Number of pairs with Bitwise OR as Odd number

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 odd.

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 Odd = 3

Input : N = 7
            A[] = {8, 6, 2, 7, 3, 4, 9}
Output :15

A Simple Solution is to check every pair and find the Bitwise-OR and count all such pairs with Bitwise OR as odd.



Below is the implementation of the above approach:

C++

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// C++ program to count pairs with odd OR
  
#include <iostream>
using namespace std;
  
// Function to count pairs with odd OR
int findOddPair(int A[], int N)
{
    int oddPair = 0;
    for (int i = 0; i < N; i++) {
        for (int j = i + 1; j < N; j++) {
  
            // find OR operation
            // check odd or odd
            if ((A[i] | A[j]) % 2 != 0)
                oddPair++;
        }
    }
  
    // return count of odd pair
    return oddPair;
}
  
// Driver Code
int main()
{
    int A[] = { 5, 6, 2, 8 };
  
    int N = sizeof(A) / sizeof(A[0]);
  
    cout << findOddPair(A, N) << endl;
  
    return 0;
}

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Java

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// Java program to count pairs
// with odd OR
class GFG
{
// Function to count pairs with odd OR
static int findOddPair(int A[], int N)
{
    int oddPair = 0;
    for (int i = 0; i < N; i++) 
    {
        for (int j = i + 1; j < N; j++)
        {
  
            // find OR operation
            // check odd or odd
            if ((A[i] | A[j]) % 2 != 0)
                oddPair++;
        }
    }
  
    // return count of odd pair
    return oddPair;
}
  
// Driver Code
public static void main(String []args)
{
    int A[] = { 5, 6, 2, 8 };
  
    int N = A.length;
  
    System.out.println(findOddPair(A, N));
}
}
  
// This code is contributed by ANKITRAI1

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Python3

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# Python3 program to count pairs with odd OR
  
   
# Function to count pairs with odd OR
def findOddPair(A, N):
      
    oddPair = 0
    for i in range(0, N):
        for j in range(i+1, N):
   
            # find OR operation
            # check odd or odd
            if ((A[i] | A[j]) % 2 != 0):
                oddPair+=1
  
    # return count of odd pair
    return oddPair
  
   
# Driver Code
def main():
      
    A = [ 5, 6, 2, 8 ]
   
    N = len(A)
   
    print(findOddPair(A, N))
  
if __name__ == '__main__':
    main()
# This code is contributed by PrinciRaj1992  

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C#

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// C#  program to count pairs 
// with odd OR
  
using System;
  
public class GFG{
      
    // Function to count pairs with odd OR 
static int findOddPair(int[] A, int N) 
    int oddPair = 0; 
    for (int i = 0; i < N; i++) 
    
        for (int j = i + 1; j < N; j++) 
        
  
            // find OR operation 
            // check odd or odd 
            if ((A[i] | A[j]) % 2 != 0) 
                oddPair++; 
        
    
  
    // return count of odd pair 
    return oddPair; 
  
// Driver Code 
    static public void Main (){
    int []A = { 5, 6, 2, 8 }; 
    int N = A.Length; 
  
    Console.WriteLine(findOddPair(A, N)); 
    
  
//This code is contributed by ajit 

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PHP

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<?php
//PHP program to count pairs with odd OR
  
// Function to count pairs with odd OR
function  findOddPair($A, $N)
{
    $oddPair = 0;
    for ($i = 0; $i < $N; $i++) {
        for ($j = $i + 1; $j < $N; $j++) {
  
            // find OR operation
            // check odd or odd
            if (($A[$i] | $A[$j]) % 2 != 0)
                $oddPair++;
        }
    }
  
    // return count of odd pair
    return $oddPair;
}
  
// Driver Code
    $A = array (5, 6, 2, 8 );
    $N = sizeof($A) / sizeof($A[0]);
  
    echo findOddPair($A, $N),"\n";
  
  
  
#This code is contributed by ajit
?>

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Output:

3

Time Complexity: O(N2)

An Efficient Solution is to count pairs with even OR and subtract them with total number of pairs to get pairs with odd Bitwise-OR. To do this, count numbers with last bit as 0. Then number of pairs with even Bitwise-OR = count * (count – 1)/2 and total number of pairs will be N*(N-1)/2.

