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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