# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

A naive approach is to check for every pair and print the count of pairs.

Below is the implementation of the above approach:

 `// 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 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 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# 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. `

 ` `

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++ 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 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 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# 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 `

 ` `

Output:
```3
```

Time Complexity: O(N)

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