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

• Difficulty Level : Medium
• Last Updated : 23 Aug, 2022

Consider the subject Data Structures for which the total number of classes held till present date is , and some students attend only out of these classes. Find the minimum number of lectures they have to attend so that their 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,

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  #include  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

 

## Javascript

 

Output:

23

Time Complexity: O(1)

Auxiliary Space: O(1)

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