Maximum students to pass after giving bonus to everybody and not exceeding 100 marks

Given an array which represents the marks of students. The passing grade is and maximum marks that a student can score is , 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 . 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 be the maximum marks of a student among all others then the maximum possible bonus marks that can be given will be . 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 <bits/stdc++.h>
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 + n);
    // 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 Aditya Kumar (adityakumar129)

C

// C Implementation of above approach.
#include <stdio.h>

int max_element(int arr[], int n)
{
    // Initialize maximum element
    int max = arr[0];
    // Traverse array elements from second and compare every
    // element with current max
    for (int i = 1; i < n; i++)
        if (arr[i] > max)
            max = arr[i];
    return max;
}

// Function to return the number of students that can pass
int check(int n, int marks[])
{
    // maximum marks
    int x = max_element(marks, n);

    // 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
int main()
{
    int n = 5;
    int marks[] = { 0, 21, 83, 45, 64 };
    printf("%d\n", check(n, marks));
    return 0;
}

// This code is contributed by Aditya Kumar (adityakumar129)

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);

Javascript

<script>

// JavaScript Implementation of above approach.

// Function to return the number
// of students that can pass
function check(n, marks)
{
    // maximum marks
    let x = Math.max(...marks);

    // maximum bonus marks that can be given
    let bonus = 100-x;
    let c = 0;
    for(let i=0; i<n;i++)
    {

        // counting the number of
        // students that can pass
        if(marks[i] + bonus >= 50)
            c += 1;
    }
    return c;
}

    // Driver code

    let n = 5;
     let marks = [0, 21, 83, 45, 64];
     document.write(check(n, marks));


</script>

Output

3

Time Complexity: O(n), since the loop runs from 0 to (n – 1).

Auxiliary Space: O(1), since no extra space has been taken.