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# Count pairs with Bitwise AND as ODD number

• Difficulty Level : Medium
• Last Updated : 21 Apr, 2021

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:
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 :
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++

 `// C++ program to count pairs``// with AND giving a odd number``#include ``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;``}`

## Java

 `// 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`

## Python

 `# 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`

## C#

 `// 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.`

## PHP

 ``

## Javascript

 ``
Output:

`3`

Time Complexity:O(N^2)
An efficient solution is to count the odd numbers. Then return count * (count â€“ 1)/2 because AND of two numbers can be odd only if only if a pair of both numbers are odd.

## C++

 `// C++ program to count pairs with Odd AND``#include ``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;``}`

## Java

 `// 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`

## Python

 `# 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`

## C#

 `// 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`

## PHP

 ``

## Javascript

 ``
Output:
`3`

Time Complexity: O(N)

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