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++ program to count pairs // with XOR giving a even number #include <iostream> using namespace std;
// Function to count number of even pairs int findevenPair( int A[], int N)
{ int i, j;
// variable for counting even pairs
int evenPair = 0;
// find all pairs
for (i = 0; i < N; i++) {
for (j = i + 1; j < N; j++) {
// find XOR operation
// check even or even
if ((A[i] ^ A[j]) % 2 == 0)
evenPair++;
}
}
// return number of even pair
return evenPair;
} // Driver Code int main()
{ int A[] = { 5, 4, 7, 2, 1 };
int N = sizeof (A) / sizeof (A[0]);
// calling function findevenPair
// and print number of even pair
cout << findevenPair(A, N) << endl;
return 0;
} |
// C program to count pairs // with XOR giving a even number #include <stdio.h> // Function to count number of even pairs int findevenPair( int A[], int N)
{ int i, j;
// variable for counting even pairs
int evenPair = 0;
// find all pairs
for (i = 0; i < N; i++) {
for (j = i + 1; j < N; j++) {
// find XOR operation
// check even or even
if ((A[i] ^ A[j]) % 2 == 0)
evenPair++;
}
}
// return number of even pair
return evenPair;
} // Driver Code int main()
{ int A[] = { 5, 4, 7, 2, 1 };
int N = sizeof (A) / sizeof (A[0]);
// calling function findevenPair
// and print number of even pair
printf ( "%d\n" ,findevenPair(A, N));
return 0;
} // This code is contributed by kothvvsaakash. |
// Java program to count pairs // with XOR giving a even number import java.io.*;
class GFG
{ // Function to count number of even pairs static int findevenPair( int []A, int N)
{ int i, j;
// variable for counting even pairs
int evenPair = 0 ;
// find all pairs
for (i = 0 ; i < N; i++)
{
for (j = i + 1 ; j < N; j++)
{
// find XOR operation
// check even or even
if ((A[i] ^ A[j]) % 2 == 0 )
evenPair++;
}
}
// return number of even pair
return evenPair;
} // Driver Code public static void main (String[] args)
{ int A[] = { 5 , 4 , 7 , 2 , 1 };
int N = A.length;
// calling function findevenPair
// and print number of even pair
System.out.println(findevenPair(A, N));
} } // This code is contributed by inder_verma.. |
# Python3 program to count pairs # with XOR giving a even number # Function to count number of even pairs def findevenPair(A, N):
# variable for counting even pairs
evenPair = 0
# find all pairs
for i in range ( 0 , N):
for j in range (i + 1 , N):
# find XOR operation
# check even or even
if ((A[i] ^ A[j]) % 2 = = 0 ):
evenPair + = 1
# return number of even pair
return evenPair;
# Driver Code def main():
A = [ 5 , 4 , 7 , 2 , 1 ]
N = len (A)
# calling function findevenPair
# and print number of even pair
print (findevenPair(A, N))
if __name__ = = '__main__' :
main()
# This code is contributed by PrinciRaj1992 |
// C# program to count pairs // with XOR giving a even number using System;
class GFG
{ // Function to count number of // even pairs static int findevenPair( int []A, int N)
{ int i, j;
// variable for counting even pairs
int evenPair = 0;
// find all pairs
for (i = 0; i < N; i++)
{
for (j = i + 1; j < N; j++)
{
// find XOR operation
// check even or even
if ((A[i] ^ A[j]) % 2 == 0)
evenPair++;
}
}
// return number of even pair
return evenPair;
} // Driver Code public static void Main ()
{ int []A = { 5, 4, 7, 2, 1 };
int N = A.Length;
// calling function findevenPair
// and print number of even pair
Console.WriteLine(findevenPair(A, N));
} } // This code is contributed // by inder_verma.. |
<?php // PHP program to count pairs // with XOR giving a even number // Function to count number // of even pairs function findevenPair(& $A , $N )
{ // variable for counting even pairs
$evenPair = 0;
// find all pairs
for ( $i = 0; $i < $N ; $i ++)
{
for ( $j = $i + 1; $j < $N ; $j ++)
{
// find XOR operation
// check even or even
if (( $A [ $i ] ^ $A [ $j ]) % 2 == 0)
$evenPair ++;
}
}
// return number of even pair
return $evenPair ;
} // Driver Code $A = array (5, 4, 7, 2, 1 );
$N = sizeof( $A );
// calling function findevenPair // and print number of even pair echo (findevenPair( $A , $N ));
// This code is contributed // by Shivi_Aggarwal ?> |
<script> // Javascript program to count pairs // with XOR giving a even number // Function to count number of even pairs function findevenPair(A, N)
{ let i, j;
// variable for counting even pairs
let evenPair = 0;
// find all pairs
for (i = 0; i < N; i++) {
