Count pairs with Bitwise XOR 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 = 5
A[] =  { 5, 4, 7, 2, 1}
Output :6
Since pair of A[] =
( 5, 4 ) = 1( 5, 7 ) = 2( 5, 2 ) = 7( 5, 1 ) = 4
( 4, 7 ) = 3( 4, 2 ) = 6( 4, 1 ) = 5
( 7, 2 ) = 5( 7, 1 ) = 6
( 2, 1 ) = 3
Total XOR ODD pair  = 6

Input : N = 7
A[] = { 7, 2, 8, 1, 0, 5, 11 }
Output :12
Since pair of A[] =
( 7, 2 ) = 5( 7, 8 ) = 15( 7, 1 ) = 6( 7, 0 ) = 7( 7, 5 ) = 2( 7, 11 ) = 12
( 2, 8 ) = 10( 2, 1 ) = 3( 2, 0 ) = 2( 2, 5 ) = 7( 2, 11 ) = 9
( 8, 1 ) = 9( 8, 0 ) = 8( 8, 5 ) = 13( 8, 11 ) = 3
( 1, 0 ) = 1( 1, 5 ) = 4( 1, 11 ) = 10
( 0, 5 ) = 5( 0, 11 ) = 11
( 5, 11 ) = 14
```

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++

 `// C++ program to count pairs ` `// with XOR 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 XOR operation ` `            ``// check odd or even ` `            ``if` `((A[i] ^ A[j]) % 2 != 0) ` `                ``oddPair++; ` `        ``} ` `        ``cout << endl; ` `    ``} ` ` `  `    ``// return number of odd pair ` `    ``return` `oddPair; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` ` `  `    ``int` `A[] = { 5, 4, 7, 2, 1 }; ` `    ``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 XOR giving a odd number ` ` `  `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 XOR 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``, ``4``, ``7``, ``2``, ``1` `}; ` `    ``int` `N = A.length; ` ` `  `    ``// calling function findOddPair ` `    ``// and print number of odd pair ` `    ``System.out.println(findOddPair(A, N));  ` `} ` `} ` ` `  `// This code is contributed  ` `// by Kirti_Mangal `

Python3

 `# Python3 program to count pairs  ` `# with XOR 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) :  ` `        ``for` `j ``in` `range``(i``+``1``, N) : ` ` `  `            ``# find XOR operation  ` `            ``# check odd or even  ` `            ``if` `((A[i] ^ A[j]) ``%` `2` `!``=` `0``):  ` `                ``oddPair``+``=``1` ` `  `    ``# return number of odd pair  ` `    ``return` `oddPair  ` ` `  `# Driver Code ` `if` `__name__``=``=``'__main__'``: ` `    ``A ``=` `[``5``, ``4``, ``7``, ``2``, ``1` `]  ` `    ``N ``=` `len``(A) ` ` `  `# calling function findOddPair  ` `# and print number of odd pair  ` `    ``print``(findOddPair(A, N))  ` ` `  `# This code is contributed by Smitha Dinesh Semwal `

C#

 `// C# program to count pairs ` `// with XOR 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 XOR 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, 4, 7, 2, 1 }; ` `    ``int` `N = A.Length; ` ` `  `    ``// calling function findOddPair ` `    ``// and print number of odd pair ` `    ``Console.WriteLine(findOddPair(A, N));  ` `} ` `} ` ` `  `// This code is contributed  ` `// by Akanksha Rai(Abby_akku) ` ` `  ` `  `# calling function findOddPair  ` `# and print number of odd pair  ` `print(findOddPair(a, n))  `

PHP

 ` `

OUTPUT:

`6`

Time Complexity: O(N^2)

An efficient solution is to count the even numbers. Then return count * (N – count).

C++

 `// C++ program to count pairs ` `// with XOR giving a odd number ` `#include ` `using` `namespace` `std; ` ` `  `// Function to count number of odd pairs ` `int` `findOddPair(``int` `A[], ``int` `N) ` `{ ` `    ``int` `i, count = 0; ` ` `  `    ``// find all pairs ` `    ``for` `(i = 0; i < N; i++) { ` `        ``if` `(A[i] % 2 == 0) ` `            ``count++; ` `    ``} ` ` `  `    ``// return number of odd pair ` `    ``return` `count * (N - count); ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `a[] = { 5, 4, 7, 2, 1 }; ` `    ``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 XOR giving a odd number ` `class` `GFG ` `{ ` `// Function to count ` `// number of odd pairs ` `static` `int` `findOddPair(``int` `A[],  ` `                       ``int` `N) ` `{ ` `    ``int` `i, count = ``0``; ` ` `  `    ``// find all pairs ` `    ``for` `(i = ``0``; i < N; i++)  ` `    ``{ ` `        ``if` `(A[i] % ``2` `== ``0``) ` `            ``count++; ` `    ``} ` ` `  `    ``// return number of odd pair ` `    ``return` `count * (N - count); ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] arg) ` `{ ` `    ``int` `a[] = { ``5``, ``4``, ``7``, ``2``, ``1` `}; ` `    ``int` `n = a.length ; ` ` `  `    ``// calling function findOddPair ` `    ``// and print number of odd pair ` `    ``System.out.println(findOddPair(a, n)); ` `} ` `} ` ` `  `// This code is contributed ` `// by Smitha `

Python3

 `# Python3 program to count pairs  ` `# with XOR giving a odd number  ` ` `  `# Function to count number of odd pairs  ` `def` `findOddPair(A, N) : ` ` `  `    ``count ``=` `0` ` `  `    ``# find all pairs  ` `    ``for` `i ``in` `range``(``0` `, N) :  ` `        ``if` `(A[i] ``%` `2` `=``=` `0``) : ` `            ``count``+``=``1` `     `  `    ``# return number of odd pair  ` `    ``return` `count ``*` `(N ``-` `count)  ` ` `  `# Driver Code ` `if` `__name__``=``=``'__main__'``: ` `    ``a ``=` `[``5``, ``4``, ``7``, ``2``, ``1``]  ` `    ``n ``=` `len``(a) ` `    ``print``(findOddPair(a,n)) ` ` `  `# this code is contributed by Smitha Dinesh Semwal `

C#

 `// C# program to count pairs ` `// with XOR giving a odd number ` `using` `System; ` ` `  `class` `GFG ` `{ ` `// Function to count ` `// number of odd pairs ` `static` `int` `findOddPair(``int` `[]A, ` `                       ``int` `N) ` `{ ` `    ``int` `i, count = 0; ` ` `  `    ``// find all pairs ` `    ``for` `(i = 0; i < N; i++)  ` `    ``{ ` `        ``if` `(A[i] % 2 == 0) ` `            ``count++; ` `    ``} ` ` `  `    ``// return number of odd pair ` `    ``return` `count * (N - count); ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main() ` `{ ` `    ``int` `[]a = { 5, 4, 7, 2, 1 }; ` `    ``int` `n = a.Length ; ` ` `  `    ``// calling function findOddPair ` `    ``// and print number of odd pair ` `    ``Console.Write(findOddPair(a, n)); ` `} ` `} ` ` `  `// This code is contributed ` `// by Smitha `

PHP

 ` `

Output:

`6`

Time Complexity: O(N)

GeeksforGeeks has prepared a complete interview preparation course with premium videos, theory, practice problems, TA support and many more features. Please refer Placement 100 for details

My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Article Tags :
Practice Tags :

2

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.