Given an array of
Examples:
Input: N = 4, A[] = { 5, 1, 3, 2 } Output: 3 Since pair of A[] are: ( 5, 1 ), ( 5, 3 ), ( 5, 2 ), ( 1, 3 ), ( 1, 2 ), ( 3, 2 ) 5 AND 1 = 1, 5 AND 3 = 1, 5 AND 2 = 0, 1 AND 3 = 1, 1 AND 2 = 0, 3 AND 2 = 2 Total even pair A( i, j ) = 3 Input : N = 6, A[] = { 5, 9, 0, 6, 7, 3 } Output : 9 Since pair of A[] = ( 5, 9 ) = 1, ( 5, 0 ) = 0, ( 5, 6 ) = 4, ( 5, 7 ) = 5, ( 5, 3 ) = 1, ( 9, 0 ) = 0, ( 9, 6 ) = 0, ( 9, 7 ) = 1, ( 9, 3 ) = 1, ( 0, 6 ) = 0, ( 0, 7 ) = 0, ( 0, 3 ) = 0, ( 6, 7 ) = 6, ( 6, 3 ) = 2, ( 7, 3 ) = 3
A Naive Approach is to check for every pair and print the count of pairs which are even.
Below is the implementation of the above approach:
// C++ program to count pair with // bitwise-AND as even number #include <iostream> using namespace std;
// Function to count number of pairs EVEN bitwise AND 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 AND operation
// to check evenpair
if ((A[i] & A[j]) % 2 == 0)
evenPair++;
}
}
// return number of even pair
return evenPair;
} // Driver Code int main()
{ int a[] = { 5, 1, 3, 2 };
int n = sizeof (a) / sizeof (a[0]);
cout << findevenPair(a, n) << endl;
return 0;
} |
// C program to count pair with // bitwise-AND as even number #include <stdio.h> // Function to count number of pairs EVEN bitwise AND 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 AND operation
// to check evenpair
if ((A[i] & A[j]) % 2 == 0)
evenPair++;
}
}
// return number of even pair
return evenPair;
} // Driver Code int main()
{ int a[] = { 5, 1, 3, 2 };
int n = sizeof (a) / sizeof (a[0]);
printf ( "%d\n" ,findevenPair(a, n));
return 0;
} // This code is contributed by kothavvsaakash. |
// Java program to count pair with // bitwise-AND as even number import java.io.*;
class GFG
{ // Function to count number of // pairs EVEN bitwise AND 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 AND operation
// to check evenpair
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 , 1 , 3 , 2 };
int n = a.length;
System.out.println(findevenPair(a, n));
} } // This code is contributed by anuj_67.. |
# Python 3 program to count pair with # bitwise-AND as even number # Function to count number of # pairs EVEN bitwise AND 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 AND operation to
# check evenpair
if ((A[i] & A[j]) % 2 = = 0 ):
evenPair + = 1
# return number of even pair
return evenPair
# Driver Code a = [ 5 , 1 , 3 , 2 ]
n = len (a)
print (findevenPair(a, n))
# This code is contributed # by PrinciRaj1992 |
// C# program to count pair with // bitwise-AND as even number using System;
class GFG
{ // Function to count number of // pairs EVEN bitwise AND 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 AND operation
// to check evenpair
if ((A[i] & A[j]) % 2 == 0)
evenPair++;
}
}
// return number of even pair
return evenPair;
} // Driver Code public static void Main ()
{ int []a = { 5, 1, 3, 2 };
int n = a.Length;
Console.WriteLine(findevenPair(a, n));
} } // This code is contributed by anuj_67.. |
<?php // PHP program to count pair with // bitwise-AND as even number // Function to count number of // pairs EVEN bitwise AND 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 AND operation
// to check evenpair
if (( $A [ $i ] & $A [ $j ]) % 2 == 0)
$evenPair ++;
}
}
// return number of even pair
return $evenPair ;
} // Driver Code $a = array (5, 1, 3, 2 );
$n = sizeof( $a );
echo findevenPair( $a , $n );
// This code is contributed by akt_mit ?> |
<script> // Javascript program to count pair with // bitwise-AND as even number // Function to count number of pairs EVEN bitwise AND 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 AND operation
// to check evenpair
if ((A[i] & A[j]) % 2 == 0)
evenPair++;
}
}
// return number of even pair
return evenPair;
} // Driver Code let a = [ 5, 1, 3, 2 ]; let n = a.length; document.write(findevenPair(a, n)); // This code is contributed by souravmahato348. </script> |
Output
3
Time Complexity: O(N2)
Auxiliary Space: O(1)
An Efficient Approach will be to observe that the bitwise AND of two numbers will be even if atleast one of the two numbers is even. So, count all the odd numbers in the array say count. Now, number of pairs with both odd elements will be count*(count-1)/2.
