Program to find minimum number of lectures to attend to maintain 75%
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 attendInput : 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 <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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