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Maximum students to pass after giving bonus to everybody and not exceeding 100 marks
  • Last Updated : 03 Jan, 2019

Given an array arr which represents the marks of n students. The passing grade is 50 and maximum marks that a student can score is 100, the task is to maximize the student that are passing the exam by giving bonus marks to the students.
Note that if a student is given bonus marks then all other students will also be given the same amount of bonus marks without any student’s marks exceeding 100. Print the total students that can pass the exam in the end.

Examples:

Input: arr[] = {0, 21, 83, 45, 64}
Output: 3
We can only add maximum of 17 bonus marks to the marks of all the students. So, the final array becomes {17, 38, 100, 62, 81}
Only 3 students will pass the exam.

Input: arr[] = {99, 50, 46, 47, 48, 49, 98}
Output: 4

Approach: Let M be the maximum marks of a student among all others then the maximum possible bonus marks that can be given will be 100 - M. Now for every student whose marks + (100 – M) ≥ 50, increment the count. Print the count in the end.



Below is the implementation of the above approach:

C++




// C++ Implementation of above approach. 
#include<iostream> 
#include<algorithm>
using namespace std;
  
// Function to return the number 
// of students that can pass 
int check(int n, int marks[])
{
    // maximum marks 
    int* x = std::max_element(marks,marks+5); 
  
    // maximum bonus marks that can be given 
    int bonus = 100-(int)(*x);
    int c = 0;
    for(int i=0; i<n;i++)
    {
  
        // counting the number of 
        // students that can pass 
        if(marks[i] + bonus >= 50)
            c += 1;
    }
    return c;
}
      
// Driver code
int main()
{
int n = 5;
int marks[] = {0, 21, 83, 45, 64}; 
cout<<check(n, marks)<<endl; 
return 0;
}
// This code is contributed by mits

Java




// Java Implementation of above approach. 
import java.util.*;
class GFG{
// Function to return the number 
// of students that can pass 
static int check(int n, List<Integer> marks)
{
    // maximum marks 
    Integer x = Collections.max(marks); 
  
    // maximum bonus marks that can be given 
    int bonus = 100-x;
    int c = 0;
    for(int i=0; i<n;i++)
    {
  
        // counting the number of 
        // students that can pass 
        if(marks.get(i) + bonus >= 50)
            c += 1;
    }
    return c;
}
      
// Driver code
public static void main(String[] args)
{
int n = 5;
 List<Integer> marks = Arrays.asList(0, 21, 83, 45, 64); 
System.out.println(check(n, marks)); 
}
}
// This code is contributed by mits

Python3




# Python3 Implementation of above approach.
  
# Function to return the number 
# of students that can pass
def check(n, marks):
  
    # maximum marks
    x = max(marks)
  
    # maximum bonus marks that can be given
    bonus = 100-x
    c = 0
    for i in range(n):
  
        # counting the number of 
        # students that can pass
        if(marks[i] + bonus >= 50):
            c += 1
  
    return c
  
# Driver code
n = 5
marks = [0, 21, 83, 45, 64]
print(check(n, marks))

C#




// C# Implementation of above approach. 
using System;
using System.Collections.Generic;
using System.Collections;
using System.Linq;
class GFG{
// Function to return the number 
// of students that can pass 
static int check(int n, List<int> marks)
{
    // maximum marks 
    int x = marks.Max(); 
  
    // maximum bonus marks that can be given 
    int bonus = 100-x;
    int c = 0;
    for(int i=0; i<n;i++)
    {
  
        // counting the number of 
        // students that can pass 
        if(marks[i] + bonus >= 50)
            c += 1;
    }
    return c;
}
      
// Driver code
public static void Main()
{
int n = 5;
List<int> marks = new List<int>(new int[]{0, 21, 83, 45, 64}); 
Console.WriteLine(check(n, marks)); 
}
}
// This code is contributed by mits

PHP




<?php
  
// PHP Implementation of above approach. 
  
// Function to return the number 
// of students that can pass 
function check($n, $marks)
{
    // maximum marks 
    $x = max($marks); 
  
    // maximum bonus marks that can be given 
    $bonus = 100-$x;
    $c = 0;
    for($i=0; $i<$n;$i++)
    {
  
        // counting the number of 
        // students that can pass 
        if($marks[$i] + $bonus >= 50)
            $c += 1;
    }
    return $c;
}
      
// Driver code 
$n = 5;
$marks = array(0, 21, 83, 45, 64); 
echo check($n, $marks); 
Output:
3

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