hackerrank(c): grading students v2

The space complexity is $O(n)$ because a new array is created to store the results of the grading.

### Solution 2 :: C

If resulting array does is dynamically memory allocated, then the third parameter `result_count` is unnecessary, as the resulting array is not hardcoded to length 100 any longer, but instead is allocated to the same length as the input grades array.

```c

#include <stdio.h>

#include <stdlib.h>

#include <math.h>

/**

 * Apply the grading logic to the grades.

 *

 * - T.C: O(n).

 * - S.C: O(n).

 */

int* grade(int len, int* grades) {

  int i, grade;

  int* res = malloc(len * sizeof(int));

  for (i = 0; i < len; ++i) {

    grade = *(grades + i);

    if (grade < 38) {

      res[i] = grade;

      continue;

    }

    int new_grade = next_mult_of(5, grade);

    res[i] = (new_grade - grade < 3) ? new_grade : grade;

  }

  return res;

}

int main(int argc, char* argv[]) {

  int i;

  int grades[] = { 3, 37, 38, 39, 40, 41, 75, 83, 84, 98 };

  int* result;

  int len = sizeof grades / sizeof(int);

  result = grade(len, grades);

  for (i = 0; i < len; ++i)

    printf("%d\n", *(result + i));

  return 0;

}

```

Again, space complexity is $O(n)$ because the computed grades are stored in a resulting array.