# Count pairs with Bitwise-AND as even number

Given an array of integers. The task is to find the number of pairs (i, j) such that A[i] & A[j] is even.

Examples:

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

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

Below is the implementation of the above approach:

## C++

 `// C++ program to count pair with ` `// bitwise-AND as even number ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to count number of pairs EVEN bitwise AND ` `int` `findevenPair(``int` `A[], ``int` `N) ` `{ ` `    ``int` `i, j; ` ` `  `    ``// variable for counting even pairs ` `    ``int` `evenPair = 0; ` ` `  `    ``// find all pairs ` `    ``for` `(i = 0; i < N; i++) { ` `        ``for` `(j = i + 1; j < N; j++) { ` ` `  `            ``// find AND operation ` `            ``// to check evenpair ` `            ``if` `((A[i] & A[j]) % 2 == 0) ` `                ``evenPair++; ` `        ``} ` `    ``} ` ` `  `    ``// return number of even pair ` `    ``return` `evenPair; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` ` `  `    ``int` `a[] = { 5, 1, 3, 2 }; ` `    ``int` `n = ``sizeof``(a) / ``sizeof``(a[0]); ` ` `  `    ``cout << findevenPair(a, n) << endl; ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to count pair with ` `// bitwise-AND as even number ` `import` `java.io.*; ` ` `  `class` `GFG ` `{ ` ` `  `// Function to count number of ` `// pairs EVEN bitwise AND ` `static` `int` `findevenPair(``int` `[]A, ` `                        ``int` `N) ` `{ ` `    ``int` `i, j; ` ` `  `    ``// variable for counting even pairs ` `    ``int` `evenPair = ``0``; ` ` `  `    ``// find all pairs ` `    ``for` `(i = ``0``; i < N; i++)  ` `    ``{ ` `        ``for` `(j = i + ``1``; j < N; j++) ` `        ``{ ` ` `  `            ``// find AND operation ` `            ``// to check evenpair ` `            ``if` `((A[i] & A[j]) % ``2` `== ``0``) ` `                ``evenPair++; ` `        ``} ` `    ``} ` ` `  `    ``// return number of even pair ` `    ``return` `evenPair; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main (String[] args) ` `{ ` `    ``int` `[]a = { ``5``, ``1``, ``3``, ``2` `}; ` `    ``int` `n = a.length; ` `     `  `    ``System.out.println(findevenPair(a, n)); ` `} ` `} ` ` `  `// This code is contributed by anuj_67.. `

## Python3

 `# Python 3 program to count pair with ` `# bitwise-AND as even number ` ` `  `# Function to count number of  ` `# pairs EVEN bitwise AND ` `def` `findevenPair(A, N): ` ` `  `    ``# variable for counting even pairs ` `    ``evenPair ``=` `0` ` `  `    ``# find all pairs ` `    ``for` `i ``in` `range``(``0``, N): ` `        ``for` `j ``in` `range``(i ``+` `1``, N): ` `             `  `            ``# find AND operation to  ` `            ``# check evenpair ` `            ``if` `((A[i] & A[j]) ``%` `2` `=``=` `0``): ` `                ``evenPair ``+``=` `1` ` `  `    ``# return number of even pair ` `    ``return` `evenPair ` ` `  `# Driver Code ` `a ``=` `[ ``5``, ``1``, ``3``, ``2` `] ` `n ``=` `len``(a) ` ` `  `print``(findevenPair(a, n)) ` ` `  `# This code is contributed  ` `# by PrinciRaj1992 `

## C#

 `// C# program to count pair with ` `// bitwise-AND as even number ` `using` `System; ` ` `  `class` `GFG ` `{ ` ` `  `// Function to count number of ` `// pairs EVEN bitwise AND ` `static` `int` `findevenPair(``int` `[]A, ` `                        ``int` `N) ` `{ ` `    ``int` `i, j; ` ` `  `    ``// variable for counting even pairs ` `    ``int` `evenPair = 0; ` ` `  `    ``// find all pairs ` `    ``for` `(i = 0; i < N; i++)  ` `    ``{ ` `        ``for` `(j = i + 1; j < N; j++) ` `        ``{ ` ` `  `            ``// find AND operation ` `            ``// to check evenpair ` `            ``if` `((A[i] & A[j]) % 2 == 0) ` `                ``evenPair++; ` `        ``} ` `    ``} ` ` `  `    ``// return number of even pair ` `    ``return` `evenPair; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main () ` `{ ` `    ``int` `[]a = { 5, 1, 3, 2 }; ` `    ``int` `n = a.Length; ` `     `  `    ``Console.WriteLine(findevenPair(a, n)); ` `} ` `} ` ` `  `// This code is contributed by anuj_67.. `

## PHP

 ` `

Output:

```3
```

An Efficient Approach will be to observe that the bitwise AND of two numbers will be even if atleast one of the two numbers is even. So, count all the odd numbers in the array say count. Now, number of pairs with both odd elements will be count*(count-1)/2.

