Count pairs with Bitwise AND as ODD 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 odd.

Examples:

Input: N = 4
A[] = { 5, 1, 3, 2 }
Output: 3
Since pair of A[] = ( 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 odd pair A( i, j ) = 3



Input : N = 6
A[] = { 5, 9, 0, 6, 7, 3 }
Output :6
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.

Below is the implementation of the above approach:

C++

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// C++ program to count pairs
// with AND giving a odd number
#include <iostream>
using namespace std;
  
// Function to count number of odd pairs
int findOddPair(int A[], int N)
{
    int i, j;
  
    // variable for counting odd pairs
    int oddPair = 0;
  
    // find all pairs
    for (i = 0; i < N; i++) {
        for (j = i + 1; j < N; j++) {
  
            // find AND operation
            // check odd or even
            if ((A[i] & A[j]) % 2 != 0)
                oddPair++;
        }
    }
    // return number of odd pair
    return oddPair;
}
// Driver Code
int main()
{
  
    int a[] = { 5, 1, 3, 2 };
    int n = sizeof(a) / sizeof(a[0]);
  
    // calling function findOddPair
    // and print number of odd pair
    cout << findOddPair(a, n) << endl;
  
    return 0;
}

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Java

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// Java program to count pairs
// with AND giving a odd number
class solution_1
      
// Function to count
// number of odd pairs
static int findOddPair(int A[], 
                       int N)
{
    int i, j;
  
    // variable for counting
    // odd pairs
    int oddPair = 0;
  
    // find all pairs
    for (i = 0; i < N; i++) 
    {
        for (j = i + 1; j < N; j++) 
        {
  
            // find AND operation
            // check odd or even
            if ((A[i] & A[j]) % 2 != 0)
                oddPair++;
        }
    }
      
    // return number 
    // of odd pair
    return oddPair;
}
  
// Driver Code
public static void main(String args[])
{
    int a[] = { 5, 1, 3, 2 };
    int n = a.length;
  
    // calling function findOddPair
    // and print number of odd pair
    System.out.println(findOddPair(a, n));
}
}
  
// This code is contributed 
// by Arnab Kundu

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Python

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# Python program to count pairs
# with AND giving a odd number
  
# Function to count number
# of odd pairs
def findOddPair(A, N):
  
    # variable for counting odd pairs
    oddPair = 0
  
    # find all pairs
    for i in range(0, N - 1): 
        for j in range(i + 1, N - 1): 
  
            # find AND operation
            # check odd or even
            if ((A[i] & A[j]) % 2 != 0):
                oddPair = oddPair + 1
          
    # return number of odd pair
    return oddPair
  
# Driver Code
a = [5, 1, 3, 2]
n = len(a) 
  
# calling function findOddPair
# and print number of odd pair
print(findOddPair(a, n))
  
# This code is contributed
# by Shivi_Aggarwal

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C#

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// C# program to count pairs
// with AND giving a odd number
using System;
class GFG
      
// Function to count
// number of odd pairs
static int findOddPair(int []A, 
                       int N)
{
    int i, j;
  
    // variable for counting
    // odd pairs
    int oddPair = 0;
  
    // find all pairs
    for (i = 0; i < N; i++) 
    {
        for (j = i + 1; j < N; j++) 
        {
  
            // find AND operation
            // check odd or even
            if ((A[i] & A[j]) % 2 != 0)
                oddPair++;
        }
    }
      
    // return number 
    // of odd pair
    return oddPair;
}
  
// Driver Code
public static void Main()
{
    int []a = { 5, 1, 3, 2 };
    int n = a.Length;
  
    // calling function findOddPair
    // and print number of odd pair
    Console.WriteLine(findOddPair(a, n));
}
}
  
// This code is contributed 
// inder_verma.