Therefore, number of elements with atleast one even element will be:
Total Pairs - Count of pair with both odd elements
Below is the implementation of the above approach:
// C++ program to count pair with // bitwise-AND as even number #include <iostream> using namespace std;
// Function to count number of pairs // with EVEN bitwise AND int findevenPair( int A[], int N)
{ int count = 0;
// count odd numbers
for ( int i = 0; i < N; i++)
if (A[i] % 2 != 0)
count++;
// count odd pairs
int oddCount = count * (count - 1) / 2;
// return number of even pair
return (N * (N - 1) / 2) - oddCount;
} // Driver Code int main()
{ int a[] = { 5, 1, 3, 2 };
int n = sizeof (a) / sizeof (a[0]);
cout << findevenPair(a, n) << endl;
return 0;
} |
// C program to count pair with // bitwise-AND as even number #include <stdio.h> // Function to count number of pairs // with EVEN bitwise AND int findevenPair( int A[], int N)
{ int count = 0;
// count odd numbers
for ( int i = 0; i < N; i++)
if (A[i] % 2 != 0)
count++;
// count odd pairs
int oddCount = count * (count - 1) / 2;
// return number of even pair
return (N * (N - 1) / 2) - oddCount;
} // Driver Code int main()
{ int a[] = { 5, 1, 3, 2 };
int n = sizeof (a) / sizeof (a[0]);
printf ( "%d\n" ,findevenPair(a, n));
return 0;
} // This code is contributed by kothavvsaakash. |
// Java program to count pair with // bitwise-AND as even number import java.io.*;
class GFG {
// Function to count number of pairs // with EVEN bitwise AND static int findevenPair( int A[], int N)
{ int count = 0 ;
// count odd numbers
for ( int i = 0 ; i < N; i++)
if (A[i] % 2 != 0 )
count++;
// count odd pairs
int oddCount = count * (count - 1 ) / 2 ;
// return number of even pair
return (N * (N - 1 ) / 2 ) - oddCount;
} // Driver Code public static void main (String[] args) {
int a[] = { 5 , 1 , 3 , 2 };
int n =a.length;
System.out.print( findevenPair(a, n));
}
} // This code is contributed by anuj_67.. |
# Python 3 program to count pair with # bitwise-AND as even number # Function to count number of pairs # with EVEN bitwise AND def findevenPair(A, N):
count = 0
# count odd numbers
for i in range ( 0 , N):
if (A[i] % 2 ! = 0 ):
count + = 1
# count odd pairs
oddCount = count * (count - 1 ) / 2
# return number of even pair
return ( int )((N * (N - 1 ) / 2 ) - oddCount)
# Driver Code a = [ 5 , 1 , 3 , 2 ]
n = len (a)
print (findevenPair(a, n))
# This code is contributed # by PrinciRaj1992 |
// C# program to count pair with // bitwise-AND as even number using System;
public class GFG{
// Function to count number of pairs // with EVEN bitwise AND static int findevenPair( int []A, int N)
{ int count = 0;
// count odd numbers
for ( int i = 0; i < N; i++)
if (A[i] % 2 != 0)
count++;
// count odd pairs
int oddCount = count * (count - 1) / 2;
// return number of even pair
return (N * (N - 1) / 2) - oddCount;
} // Driver Code static public void Main (){
int []a = { 5, 1, 3, 2 };
int n =a.Length;
Console.WriteLine( findevenPair(a, n));
}
} // This code is contributed by ajit |
<?php // PHP program to count pair with // bitwise-AND as even number // Function to count number of pairs // with EVEN bitwise AND function findevenPair( $A , $N )
{ $count = 0;
// count odd numbers
for ( $i = 0; $i < $N ; $i ++)
if ( $A [ $i ] % 2 != 0)
$count ++;
// count odd pairs
$oddCount = $count * ( $count - 1) / 2;
// return number of even pair
return ( $N * ( $N - 1) / 2) - $oddCount ;
} // Driver Code $a = array (5, 1, 3, 2);
$n = sizeof( $a );
echo findevenPair( $a , $n ) . "\n" ;
// This code is contributed // by Akanksha Rai ?> |
<script> // Javascript program to count pair with // bitwise-AND as even number // Function to count number of pairs // with EVEN bitwise AND function findevenPair(A, N)
{ let count = 0;
// Count odd numbers
for (let i = 0; i < N; i++)
if (A[i] % 2 != 0)
count++;
// Count odd pairs
let oddCount = parseInt((count *
(count - 1)) / 2);
// Return number of even pair
return parseInt((N * (N - 1)) / 2) - oddCount;
} // Driver Code let a = [ 5, 1, 3, 2 ]; let n = a.length; document.write(findevenPair(a, n)); // This code is contributed by subhammahato348 </script> |
Output
3
Time Complexity: O(N)
Auxiliary Space: O(1)
Another Method: we observe that bitwise AND of any two numbers will be even if any one of them will be even, no matter what is another number, if one number is even bitwise ANd of that number with any number will be even.
