Counting the Number of Even and Odd Elements in an Array in C++

Understanding Even and Odd Count

This task involves counting the number of even and odd elements in an array.

We will explore three different methods to achieve this in C++.

Method 1: Using Loop

This method iterates through the array and counts even and odd elements.

#include <iostream>
using namespace std;

void countEvenOdd(int arr[], int n) {
    int evenCount = 0, oddCount = 0;
    for (int i = 0; i < n; i++) {
        if (arr[i] % 2 == 0)
            evenCount++;
        else
            oddCount++;
    }
    cout << "Even Count: " << evenCount << ", Odd Count: " << oddCount << endl;
}

int main() {
    int arr[] = {1, 2, 3, 4, 5, 6, 7, 8};
    int n = sizeof(arr) / sizeof(arr[0]);
    countEvenOdd(arr, n);
    return 0;
}
            

Method 2: Using Recursion

This method counts even and odd numbers using recursion.

#include <iostream>
using namespace std;

void countEvenOddRec(int arr[], int n, int index, int &evenCount, int &oddCount) {
    if (index == n) return;
    if (arr[index] % 2 == 0)
        evenCount++;
    else
        oddCount++;
    countEvenOddRec(arr, n, index + 1, evenCount, oddCount);
}

int main() {
    int arr[] = {1, 2, 3, 4, 5, 6, 7, 8};
    int n = sizeof(arr) / sizeof(arr[0]);
    int evenCount = 0, oddCount = 0;
    countEvenOddRec(arr, n, 0, evenCount, oddCount);
    cout << "Even Count: " << evenCount << ", Odd Count: " << oddCount << endl;
    return 0;
}
            

Method 3: Using Bitwise Operations

This method uses bitwise operations to determine even and odd numbers.

#include <iostream>
using namespace std;

void countEvenOddBitwise(int arr[], int n) {
    int evenCount = 0, oddCount = 0;
    for (int i = 0; i < n; i++) {
        if ((arr[i] & 1) == 0)
            evenCount++;
        else
            oddCount++;
    }
    cout << "Even Count: " << evenCount << ", Odd Count: " << oddCount << endl;
}

int main() {
    int arr[] = {1, 2, 3, 4, 5, 6, 7, 8};
    int n = sizeof(arr) / sizeof(arr[0]);
    countEvenOddBitwise(arr, n);
    return 0;
}