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# 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:
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:

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 ``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 ` `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 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= ``50``)``            ``c += ``1``;``    ``}``    ``return` `c;``}``    ` `// Driver code``public` `static` `void` `main(String[] args)``{``int` `n = ``5``;`` ``List 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= 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

 `= 50)``            ``\$c` `+= 1;``    ``}``    ``return` `\$c``;``}``    ` `// Driver code``\$n` `= 5;``\$marks` `= ``array``(0, 21, 83, 45, 64);``echo` `check(``\$n``, ``\$marks``);`

## Javascript

 ``

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.

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