Therefore, number of elements with atleast one even element will be:

```Total Pairs - Count of pair with both odd elements
```

Below is the implementation of the above approach:

## C++

 `// C++ program to count pair with ` `// bitwise-AND as even number ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to count number of pairs ` `// with EVEN bitwise AND ` `int` `findevenPair(``int` `A[], ``int` `N) ` `{ ` `    ``int` `count = 0; ` ` `  `    ``// count odd numbers ` `    ``for` `(``int` `i = 0; i < N; i++) ` `        ``if` `(A[i] % 2 != 0) ` `            ``count++; ` ` `  `    ``// count odd pairs ` `    ``int` `oddCount = count * (count - 1) / 2; ` ` `  `    ``// return number of even pair ` `    ``return` `(N * (N - 1) / 2) - oddCount; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` ` `  `    ``int` `a[] = { 5, 1, 3, 2 }; ` `    ``int` `n = ``sizeof``(a) / ``sizeof``(a[0]); ` ` `  `    ``cout << findevenPair(a, n) << endl; ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to count pair with ` `// bitwise-AND as even number ` ` `  ` `  `import` `java.io.*; ` ` `  `class` `GFG { ` `  `  ` `  ` `  `// Function to count number of pairs ` `// with EVEN bitwise AND ` `static` `int` `findevenPair(``int` `A[], ``int` `N) ` `{ ` `    ``int` `count = ``0``; ` ` `  `    ``// count odd numbers ` `    ``for` `(``int` `i = ``0``; i < N; i++) ` `        ``if` `(A[i] % ``2` `!= ``0``) ` `            ``count++; ` ` `  `    ``// count odd pairs ` `    ``int` `oddCount = count * (count - ``1``) / ``2``; ` ` `  `    ``// return number of even pair ` `    ``return` `(N * (N - ``1``) / ``2``) - oddCount; ` `} ` ` `  `// Driver Code ` ` `  `    ``public` `static` `void` `main (String[] args) { ` `            ``int` `a[] = { ``5``, ``1``, ``3``, ``2` `}; ` `    ``int` `n =a.length; ` ` `  `    ``System.out.print( findevenPair(a, n)); ` `    ``} ` `} ` `// This code is contributed by anuj_67.. `

## Python3

 `# Python 3 program to count pair with ` `# bitwise-AND as even number ` ` `  `# Function to count number of pairs ` `# with EVEN bitwise AND ` `def` `findevenPair(A, N): ` `    ``count ``=` `0` ` `  `    ``# count odd numbers ` `    ``for` `i ``in` `range``(``0``, N): ` `        ``if` `(A[i] ``%` `2` `!``=` `0``): ` `            ``count ``+``=` `1` ` `  `    ``# count odd pairs ` `    ``oddCount ``=` `count ``*` `(count ``-` `1``) ``/` `2` ` `  `    ``# return number of even pair ` `    ``return` `(``int``)((N ``*` `(N ``-` `1``) ``/` `2``) ``-` `oddCount) ` ` `  `# Driver Code ` `a ``=` `[``5``, ``1``, ``3``, ``2` `] ` `n ``=` `len``(a) ` ` `  `print``(findevenPair(a, n)) ` ` `  `# This code is contributed  ` `# by PrinciRaj1992  `

## C#

 `// C# program to count pair with ` `// bitwise-AND as even number ` ` `  `using` `System; ` ` `  `public` `class` `GFG{ ` `     `  `// Function to count number of pairs ` `// with EVEN bitwise AND ` `static` `int` `findevenPair(``int` `[]A, ``int` `N) ` `{ ` `    ``int` `count = 0; ` ` `  `    ``// count odd numbers ` `    ``for` `(``int` `i = 0; i < N; i++) ` `        ``if` `(A[i] % 2 != 0) ` `            ``count++; ` ` `  `    ``// count odd pairs ` `    ``int` `oddCount = count * (count - 1) / 2; ` ` `  `    ``// return number of even pair ` `    ``return` `(N * (N - 1) / 2) - oddCount; ` `} ` ` `  `// Driver Code ` ` `  `    ``static` `public` `void` `Main (){ ` `    ``int` `[]a = { 5, 1, 3, 2 }; ` `    ``int` `n =a.Length; ` `    ``Console.WriteLine( findevenPair(a, n)); ` `    ``} ` `} ` `// This code is contributed by ajit `

## PHP

 ` `

Output:

```3
```

Time Complexity: O(N)

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