = 6, 6 OR 8 = 14, 2 OR 8 = 10
Total pair A( i, j ) = 6 and Even = 3
Input : N = 7
A[] = {8, 6, 2, 7, 3, 4, 9}
Output :6
Implementation: A simple solution is to check every pair.
C++
#include <iostream>
using namespace std;
int findEvenPair( int A[], int N)
{
int evenPair = 0;
for ( int i = 0; i < N; i++) {
for ( int j = i + 1; j < N; j++) {
if ((A[i] | A[j]) % 2 == 0)
evenPair++;
}
}
return evenPair;
}
int main()
{
int A[] = { 5, 6, 2, 8 };
int N = sizeof (A) / sizeof (A[0]);
cout << findEvenPair(A, N) << endl;
return 0;
}
|
Java
import java.io.*;
class GFG
{
static int findEvenPair( int A[],
int N)
{
int evenPair = 0 ;
for ( int i = 0 ; i < N; i++)
{
for ( int 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 , 6 , 2 , 8 };
int N = A.length;
System.out.println(findEvenPair(A, N));
}
}
|
Python 3
def findEvenPair(A, N) :
evenPair = 0
for i in range (N) :
for j in range (i + 1 , N):
if (A[i] | A[j]) % 2 = = 0 :
evenPair + = 1
return evenPair
if __name__ = = "__main__" :
A = [ 5 , 6 , 2 , 8 ]
N = len (A)
print (findEvenPair(A, N))
|
C#
using System;
class GFG
{
static int findEvenPair( int []A,
int N)
{
int evenPair = 0;
for ( int i = 0; i < N; i++)
{
for ( int j = i + 1; j < N; j++)
{
if ((A[i] | A[j]) % 2 == 0)
evenPair++;
}
}
return evenPair;
}
public static void Main ()
{
int []A = { 5, 6, 2, 8 };
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, 6, 2, 8);
$N = count ( $A );
echo findEvenPair( $A , $N );
?>
|
Javascript
<script>
function findEvenPair(A, N)
{
let evenPair = 0;
for (let i = 0; i < N; i++) {
for (let j = i + 1; j < N; j++) {
if ((A[i] | A[j]) % 2 == 0)
evenPair++;
}
}
return evenPair;
}
let A = [5, 6, 2, 8 ];
let N = A.length;
document.write(findEvenPair(A, N));
</script>
|
Complexity Analysis:
Time Complexity: O(N^2), since two nested loops are used.
Auxiliary Space: O(1), as constant extra space used by the algorithm.
Implementation: A efficient solution is to count numbers with last bit as 0. Then return count * (count – 1)/2. Note that OR of two numbers can be even only if their last bits are 0.
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] & 1))
count++;
return count * (count - 1) / 2;
}
int main()
{
int A[] = { 5, 6, 2, 8 };
int N = sizeof (A) / sizeof (A[0]);
cout << findEvenPair(A, N) << endl;
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] & 1 ) > 0 )))
count++;
return count * (count - 1 ) / 2 ;
}
public static void main (String[] args)
{
int A[] = { 5 , 6 , 2 , 8 };
int N = A.length;
System.out.println(findEvenPair(A, N));
}
}
|
Python 3
def findEvenPair(A, N):
count = 0
for i in range (N):
if ( not (A[i] & 1 )):
count + = 1
return count * (count - 1 ) / / 2
if __name__ = = "__main__" :
A = [ 5 , 6 , 2 , 8 ]
N = len (A)
print (findEvenPair(A, N))
|
C#
using System;
class GFG
{
static int findEvenPair( int []A, int N)
{
int count = 0;
for ( int i = 0; i < N; i++)
if ((!((A[i] & 1) > 0)))
count++;
return count * (count - 1) / 2;
}
public static void Main (String[] args)
{
int []A = { 5, 6, 2, 8 };
int N = A.Length;
Console.WriteLine(findEvenPair(A, N));
}
}
|
PHP
<?php
function findEvenPair(& $A , $N )
{
$count = 0;
for ( $i = 0; $i < $N ; $i ++)
if (!( $A [ $i ] & 1))
$count ++;
return $count * ( $count - 1) / 2;
}
$A = array (5, 6, 2, 8 );
$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] & 1) > 0)))
count++;
return count * (count - 1) / 2;
}
let A=[5, 6, 2, 8 ];
let N = A.length;
document.write(findEvenPair(A, N));
</script>
|
Complexity Analysis:
Time Complexity: O(N), where N is the size of the given array.
Auxiliary Space: O(1), as constant extra space used by the algorithm.