Therefore, pairs with ODD Bitwise-OR will be:

Total Pairs - Pairs with EVEN Bitwise-OR

Below is the implementation of the above approach:

C++

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// C++ program to count pairs with odd OR
#include <iostream>
using namespace std;
  
// Function to count pairs with odd OR
int countOddPair(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++;
  
    // Even pair count
    int evenPairCount = count * (count - 1) / 2;
  
    // Total pairs
    int totPairs = N * (N - 1) / 2;
  
    // Return Odd pair count
    return totPairs - evenPairCount;
}
  
// Driver main
int main()
{
    int A[] = { 5, 6, 2, 8 };
    int N = sizeof(A) / sizeof(A[0]);
  
    cout << countOddPair(A, N) << endl;
  
    return 0;
}

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Java

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// Java program to count pairs with odd OR
  
public class GFG {
  
// Function to count pairs with odd OR
    static int countOddPair(int A[], int N) {
        // Count total even numbers in
        // array
  
        int count = 0;
        for (int i = 0; i < N; i++) {
            if ((A[i] % 2 != 1)) {
                count++;
            }
        }
  
        // Even pair count
        int evenPairCount = count * (count - 1) / 2;
  
        // Total pairs
        int totPairs = N * (N - 1) / 2;
  
        // Return Odd pair count
        return totPairs - evenPairCount;
    }
  
// Driver main
    public static void main(String[] args) {
        int A[] = {5, 6, 2, 8};
        int N = A.length;
  
        System.out.println(countOddPair(A, N));
  
    }
}

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Python3

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# Python 3program to count pairs with odd OR
  
# Function to count pairs with odd OR
def countOddPair(A, N):
      
    # Count total even numbers in
    # array
    count = 0
    for i in range(0, N):
        if (A[i] % 2 != 1):
            count+=1
  
    # Even pair count
    evenPairCount = count * (count - 1) / 2
  
    # Total pairs
    totPairs = N * (N - 1) / 2
  
    # Return Odd pair count
    return (int)(totPairs - evenPairCount)
      
# Driver Code
A = [ 5, 6, 2, 8 ]
  
N = len(A)
  
print(countOddPair(A, N))
  
# This code is contributed by PrinciRaj1992 

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C#

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// C# program to count pairs with odd OR
using System; 
  
public class GFG { 
  
// Function to count pairs with odd OR
    static int countOddPair(int []A, int N) {
        // Count total even numbers in
        // array
   
        int count = 0;
        for (int i = 0; i < N; i++) {
            if ((A[i] % 2 != 1)) {
                count++;
            }
        }
   
        // Even pair count
        int evenPairCount = count * (count - 1) / 2;
   
        // Total pairs
        int totPairs = N * (N - 1) / 2;
   
        // Return Odd pair count
        return totPairs - evenPairCount;
    }
   
// Driver main
    public static void Main() {
        int []A = {5, 6, 2, 8};
        int N = A.Length;
   
        Console.WriteLine(countOddPair(A, N));
   
    }
}
/*This code is contributed by PrinciRaj1992*/

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PHP

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<?php
// PHP program to count pairs with odd OR 
// Function to count pairs with odd OR 
function countOddPair( $A, $N
    // Count total even numbers 
    // in array 
    $count = 0; 
    for ($i = 0; $i < $N; $i++) 
        if (!($A[$i] & 1)) 
            $count++; 
              
    // Even pair count 
    $evenPairCount = $count *
                    ($count - 1) / 2; 
  
    // Total pairs 
    $totPairs = $N * ($N - 1) / 2; 
  
    // Return Odd pair count 
    return ($totPairs - $evenPairCount); 
  
// Driver main 
$A = array( 5, 6, 2, 8 ); 
$N = sizeof($A); 
  
echo countOddPair($A, $N),"\n"
  
// This code is contributed by Sach_Code
?>

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Output:

3

Time Complexity: O(N)



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