### Problem Statement :

```HackerLand University has the following grading policy:

1. Every student receives a grade in the inclusive range from 1 to 100.

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.

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

### Solution :

```                            ```Solution in C :

In C :

for(int i = 0;  i< grades_count ; i++)
{
else
{
while((num % 5) !=0)
{
num = num +1;
}

if((num-num2) < 3) my_grades[i] = num;

}
}

}

In C ++ :

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <string>
#include <bitset>
#include <cstdio>
#include <limits>
#include <vector>
#include <climits>
#include <cstring>
#include <cstdlib>
#include <fstream>
#include <numeric>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <unordered_map>

using namespace std;

int main(){
int n;
cin >> n;
for(int a0 = 0; a0 < n; a0++){
int rem = grade % 5;
if (rem >= 3) grade += 5 - rem;
}
}
return 0;
}

In Java :

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
for(int a0 = 0; a0 < n; a0++){
if (grade % 5 >= 3) {
}
}
}
}
}

In Python3 :

#!/bin/python3

import sys

n = int(input().strip())
for a0 in range(n):
x = int(input().strip())

if x >= 38:
if x % 5 > 2:
while x % 5 != 0: x += 1
print(x)```
```

## Tree: Inorder Traversal

In this challenge, you are required to implement inorder traversal of a tree. Complete the inorder function in your editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's inorder traversal as a single line of space-separated values. Input Format Our hidden tester code passes the root node of a binary tree to your \$inOrder* func

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## Binary Search Tree : Insertion

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