Count pairs with Bitwise-AND 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: 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++

// 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; 
} 

Java

// 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.. 

Python3

# 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#

// 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 
// 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 
?> 

Output:

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++

// 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; 
} 

Java

// 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.. 

Python3

# 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#

// 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 
// 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 
?> 

Output:

Time Complexity: O(N)

My Personal Notes

C++

Java

Python3

C#

PHP

C++

Java

Python3

C#

PHP

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