Grading Students


Problem Statement :


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


Solution :



title-img 


                            Solution in C :

In C :

int* gradingStudents(int grades_count, int* grades, int* result_count) {
    
   static int my_grades[10];
    for(int i = 0;  i< grades_count ; i++)
    {
        if(grades[i] < 38) my_grades[i] = grades[i];
        else if((grades[i] % 5)==0) my_grades[i] = grades[i];
        else
         {
             int num = grades[i];
             int num2 = grades[i];
             while((num % 5) !=0)
             {
                num = num +1;
             } 
             
        
             if((num-num2) < 3) my_grades[i] = num;
             else my_grades[i] = num2;

        }
    }
    

    *result_count = grades_count;
    return my_grades;
}




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 grade;
        cin >> grade;
        if (grade >= 38) {
            int rem = grade % 5;
            if (rem >= 3) grade += 5 - rem;
        }
        cout << grade << endl;
    }
    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++){
            int grade = in.nextInt();
            if (grade >= 38) {
                if (grade % 5 >= 3) {
                    grade += 5 - (grade % 5);
                }
            }
            System.out.println(grade);
        }
    }
}




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)








View More Similar Problems

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

View Solution →

Tree: Height of a Binary Tree

The height of a binary tree is the number of edges between the tree's root and its furthest leaf. For example, the following binary tree is of height : image Function Description Complete the getHeight or height function in the editor. It must return the height of a binary tree as an integer. getHeight or height has the following parameter(s): root: a reference to the root of a binary

View Solution →

Tree : Top View

Given a pointer to the root of a binary tree, print the top view of the binary tree. The tree as seen from the top the nodes, is called the top view of the tree. For example : 1 \ 2 \ 5 / \ 3 6 \ 4 Top View : 1 -> 2 -> 5 -> 6 Complete the function topView and print the resulting values on a single line separated by space.

View Solution →

Tree: Level Order Traversal

Given a pointer to the root of a binary tree, you need to print the level order traversal of this tree. In level-order traversal, nodes are visited level by level from left to right. Complete the function levelOrder and print the values in a single line separated by a space. For example: 1 \ 2 \ 5 / \ 3 6 \ 4 F

View Solution →

Binary Search Tree : Insertion

You are given a pointer to the root of a binary search tree and values to be inserted into the tree. Insert the values into their appropriate position in the binary search tree and return the root of the updated binary tree. You just have to complete the function. Input Format You are given a function, Node * insert (Node * root ,int data) { } Constraints No. of nodes in the tree <

View Solution →

Tree: Huffman Decoding

Huffman coding assigns variable length codewords to fixed length input characters based on their frequencies. More frequent characters are assigned shorter codewords and less frequent characters are assigned longer codewords. All edges along the path to a character contain a code digit. If they are on the left side of the tree, they will be a 0 (zero). If on the right, they'll be a 1 (one). Only t

View Solution →