C Program to Count Odd and Even Elements in an Array.

Problem Statement: Given an array of integers. Your task is to write a C program to count the number of odd and even elements in the array.

  • Even Number: A number that is completely divisible by 2.
  • Odd Number: A number that is not completely divisible by 2.

Example: 

Input: arr[] = {2, 4, 5, 7, 10}
Output: Even Count = 3
        Odd Count = 2

Input: arr[] = {12, 13, 8, 2, 10}
Output: Even Count = 4
        Odd Count = 1

Algorithm to Count Even and Odd Elements.

Below are the steps that need to follow:

Step 1: Initialize two variables, evenCount and oddCount, to keep track of the count of even and odd elements respectively. Set both counts to zero.
Step 2: Traverse through the array one element at a time.
Step 3: For each element, check if it is even or odd.
Step 4: If the element is even, increment the evenCount by 1.
Step 5: If the element is odd, increment the oddCount by 1.
Step 6: After traversing through the entire array, output the values of evenCount and oddCount.

Program to Count numbers of odd and even elements in an Array.

// C program to count even and odd numbers in the array
#include <stdio.h>

int main() {
    int arr[] = {2, 4, 5, 7, 10};
    //size of given array
    int size = sizeof(arr) / sizeof(arr[0]);
    int evenCount = 0, oddCount = 0;
    
    //check even and odd
    for (int i = 0; i < size; i++) {
        if (arr[i] % 2 == 0) {
            evenCount++;
        } else {
            oddCount++;
        }
    }

    printf("Number of even elements: %d\n", evenCount);
    printf("Number of odd elements: %d\n", oddCount);

    return 0;
}
Output:
Number of even elements: 3
Number of odd elements: 2

Time Complexity: The time complexity of this algorithm is O(n) because it needs to traverse through all the elements of the array once.

Space Complexity: The space complexity of this algorithm is O(1) as it uses a constant amount of extra space to store the loop variable and other variables.

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