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

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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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Last Updated : 23 Aug, 2022
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