for (j = i + 1; j < N; j++) {
// find XOR operation
// check even or even
if ((A[i] ^ A[j]) % 2 == 0)
evenPair++;
}
}
// return number of even pair
return evenPair;
} // Driver Code let A = [ 5, 4, 7, 2, 1 ]; let N = A.length; // calling function findevenPair // and print number of even pair document.write(findevenPair(A, N)); // This code is contributed by souravmahato348. </script> |
4
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++ program to count pairs // with XOR giving a even number #include <iostream> using namespace std;
// Function to count number of even pairs int findEvenPair( int A[], int N)
{ int count = 0;
// find all pairs
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 number of even pair
return totalPairs - oddEvenPairs;
} // Driver Code int main()
{ int a[] = { 5, 4, 7, 2, 1 };
int n = sizeof (a) / sizeof (a[0]);
// calling function findEvenPair
// and print number of even pair
cout << findEvenPair(a, n) << endl;
return 0;
} |
// C program to count pairs // with XOR giving a even number #include <stdio.h> // Function to count number of even pairs int findEvenPair( int A[], int N)
{ int count = 0;
// find all pairs
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 number of even pair
return totalPairs - oddEvenPairs;
} // Driver Code int main()
{ int a[] = { 5, 4, 7, 2, 1 };
int n = sizeof (a) / sizeof (a[0]);
// calling function findEvenPair
// and print number of even pair
printf ( "%d\n" ,findEvenPair(a, n));
return 0;
} // This code is contributed by kothvvsaakash. |
// Java program to count pairs // with XOR giving a even number import java.io.*;
class GFG {
// Function to count number of even pairs
static int findEvenPair( int A[], int N)
{ int count = 0 ;
// find all pairs
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 number of even pair
return totalPairs - oddEvenPairs;
} // Driver Code public static void main (String[] args) {
int a[] = { 5 , 4 , 7 , 2 , 1 };
int n = a.length;
// calling function findEvenPair
// and print number of even pair
System.out.println(findEvenPair(a, n));
}
//This code is contributed by akt_mit } |
# python program to count pairs # with XOR giving a even number # Function to count number of even pairs def findEvenPair(A, N):
count = 0
# find all pairs
for i in range ( 0 ,N):
if (A[i] % 2 ! = 0 ):
count + = 1
totalPairs = (N * (N - 1 ) / 2 )
oddEvenPairs = count * (N - count)
# return number of even pair
return ( int )(totalPairs - oddEvenPairs)
# Driver Code def main():
a = [ 5 , 4 , 7 , 2 , 1 ]
n = len (a)
# calling function findEvenPair
# and print number of even pair
print (findEvenPair(a, n))
if __name__ = = '__main__' :
main()
# This code is contributed by 29AjayKumar |
// C# program to count pairs // with XOR giving a even number using System;
public class GFG {
// Function to count number of even pairs
static int findEvenPair( int []A, int N)
{
int count = 0;
// find all pairs
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 number of even pair
return totalPairs - oddEvenPairs;
}
// Driver Code
public static void Main() {
int []a = { 5, 4, 7, 2, 1 };
int n = a.Length;
// calling function findEvenPair
// and print number of even pair
Console.Write(findEvenPair(a, n));
}
} // This code is contributed by 29AjayKumar |
<?php // PHP program to count pairs // with XOR giving a even number // Function to count number of even pairs function findEvenPair( $A , $N )
{ $count = 0;
// find all pairs
for ( $i = 0; $i < $N ; $i ++)
{
if ( $A [ $i ] % 2 != 0)
$count ++;
}
$totalPairs = ( $N * ( $N - 1) / 2);
$oddEvenPairs = $count * ( $N - $count );
// return number of even pair
return $totalPairs - $oddEvenPairs ;
} // Driver Code $a = array (5, 4, 7, 2, 1);
$n = sizeof( $a );
// calling function findEvenPair // and print number of even pair echo findEvenPair( $a , $n ) . "\n" ;
// This code is contributed // by Akanksha Rai ?> |
<script> // Javascript program to count pairs // with XOR giving a even number // Function to count number of even pairs function findEvenPair(A, N)
{ let count = 0;
// find all pairs
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 number of even pair
return totalPairs - oddEvenPairs;
} // Driver Code let a = [ 5, 4, 7, 2, 1 ];
let n = a.length;
// calling function findEvenPair
// and print number of even pair
document.write(findEvenPair(a, n));
</script> |
4
Time Complexity: O(n)
Auxiliary Space: O(1)