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 <stdio.h>

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++;
    }
    printf("Even Count: %d, Odd Count: %d\n", evenCount, oddCount);
}

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 <stdio.h>

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);
    printf("Even Count: %d, Odd Count: %d\n", evenCount, oddCount);
    return 0;
}
            

Method 3: Using Bitwise Operations

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

#include <stdio.h>

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++;
    }
    printf("Even Count: %d, Odd Count: %d\n", evenCount, oddCount);
}

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