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 

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: 

3

Time Complexity: O(MAX)
Auxiliary Space: O(MAX), where MAX is 100001.

Another Approach:

The above approach works when MAX is limited to 10⁵. When the limits of MAX are extended up to 10⁹, we cannot use above approach due to Memory Limit and Time Limit Constraints.

So, we think in a different dimension, towards sorting and cumulative sum. Store the lecture time (start time and end time) in chronological order, +1 denoting the start time of a lecture and -1 denoting the end time of a lecture. Then apply the concept of cumulative sum, this gives the maximum number of lectures being conducted at a time. This gives the bare minimum number of halls that are required.

Algorithm:

• Initialize a vector of pair, Time, the first value of which indicates the entry or exit time of lecture and the second value denotes whether the lecture starts or ends. 
• Traverse the lectures vector and store the values in the vector Time.
• Sort the vector Time.
• Maintain two variables answer = 0, and sum = 0. answer denotes the final answer to be returned and sum denotes the number of lectures going on at a particular time.
• Traverse the Time vector, add the second value of the pair into sum and update the answer variable.
• Return answer.

Below is the implementation of the above approach: 

3

Time Complexity: O(n*log(n))
Auxiliary Space: O(n)

Note that instead of vector of pair, one can use map or priority queue, the time complexity and space complexity would remain the same.