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PHP

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<?php
// PHP program to count pairs
// with AND giving a odd number
  
// Function to count 
// number of odd pairs
function findOddPair(&$A, $N)
{
      
    // variable for counting
    // odd pairs
    $oddPair = 0;
  
    // find all pairs
    for ($i = 0; $i < $N; $i++)
    {
        for ($j = $i + 1; $j < $N; $j++) 
        {
  
            // find AND operation
            // check odd or even
            if (($A[$i] & $A[$j]) % 2 != 0)
                $oddPair = $oddPair + 1;
        }
    }
      
    // return number of odd pair
    return $oddPair;
}
  
// Driver Code
$a = array(5, 1, 3, 2);
$n = sizeof($a);
  
// calling function findOddPair
// and print number of odd pair
echo(findOddPair($a, $n));
      
// This code is contributed
// by Shivi_Aggarwal
?>

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Output:

3

Time Complexity:O(N^2)

An efficient solution is to count the odd numbers. Then return count * (count – 1)/2 because AND of two numbers can be odd only if only if a pair of both numbers are odd.

C++

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// C++ program to count pairs with Odd AND
#include <iostream>
using namespace std;
  
int findOddPair(int A[], int N)
{
    // Count total odd numbers in
    int count = 0;
    for (int i = 0; i < N; i++)
        if ((A[i] % 2 == 1))
            count++;
  
    // return count of even pair
    return count * (count - 1) / 2;
}
  
// Driver main
int main()
{
    int a[] = { 5, 1, 3, 2 };
    int n = sizeof(a) / sizeof(a[0]);
  
    // calling function findOddPair
    // and print number of odd pair
    cout << findOddPair(a, n) << endl;
    return 0;
}

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Java

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// Java program to count 
// pairs with Odd AND
class solution_1
static int findOddPair(int A[], 
                       int N)
{
    // Count total odd numbers in
    int count = 0;
    for (int i = 0; i < N; i++)
        if ((A[i] % 2 == 1))
            count++;
  
    // return count of even pair
    return count * (count - 1) / 2;
}
  
// Driver Code
public static void main(String args[])
{
    int a[] = { 5, 1, 3, 2 };
    int n = a.length;
  
    // calling function findOddPair
    // and print number of odd pair
    System.out.println(findOddPair(a, n));
  
}
}
  
// This code is contributed
// by Arnab Kundu

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Python

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# Python program to count 
# pairs with Odd AND 
def findOddPair(A, N): 
  
    # Count total odd numbers 
    count = 0
    for i in range(0, N - 1): 
        if ((A[i] % 2 == 1)): 
            count = count+1
  
    # return count of even pair 
    return count * (count - 1) / 2
  
# Driver Code 
a = [5, 1, 3, 2
n = len(a) 
  
# calling function findOddPair 
# and print number of odd pair 
print(int(findOddPair(a, n))) 
      
# This code is contributed
# by Shivi_Aggarwal

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C#

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// C# program to count 
// pairs with Odd AND 
using System;
  
class GFG
{
public static int findOddPair(int[] A, 
                              int N)
{
    // Count total odd numbers in 
    int count = 0;
    for (int i = 0; i < N; i++)
    {
        if ((A[i] % 2 == 1))
        {
            count++;
        }
    }
  
    // return count of even pair 
    return count * (count - 1) / 2;
}
  
// Driver Code 
public static void Main(string[] args)
{
    int[] a = new int[] {5, 1, 3, 2};
    int n = a.Length;
  
    // calling function findOddPair 
    // and print number of odd pair 
    Console.WriteLine(findOddPair(a, n));
}
}
  
// This code is contributed 
// by Shrikant13

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PHP

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<?php
// PHP program to count 
// pairs with Odd AND 
function findOddPair(&$A, $N
    // Count total odd numbers 
    $count = 0; 
    for ($i = 0; $i < $N; $i++) 
        if (($A[$i] % 2 == 1)) 
            $count++; 
  
    // return count of even pair 
    return $count * ($count - 1) / 2; 
  
// Driver Code 
$a = array(5, 1, 3, 2 ); 
$n = sizeof($a); 
  
// calling function findOddPair 
// and print number of odd pair 
echo(findOddPair($a, $n)); 
  
// This code is contributed
// by Shivi_Aggarwal
?>

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Output:

3

Time Complexity: O(N)



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