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Program to find minimum number of lectures to attend to maintain 75%

Last Updated : 23 Aug, 2022
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Consider the subject Data Structures for which the total number of classes held till present date is M  , and some students attend only N    out of these classes. Find the minimum number of lectures they have to attend so that their 75\%    attendance is maintained. 

Examples:  

Input : M = 7 and N = 6 
Output : 0 lectures to attend 
As 7 classes have been held till present, out of which 6 classes have been attended, which is greater than 
75%, so no more lectures to attend


Input : M = 9 and N = 1 
Output : 23 lectures to attend 
Out of 9 classes, only 1 class is attended. After 23 more classes, a total of 1+23 = 24 classes 
have been attended and the total number of classes held = 9+23 = 32. So 24/32 = 75%. Hence 23 is 
the minimum value.

Solution: 
Using the formula, 
Ceil\left (\frac{(0.75*M)-N}{0.25} \right )    
Before applying the formula, first, check whether N by M has 75% or not. If not, then apply the formula 

C++
// C++ Program to find minimum number of lectures to attend
// to maintain 75% attendance

#include <cmath>
#include <iostream>
using namespace std;

// Function to compute minimum lecture
int minimumLectures(int m, int n)
{
    int ans = 0;

    // Formula to compute
    if (n < (int)ceil(0.75 * m))
        ans = (int)ceil(((0.75 * m) - n) / 0.25);
    else
        ans = 0;

    return ans;
}

// Driver function
int main()
{
    int M = 9, N = 1;
    cout << minimumLectures(M, N);
    return 0;
}
Java
// Java Program to find minimum number of lectures to attend
// to maintain 75% attendance

public class GFG {

    // Method to compute minimum lecture
    static int minimumLectures(int m, int n)
    {
        int ans = 0;

        // Formula to compute
        if (n < (int)Math.ceil(0.75 * m))
            ans = (int)Math.ceil(((0.75 * m) - n) / 0.25);
        else
            ans = 0;

        return ans;
    }

    // Driver Code
    public static void main(String[] args)
    {
        int M = 9, N = 1;
        System.out.println(minimumLectures(M, N));
    }
}
Python
# Python Program to find minimum number of lectures to attend
# to maintain 75 % attendance

import math

# Function to compute minimum lecture
def minimumLecture(m, n):
    ans = 0

    # Formula to compute
    if(n < math.ceil(0.75 * m)):
        ans = math.ceil(((0.75 * m) - n) / 0.25)
    else:
        ans = 0
    return ans

# Driver Code

M = 9
N = 1

print(minimumLecture(M, N))
C#
// C# Program to find minimum
// number of lectures to attend 
// to maintain 75% attendance 
using System;

class GFG 
{ 

// Method to compute minimum lecture 
static int minimumLectures(int m, int n) 
{ 
    int ans = 0; 

    // Formula to compute 
    if (n < (int)Math.Ceiling(0.75 * m)) 
        ans = (int)Math.Ceiling(((0.75 * m) - 
                                 n) / 0.25); 
    else
        ans = 0; 

    return ans; 
} 

// Driver Code 
public static void Main() 
{ 
    int M = 9, N = 1; 
    Console.WriteLine(minimumLectures(M, N)); 
} 
} 

// This code is contributed 
// by anuj_67
PHP
<?php
// PHP Program to find minimum
// number of lectures to attend
// to maintain 75% attendance

// Function to compute minimum lecture
function minimumLectures($m, $n)
{
    $ans = 0;

    // Formula to compute
    if ($n < ceil(0.75 * $m))
        $ans = (int)ceil(((0.75 * $m) -
                          $n) / 0.25);
    else
        $ans = 0;

    return $ans;
}

// Driver Code
$M = 9; $N = 1;
echo minimumLectures($M, $N);

// This code is contributed 
// by anuj_67
?>
JavaScript
<script>
    // Javascript Program to find minimum
    // number of lectures to attend 
    // to maintain 75% attendance 
    
    // Method to compute minimum lecture 
    function minimumLectures(m, n) 
    { 
        let ans = 0; 

        // Formula to compute 
        if (n < Math.ceil(0.75 * m)) 
            ans = Math.ceil(((0.75 * m) - n) / 0.25); 
        else
            ans = 0; 

        return ans; 
    } 
    
    let M = 9, N = 1; 
    document.write(minimumLectures(M, N)); 
    
</script>

Output: 

23

Time Complexity: O(1)

Auxiliary Space: O(1) 


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