Grading student hackerrank solution with explanation, first lets see what the problem is :

Grading Students Hacker Rank Problem:

HackerLand University has the following grading policy:

   1. Every student receives a grade in the inclusive range from 1 to 100.
   2. Any grade less than 30 is a failing grade. 

     

Sam is a professor at the university and likes to round each student's grade
 according to these rules:

  1.If the difference between the grade and the next multiple of 5 is less
 than 3 , round up to the next 
multiple of 5.
  2.If the value of grade  is less than 38 , no rounding occurs as the result 
will still be a failing grade.

Examples

  1. grade = 84 round to85 (85 - 84 is less than 3)
  2. grade = 29 do not round (result is less than 40)
  3. grade = do not round (60 - 57 is 3 or higher)

Given the initial value of  grade for each of Sam's n students, write code 
to automate the rounding 
process. 

Function Description

Complete the function gradingStudents in the editor below.

gradingStudents has the following parameter(s):

    int grades[n]: the grades before rounding

Returns

    int[n]: the grades after rounding as appropriate

Input Format

The first line contains a single integer,n , the number of students.
Each line i of the n subsequent lines contains a single integer, grades[i].

Constraints
  
  1<= n < =60
  0 <= grades[i] <= 100

Grading students Hackerrank solution JavaScript:

function gradingStudents(grades) {
   grades.forEach((grade, i) => {
     let  count = 5 - grade % 5;
      if (grade >= 38 && count < 3) 
      {grades[i] += count;
      }  
  });
  return grades;
}

Grading students Hackerrank solution Python:

def gradingStudents(grades):
    for i in range(len(grades)):
        if grades[i] < 38: continue
        diff = grades[i] % 5
        if 5 - diff < 3:
            grades[i] += 5 - diff       
    return grades

Grading students Hackerrank solution C#:

public static List GradingStudents(List grades)
{
    for (int i = 0; i < grades.Count; i++) { int diff = 5 - grades[i] % 5; if (grades[i] 
>= 38 && diff < 3)
            grades[i] += diff;
    } 
    return grades;
}

Explanation of Code:

  1. First, we need to find whether grade is less greater than 38 or not.
  2. Then we need to find if the number on diving by 5 gives a number less than 3 or not using given line 5 – grade[i]Β  % 5 and made it equal to some temp variable.
  3. If both conditions satisfied then add the temp variable to grades else return the original grades

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