Count pairs with Bitwise OR as Even number

Last Updated : 05 Sep, 2022

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++

// C++ program to count pairs with even OR
#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++) {

            // find OR operation
            // check odd or even
            if ((A[i] | A[j]) % 2 == 0)
                evenPair++;
        }
    }
    // return count of even pair
    return evenPair;
}
// Driver main
int main()
{
    int A[] = { 5, 6, 2, 8 };
    int N = sizeof(A) / sizeof(A[0]);
    cout << findEvenPair(A, N) << endl;
    return 0;
}

Java

// Java program to count
// pairs with even OR
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++)
        {

            // find OR operation
            // check odd or even
            if ((A[i] | A[j]) % 2 == 0)
                evenPair++;
        }
    }
    
    // return count of even pair
    return evenPair;
}

// Driver Code
public static void main (String[] args)
{
    int A[] = { 5, 6, 2, 8 };
    int N = A.length;
    System.out.println(findEvenPair(A, N));
}
}

// This code is contributed
// by inder_verma.

Python 3

# Python 3 program to count pairs
# with even OR

def findEvenPair(A, N) :
    evenPair = 0

    for i in range(N) :
        for j in range(i+1, N):

            # find OR operation 
            # check odd or even 
            if (A[i] | A[j]) % 2 == 0 :
                evenPair += 1

    # return count of even pair
    return evenPair

# Driver Code
if __name__ == "__main__" :

    A = [ 5, 6, 2, 8]
    N = len(A)

    # function calling
    print(findEvenPair(A, N))

# This code is contributed by ANKITRAI1

// C# program to count
// pairs with even OR
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++)
        {

            // find OR operation
            // check odd or even
            if ((A[i] | A[j]) % 2 == 0)
                evenPair++;
        }
    }
    
    // return count of even pair
    return evenPair;
}

// Driver Code
public static void Main ()
{
    int []A = { 5, 6, 2, 8 };
    int N = A.Length;
    Console.WriteLine(findEvenPair(A, N));
}
}

// This code is contributed
// by inder_verma.

PHP

<?php
// PHP program to count 
// pairs with even OR
function findEvenPair($A, $N)
{
    $evenPair = 0;
    for ($i = 0; $i < $N; $i++) 
    {
        for ($j = $i + 1; $j < $N; $j++) 
        {

            // find OR operation
            // check odd or even
            if (($A[$i] | $A[$j]) % 2 == 0)
                $evenPair++;
        }
    }
    // return count of even pair
    return $evenPair;
}

// Driver Code
$A = array(5, 6, 2, 8);
$N = count($A);
echo findEvenPair($A, $N);

// This code is contributed
// by inder_verma.
?>

JavaScript

<script>
// JavaScript program to count pairs with even OR

function findEvenPair(A, N)
{
    let evenPair = 0;
    for (let i = 0; i < N; i++) {
        for (let j = i + 1; j < N; j++) {

            // find OR operation
            // check odd or even
            if ((A[i] | A[j]) % 2 == 0)
                evenPair++;
        }
    }
    // return count of even pair
    return evenPair;
}
// Driver main

    let A = [5, 6, 2, 8 ];
    let N = A.length;
    document.write(findEvenPair(A, N));
    
// This code contributed by Rajput-Ji 

</script>

Output

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++

// C++ program to count pairs with even OR
#include <iostream>
using namespace std;

int findEvenPair(int A[], int N)
{ 
    // Count total even numbers in
    // array.
    int count = 0;
    for (int i = 0; i < N; i++) 
       if (!(A[i] & 1))
            count++;
            
    // return count of even pair
    return count * (count - 1) / 2;
}

// Driver main
int main()
{
    int A[] = { 5, 6, 2, 8 };
    int N = sizeof(A) / sizeof(A[0]);
    cout << findEvenPair(A, N) << endl;
    return 0;
}

Java

// Java program to count
// pairs with even OR
import java.io.*;

class GFG 
{
static int findEvenPair(int A[], int N)
{ 
    // Count total even numbers in
    // array.
    int count = 0;
    for (int i = 0; i < N; i++) 
    if ((!((A[i] & 1) > 0)))
            count++;
            
    // return count of even pair
    return count * (count - 1) / 2;
}

// Driver Code
public static void main (String[] args) 
{
    int A[] = { 5, 6, 2, 8 };
    int N = A.length;
    System.out.println(findEvenPair(A, N));
}
}

// This code is contributed
// by inder_verma.

Python 3

# Python 3 program to count 
# pairs with even OR

def findEvenPair(A, N):

    # Count total even numbers 
    # in array.
    count = 0
    for i in range(N): 
        if (not (A[i] & 1)):
            count += 1
          
    # return count of even pair
    return count * (count - 1) // 2

# Driver Code
if __name__ == "__main__":
    A = [ 5, 6, 2, 8 ]
    N = len(A)
    print(findEvenPair(A, N))

# This code is contributed
# by ChitraNayal

// C# program to count
// pairs with even OR
using System;

class GFG 
{
static int findEvenPair(int []A, int N)
{ 
    // Count total even numbers 
    // in array.
    int count = 0;
    for (int i = 0; i < N; i++) 
    if ((!((A[i] & 1) > 0)))
            count++;
            
    // return count of even pair
    return count * (count - 1) / 2;
}

// Driver Code
public static void Main (String[] args) 
{
    int []A = { 5, 6, 2, 8 };
    int N = A.Length;
    Console.WriteLine(findEvenPair(A, N));
}
}

// This code is contributed 
// by Kirti_Mangal

PHP

<?php 
// PHP program to count pairs
// with even OR

function findEvenPair(&$A, $N)
{ 
    // Count total even numbers in
    // array.
    $count = 0;
    for ($i = 0; $i < $N; $i++) 
        if (!($A[$i] & 1))
            $count++;
            
    // return count of even pair
    return $count * ($count - 1) / 2;
}

// Driver Code
$A = array(5, 6, 2, 8 );
$N = sizeof($A);
echo findEvenPair($A, $N)."\n";

// This code is contributed 
// by ChitraNayal
?>

JavaScript

<script>
// Javascript program to count
// pairs with even OR

    function findEvenPair(A,N)
    {
        // Count total even numbers in
    // array.
    let count = 0;
    for (let i = 0; i < N; i++)
        if ((!((A[i] & 1) > 0)))
            count++;
             
    // return count of even pair
    return count * (count - 1) / 2;
    }
    
    // Driver Code
    let A=[5, 6, 2, 8 ];
    let N = A.length;
    document.write(findEvenPair(A, N));
    
    
// This code is contributed by rag2127
</script>

Output

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.

Count pairs with Bitwise XOR as ODD number

Rajput-Ji

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Count pairs with Bitwise OR as Even number

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