so if our array has an even number count as evenNumCount then the total no. of pair whose bitwise AND is even will be
(n-1)*evenNumCount - (evenNumCount*(evenNumCount-1))/2
here second term (evenNumCount*(evenNumCount-1))/2 will be subtracted from actual ans as we counted (even, even) pair twice.
// C++ program to count pair with bitwise-AND as even number #include <iostream> using namespace std;
// Function to count number of pairs with EVEN bitwise AND int findevenPair( int A[], int N)
{ int evenNumCount = 0;
// count odd numbers
for ( int i = 0; i < N; i++)
if (A[i] % 2 == 0)
evenNumCount++;
// return number of even pair
return (N - 1) * evenNumCount - ((evenNumCount * (evenNumCount - 1)) / 2);
} // Driver Code int main()
{ int a[] = { 5, 1, 3, 2 };
int n = sizeof (a) / sizeof (a[0]);
cout << findevenPair(a, n) << endl;
return 0;
} |
// Java program to count pair with bitwise-AND as even // number import java.util.*;
public class Main {
// Function to count number of pairs with EVEN bitwise
// AND
static int findevenPair( int A[], int N)
{
int evenNumCount = 0 ;
// count odd numbers
for ( int i = 0 ; i < N; i++)
if (A[i] % 2 == 0 )
evenNumCount++;
// return number of even pair
return (N - 1 ) * evenNumCount
- ((evenNumCount * (evenNumCount - 1 )) / 2 );
}
// Driver Code
public static void main(String[] args)
{
int a[] = { 5 , 1 , 3 , 2 };
int n = a.length;
System.out.println(findevenPair(a, n));
}
} |
using System;
public class MainClass {
// Function to count number of pairs with EVEN bitwise
// AND
public static int FindevenPair( int [] A, int N)
{
int evenNumCount = 0;
// count odd numbers
for ( int i = 0; i < N; i++)
if (A[i] % 2 == 0)
evenNumCount++;
// return number of even pair
return (N - 1) * evenNumCount
- ((evenNumCount * (evenNumCount - 1)) / 2);
}
// Driver Code
public static void Main()
{
int [] a = { 5, 1, 3, 2 };
int n = a.Length;
Console.WriteLine(FindevenPair(a, n));
}
} |
// JavaScript program to count pair with bitwise-AND as even number function findevenPair(A, N) {
let evenNumCount = 0;
// count odd numbers
for (let i = 0; i < N; i++) {
if (A[i] % 2 === 0) {
evenNumCount++;
}
}
// return number of even pair
return (N - 1) * evenNumCount - ((evenNumCount * (evenNumCount - 1)) / 2);
} // Driver Code const a = [5, 1, 3, 2]; const n = a.length; console.log(findevenPair(a, n)); |
# Function to count number of pairs with EVEN bitwise AND def findevenPair(A, N):
evenNumCount = 0
# count odd numbers
for i in range (N):
if A[i] % 2 = = 0 :
evenNumCount + = 1
# return number of even pair
return (N - 1 ) * evenNumCount - ((evenNumCount * (evenNumCount - 1 )) / / 2 )
# Driver Code a = [ 5 , 1 , 3 , 2 ]
n = len (a)
print (findevenPair(a, n))
|
Output
3
Time Complexity: O(N)
Auxiliary Space: O(1)