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

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:

• Bitwise XOR of a pair is odd only if one element is odd and the other is even.
• Therefore, removing such a pair from the array and adding their Bitwise OR into the array doesn't affect the total count of odd elements in the array. But the count of even array elements reduces by 1.

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++ program for the above approach

#include <bits/stdc++.h>
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[0]);
    countPairs(arr, N);

    return 0;
}

// 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 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# 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

<script>

// JavaScript 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
function countPairs(arr, N)
{
    // Stores the count of
    // even array elements
    let even = 0;

    // Traverse the given array
    for (let 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) {
        document.write(even);
        return;
    }

    // Otherwise, print 0
    document.write(0);
}

// Driver Code
let arr = [ 5, 4, 7, 2 ];
let N = arr.length;
countPairs(arr, N);

</script>

2

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