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# Count pairs with odd Bitwise XOR that can be removed and replaced by their Bitwise OR

• Last Updated : 06 Apr, 2021

Given an array arr[] consisting of N integers, the task is to count the number of pairs whose Bitwise XOR is odd, that can be removed and replaced by their Bitwise OR values until no such pair exists in the array.

Examples:

Input: arr[] = {5, 4, 7, 2}
Output: 2
Explanation:
Pair (5, 4): Bitwise XOR of 5 and 4 is 1. Remove this pair and add their Bitwise OR (= 5) into the array. Therefore, the modified array is {5, 7, 2}.
Pair (5, 2): Bitwise XOR of 5 and 2 is 7. Remove this pair and add their Bitwise OR (= 7) into the array. Therefore, the modified array is {7, 7}.

Therefore, the count of such pairs that can be removed is 2.

Input: arr[] = {2, 4, 6}
Output: 0

Approach: The given problem can be solved based on the following observations:

Therefore, the idea is to find the count of even elements present in the given array. If the count of even elements is N, then 0 moves are required. Otherwise, print the value of count as the resultant count of pairs required to be removed.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ``using` `namespace` `std;` `// Function to count the number of pairs``// required to be removed from the array``// and replaced by their Bitwise OR values``void` `countPairs(``int` `arr[], ``int` `N)``{``    ``// Stores the count of``    ``// even array elements``    ``int` `even = 0;` `    ``// Traverse the given array``    ``for` `(``int` `i = 0; i < N; i++) {` `        ``// Increment the count``        ``// of even array elements``        ``if` `(arr[i] % 2 == 0)``            ``even++;``    ``}` `    ``// If the array contains at``    ``// least one odd array element``    ``if` `(N - even >= 1) {``        ``cout << even;``        ``return``;``    ``}` `    ``// Otherwise, print 0``    ``cout << 0;``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 5, 4, 7, 2 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);``    ``countPairs(arr, N);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.io.*;``import` `java.lang.*;``import` `java.util.*;` `class` `GFG{` `// Function to count the number of pairs``// required to be removed from the array``// and replaced by their Bitwise OR values``static` `void` `countPairs(``int` `arr[], ``int` `N)``{``    ` `    ``// Stores the count of``    ``// even array elements``    ``int` `even = ``0``;` `    ``// Traverse the given array``    ``for``(``int` `i = ``0``; i < N; i++)``    ``{``        ` `        ``// Increment the count``        ``// of even array elements``        ``if` `(arr[i] % ``2` `== ``0``)``            ``even++;``    ``}` `    ``// If the array contains at``    ``// least one odd array element``    ``if` `(N - even >= ``1``)``    ``{``        ``System.out.println(even);``        ``return``;``    ``}` `    ``// Otherwise, print 0``    ``System.out.println(``0``);``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int` `arr[] = { ``5``, ``4``, ``7``, ``2` `};``    ``int` `N = arr.length;` `    ``countPairs(arr, N);``}``}` `// This code is contributed by Kingash`

## Python3

 `# Python3 program for the above approach` `# Function to count the number of pairs``# required to be removed from the array``# and replaced by their Bitwise OR values``def` `countPairs(arr, N):` `    ``# Stores the count of``    ``# even array elements``    ``even ``=` `0` `    ``# Traverse the given array``    ``for` `i ``in` `range``(N):` `        ``# Increment the count``        ``# of even array elements``        ``if` `(arr[i] ``%` `2` `=``=` `0``):``            ``even ``+``=` `1``    ` `    ``# If the array contains at``    ``# least one odd array element``    ``if` `(N ``-` `even >``=` `1``):``        ``print``(even)``        ``return` `    ``# Otherwise, print 0``    ``print``(``0``)` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:` `    ``arr ``=` `[ ``5``, ``4``, ``7``, ``2` `]``    ``N ``=` `len``(arr)``    ` `    ``countPairs(arr, N)` `# This code is contributed by AnkThon`

## C#

 `// C# program for the above approach``using` `System;` `class` `GFG{``    ` `// Function to count the number of pairs``// required to be removed from the array``// and replaced by their Bitwise OR values``static` `void` `countPairs(``int``[] arr, ``int` `N)``{``    ` `    ``// Stores the count of``    ``// even array elements``    ``int` `even = 0;` `    ``// Traverse the given array``    ``for``(``int` `i = 0; i < N; i++)``    ``{``        ` `        ``// Increment the count``        ``// of even array elements``        ``if` `(arr[i] % 2 == 0)``            ``even++;``    ``}` `    ``// If the array contains at``    ``// least one odd array element``    ``if` `(N - even >= 1)``    ``{``        ``Console.WriteLine(even);``        ``return``;``    ``}` `    ``// Otherwise, print 0``    ``Console.WriteLine(0);``}` `// Driver code``static` `void` `Main()``{``    ``int``[] arr = { 5, 4, 7, 2 };``    ``int` `N = arr.Length;` `    ``countPairs(arr, N);``}``}` `// This code is contributed by sanjoy_62`

## Javascript

 ``
Output:
`2`

Time Complexity: O(N)
Auxiliary Space: O(1)

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