Minimum halls required for class scheduling

Given N lecture timings, with their start time and end time (both inclusive), the task is to find the minimum number of halls required to hold all the classes such that a single hall can be used for only one lecture at a given time. Note that the maximum end time can be 10⁵.

Examples:

Input: lectures[][] = {{0, 5}, {1, 2}, {1, 10}}
Output: 3
All lectures must be held in different halls because
at time instance 1 all lectures are ongoing.

Input: lectures[][] = {{0, 5}, {1, 2}, {6, 10}}
Output: 2

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

Assuming that time T starts with 0. The task is to find the maximum number of lectures that are ongoing at a particular instance of time. This will give the minimum number of halls required to schedule all the lectures.
To find the number of lectures ongoing at any instance of time. Maintain a prefix_sum[] which will store the number of lectures ongoing at any instance of time t. For any lecture with timings between [s, t], do prefix_sum[s]++ and prefix_sum[t + 1]–.
Afterward, the cumulative sum of this prefix array will give the count of lectures going on at any instance of time.
The maximum value for any time instant t in the array is the minimum number of halls required.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach 
#include <bits/stdc++.h> 
using namespace std; 
  
#define MAX 100001 
  
// Function to return the minimum 
// number of halls required 
int minHalls(int lectures[][2], int n) 
{ 
  
    // Array to store the number of 
    // lectures ongoing at time t 
    int prefix_sum[MAX] = { 0 }; 
  
    // For every lecture increment start 
    // point s decrement (end point + 1) 
    for (int i = 0; i < n; i++) { 
        prefix_sum[lectures[i][0]]++; 
        prefix_sum[lectures[i][1] + 1]--; 
    } 
  
    int ans = prefix_sum[0]; 
  
    // Perform prefix sum and update 
    // the ans to maximum 
    for (int i = 1; i < MAX; i++) { 
        prefix_sum[i] += prefix_sum[i - 1]; 
        ans = max(ans, prefix_sum[i]); 
    } 
  
    return ans; 
} 
  
// Driver code 
int main() 
{ 
    int lectures[][2] = { { 0, 5 }, 
                          { 1, 2 }, 
                          { 1, 10 } }; 
    int n = sizeof(lectures) / sizeof(lectures[0]); 
  
    cout << minHalls(lectures, n); 
  
    return 0; 
} 

Java

// Java implementation of the approach 
import java.util.*; 
  
class GFG  
{ 
static int MAX = 100001; 
  
// Function to return the minimum 
// number of halls required 
static int minHalls(int lectures[][], int n) 
{ 
  
    // Array to store the number of 
    // lectures ongoing at time t 
    int []prefix_sum = new int[MAX]; 
  
    // For every lecture increment start 
    // point s decrement (end point + 1) 
    for (int i = 0; i < n; i++)  
    { 
        prefix_sum[lectures[i][0]]++; 
        prefix_sum[lectures[i][1] + 1]--; 
    } 
  
    int ans = prefix_sum[0]; 
  
    // Perform prefix sum and update 
    // the ans to maximum 
    for (int i = 1; i < MAX; i++)  
    { 
        prefix_sum[i] += prefix_sum[i - 1]; 
        ans = Math.max(ans, prefix_sum[i]); 
    } 
    return ans; 
} 
  
// Driver code 
public static void main(String[] args) 
{ 
    int lectures[][] = {{ 0, 5 }, 
                        { 1, 2 }, 
                        { 1, 10 }}; 
    int n = lectures.length; 
  
    System.out.println(minHalls(lectures, n)); 
} 
} 
  
// This code is contributed by PrinciRaj1992 

Python3

# Python3 implementation of the approach  
MAX = 100001
  
# Function to return the minimum  
# number of halls required  
def minHalls(lectures, n) : 
  
    # Array to store the number of  
    # lectures ongoing at time t  
    prefix_sum = [0] * MAX; 
      
    # For every lecture increment start 
    # point s decrement (end point + 1) 
    for i in range(n) : 
        prefix_sum[lectures[i][0]] += 1; 
        prefix_sum[lectures[i][1] + 1] -= 1; 
          
    ans = prefix_sum[0]; 
      
    # Perform prefix sum and update 
    # the ans to maximum 
    for i in range(1, MAX) : 
        prefix_sum[i] += prefix_sum[i - 1]; 
        ans = max(ans, prefix_sum[i]); 
          
    return ans;  
  
# Driver code  
if __name__ == "__main__" :  
  
    lectures = [[ 0, 5 ],  
                [ 1, 2 ],  
                [ 1, 10 ]];  
                  
    n = len(lectures);  
  
    print(minHalls(lectures, n));  
  
# This code is contributed by AnkitRai01 

C#

// C# implementation of the approach 
using System; 
      
class GFG  
{ 
static int MAX = 100001; 
  
// Function to return the minimum 
// number of halls required 
static int minHalls(int [,]lectures, int n) 
{ 
  
    // Array to store the number of 
    // lectures ongoing at time t 
    int []prefix_sum = new int[MAX]; 
  
    // For every lecture increment start 
    // point s decrement (end point + 1) 
    for (int i = 0; i < n; i++)  
    { 
        prefix_sum[lectures[i,0]]++; 
        prefix_sum[lectures[i,1] + 1]--; 
    } 
  
    int ans = prefix_sum[0]; 
  
    // Perform prefix sum and update 
    // the ans to maximum 
    for (int i = 1; i < MAX; i++)  
    { 
        prefix_sum[i] += prefix_sum[i - 1]; 
        ans = Math.Max(ans, prefix_sum[i]); 
    } 
    return ans; 
} 
  
// Driver code 
public static void Main(String[] args) 
{ 
    int [,]lectures = {{ 0, 5 }, 
                       { 1, 2 }, 
                       { 1, 10 }}; 
    int n = lectures.GetLength(0); 
  
    Console.WriteLine(minHalls(lectures, n)); 
} 
} 
  
// This code is contributed by 29AjayKumar 

Output:

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

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