Example:
Input array: 1 2 3 4 5 6 7 8 9
Output:
Total even elements: 4
Total odd elements: 5
Required knowledge
Basic C programming, If else, Loop, ArrayLogic to count total even/odd elements in array
Counting total number of even odd is a very simple process. You just need to have a good understanding of even/odd numbers and checking even/odd numbers in C programming. Apart from that you must know how to iterate through the array elements using loops.Below is a basic logic to count total even or odd elements in an array.
- Run a loop from the starting to end index of the array 0-N (Where N is the total number of elements in array).
- Check the current element of array. If it is even then, increment the even count by 1. Otherwise increment the odd count.
- Print the values of even and odd after the termination of loop.
Program to count total even/odd elements in array
/**
* C program to count total number of even and odd elements in an array
*/
#include <stdio.h>
#define MAX_SIZE 100 //Maximum size of the array
int main()
{
int arr[MAX_SIZE];
int i, N, even, odd;
/*
* Reads size and elements in array
*/
printf("Enter size of the array: ");
scanf("%d", &N);
printf("Enter %d elements in array: ", N);
for(i=0; i<N; i++)
{
scanf("%d", &arr[i]);
}
/* Assuming that there are 0 even and odd elements */
even = 0;
odd = 0;
for(i=0; i<N; i++)
{
/* If the current element of array is even then increment even count */
if(arr[i]%2 == 0)
{
even++;
}
else
{
odd++;
}
}
printf("Total even elements: %d\n", even);
printf("Total odd elements: %d\n", odd);
return 0;
}
Output
Enter size of the array: 10
Enter 10 elements in array: 5 6 4 12 19 121 1 7 9 63
Total even elements: 3
Total odd elements: 7
Enter 10 elements in array: 5 6 4 12 19 121 1 7 9 63
Total even elements: 3
Total odd elements: 7
Happy coding ;)
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