Nice solution. I am writing in C#, myself. I def. had to start with a more fleshed-out solution and work my way down to this. I wonder if a lengthier solution would be more communicative of the imposed requirements. I pushed mine up as a one-liner, though.

Great Solution.........

Got some fun with Java 8:

Arrays.stream(new int[n]).map(i-> in.nextInt()).map(g -> (g >= 38 && g % 5 >=3) ? g+5-g%5 : g).forEach(System.out::println);

Can anyone explain me how it is printing grade for the values like 67.

A simple and perhaps readable solution.. You can use n instead of temp as it's dummy..

int main() 
{
    int n,temp;
    cin>>temp;
    while(cin>>n)
    {
        if(n<38)
            cout<<n<<endl;
        else if(n%5 >= 3)
            cout<<n+ (5-n%5)<<endl;
        else
            cout<<n<<endl;

    }
    return 0;
}

Python3 one-liner:

grades = [int(input()) for __ in range(int(input()))]

[print(g+5 - g%5 if g%5 > 2 and g>37 else g) for g in grades]

I knew I would find something much shorter than my code here but wow! Nice work!

Your code does not working out in the editor.

but this code should round off 67 as 67%5 is 2 which is less 3 . Help me undertand this please.

Wow! Clearly beatiful and cleanest way I've seen so far

Neat !

i dont understan java .. bt i understood the logic .. thank you:)

I've got a pretty sick one-liner too for C++ can your boy get some upvotes =D

for (int i = 0; i < grades.size(); i++){int nearest5Multiple = 5*((grades[i]/5)+1);if (grades[i] >= 38 &&(nearest5Multiple-grades[i]) < 3 ){grades[i] = nearest5Multiple;}}return grades;

it looks good,but keeping result of (grade%5) in a variable could save you from calculating it two times :)

actually not , as it will work for 41,42 . as 41 % 5 =1 , but we want to have the for 43,44. that goes for all the cases.

Great.

Python - 2 liner

grades = [int(input()) for x in range(int(input()))]
print('\n'.join(map(str,[(i-(i%5)+5) if (i>=38 and i%5>2) else i for i in grades])))

It works

This is my solution

static int[] solve(int[] grades){
        // Complete this function
        for(int i=0; i<grades.length; i++){
            if(grades[i]>=38 && grades[i]%5>2)
                grades[i] = grades[i] + (5 - (grades[i]%5));
            
        }
        return grades;

if (grade > 37){ int newGrade = grade; for (int i=1; i<=2; i++){ newGrade++; if(newGrade%10 == 5 || newGrade%10 == 0){ grade = newGrade; return grade; } } } return grade;

won't work for grade =73 since 73%5 is 3

I achived useing my own implementaion upto 20 lines to achieve this code.But your solution is just awe.I will adopt for this technique.Thank you.

nice logic ..

Even my mobile has 4GB ram.

Python 3 , 3-liner

for _ in range(int(input())):
    g=int(input())
    print(g if g < 38 or g % 5 < 3 else g+(5-(g % 5)))

awesome.

awesome use of the modulo operation

this is such a clever solution great mind...

it is not working it is giving me error unsupported operand type for %: list and int. can any help me? there compiler is not working correctly, i have run this code in python 2 in my system it is working fine but not in here

new in python can u help me with this code

#!/bin/python

import sys

def solve(grades):
    r=[]
    y=0
    d=0
    # Complete this function
    for i in range(0,len(grades)-1):
        if(grades[i]>=38):
            y=grades[i]/5
            d=5*(y+1)-grades[i]
            if(d<3):
                r[i]=5*(y+1)
            else:
                r[i]=grades[i]
        else:
            r[i]=grades[i]
    return r
               

n = int(raw_input().strip())
grades = []
grades_i = 0
for grades_i in xrange(n):
    grades_t = int(raw_input().strip())
    grades.append(grades_t)
result = solve(grades)
print "\n".join(map(str, result))

Great solution!

neat

heres the same solution in PHP

function solve($grades){

for ($i=0;$i <= sizeof($grades); $i++){
    if ($grades[$i] >= 38){
        if ($grades[$i] % 5 == 3){
            $grades[$i] += 2;
        }
        else if ($grades[$i] % 5 == 4){
            $grades[$i] += 1;
        }
    }
}
      
    return $grades;
}

A difficult and intricate solution with only one line and no conditionals (nor variables).

print . ((uncurry (+) .) . ((. ((*) . fromEnum)) . flip second) <*> (uncurry ((. (>) 3) . (&&) . (<) 37))) . ((,) <*> (flip mod 5 . negate))

fantastic. Such simple logic!!!

You can just group 1st and 2nd if condition into single if condition

1)int* solve(int grades_size, int* grades, int *result_size) -> In this line, why are we using "result size" variable? 2)*result_size = grades_size; -> What is the meaning of this line?

My solution in C#

static int[] gradingStudents(int[] grades) {

     List<int> resp = new List<int>();
     foreach (int value in grades){
         if(value >= 38){
             if(value % 5 == 0){
                 resp.Add(value);
             } else {
                 int multiFive = value;
                 while(multiFive % 5 != 0){
                     multiFive++;
                 }
                 if((multiFive - value) < 3){
                     resp.Add(multiFive);
                 } else {
                     resp.Add(value);
                 }
             }

         } else {
             resp.Add(value);
         }

     }
    return resp.ToArray();
}