Meeting Rooms II Leetcode Problem 253 [Python Solution]

If you’ve ever faced the challenge of scheduling meetings, you probably know how important it is to ensure that there are enough meeting rooms available to accommodate all your appointments.

The Meeting Rooms II problem on LeetCode (Problem 253) tackles this exact issue.

Given an array of meeting time intervals, our goal is to find the minimum number of conference rooms required to schedule these meetings without any overlaps.

In this blog post, we’ll delve into this problem, explore different approaches to solving it, and provide a Python solution that you can use to tackle it efficiently.

We’ll also discuss the time and space complexity of our solution and explain the reasoning behind our approach.

Problem Overview

Let’s start by understanding the problem in more detail.

We’re given a list of meeting time intervals, each represented as a pair of start and end times: [(s1, e1), (s2, e2), ...].

The important constraint is that the start time (si) is always less than the end time (ei).

Our task is to determine the minimum number of conference rooms required to accommodate all the meetings.

The key insight here is that we need to find the maximum number of overlapping meetings at any point in time.

If we can find this value, we’ll know how many conference rooms are needed.

Example 1:

Let’s illustrate the problem with an example:

Input: intervals = [(0, 30), (5, 10), (15, 20)]
Output: 2

In this scenario, we have three meetings:

Meeting 1: Start at 0, end at 30.
Meeting 2: Start at 5, end at 10.
Meeting 3: Start at 15, end at 20. To schedule these meetings, we need a minimum of two conference rooms.

The reason is that Meetings 2 and 3 can’t overlap with Meeting 1, so they must be scheduled in separate rooms.

Example 2:

Here’s another example:

Input: intervals = [(2, 7)]
Output: 1

In this case, we have only one meeting, and therefore, we need just one conference room.

The meeting starts at 2 and ends at 7, so there is no overlap with other meetings.

Now that we understand the problem, let’s explore how we can approach it efficiently.

Efficient Approach to Meeting Rooms II

To efficiently solve this problem, we need a well-thought-out strategy.

The basic idea is to keep track of the start and end times of meetings while traversing the intervals.

This approach allows us to determine how many meetings are happening simultaneously at any given moment.

We’ll start by sorting all the start times and end times in separate arrays.

Then, we’ll iterate through these arrays, maintaining a count of meetings in progress and keeping track of the maximum count.

This maximum count will represent the minimum number of conference rooms required.

But before we dive into the Python code solution, let’s understand the reasoning behind this approach.

Reasoning Behind Our Efficient Approach

The key insight here is to visualize the meetings as points on a timeline.

We’re interested in moments when meetings start and end, and we want to know how many meetings are happening at the same time.

Start by sorting all the start times and end times in separate arrays.

Sorting is crucial for efficiently comparing and tracking the points on the timeline.

Initialize two pointers, one for the start times and one for the end times, both initially set to 0.
Maintain a count variable to keep track of the number of meetings in progress.

Initialize it to 0.

Iterate through the start times and end times, always choosing the smaller value between them to advance our position on the timeline.
If we encounter a start time, it means a new meeting is beginning, so we increment the count.
If we encounter an end time, it means a meeting is ending, so we decrement the count.
At each step, update the result by taking the maximum of the current result and the count.

This ensures we keep track of the maximum number of overlapping meetings.

Continue this process until you’ve processed all start and end times.

The final result will represent the minimum number of conference rooms required, as it’s the maximum number of overlapping meetings at any point in time.

This approach is efficient, with a time complexity of O(n log n) due to the initial sorting of the start and end times.

Now, let’s put this approach into Python code.

Python Code Solution

Here’s the Python solution for the Meeting Rooms II problem:

View on GitHub ➹

def minMeetingRooms(intervals):
    start_times = []
    end_times = []

    # Populate start and end time arrays
    for start, end in intervals:
        start_times.append((start, 1))
        end_times.append((end, -1))

    # Sort the combined array based on time and event type
    time = start_times + end_times
    time.sort(key=lambda x: (x[0], x[1]))

    count = 0
    max_count = 0

    for t in time:
        count += t[1]
        max_count = max(max_count, count)

    return max_count

This Python function efficiently calculates the minimum number of conference rooms required for a given set of meeting intervals.

It follows the approach we discussed earlier, maintaining the count and updating the maximum count as we traverse the sorted timeline.

Time and Space Complexity

Let’s analyze the time and space complexity of our solution.

Time Complexity: The most time-consuming part of our solution is the initial sorting of the combined array of start and end times, which has a time complexity of O(n log n).

The subsequent loop through this array is linear, resulting in an overall time complexity of O(n log n).

Space Complexity: We create two arrays, start_times and end_times, each with a size of n, where n is the number of meeting intervals.

Additionally, we create the time array, which has a size of 2n.

Therefore, the space complexity is O(n).

Constraints to Consider

When dealing with this problem, it’s essential to consider the constraints provided in the prompt:

The start time (si) is always less than the end time (ei).
Meeting intervals are represented as pairs of start and end times.
The number of meeting intervals, ‘n’, can be as large as 10^4. Therefore, your solution should be efficient and avoid unnecessary operations.

Edge Cases a Valid Solution Must Consider

To ensure your solution covers all cases, let’s explore some edge cases that a valid solution should consider:

Empty Input: Handle cases where there are no meeting intervals.

In this case, the expected output is 0.

Non-overlapping Meetings: Test scenarios where all meetings are non-overlapping.

The result should be 1, as only one meeting room is needed.

All Meetings Overlapping: Consider cases where all meetings overlap.

In such situations, the number of meeting rooms required is equal to the total number of meetings.

By addressing these edge cases, your solution will be more robust and provide accurate results.

Conclusion

In this blog post, we explored the Meeting Rooms II problem (LeetCode Problem

253) and provided an efficient Python solution to find the minimum number of conference rooms required to schedule a set of meeting intervals without any overlaps.

We explained the reasoning behind our approach and discussed the time and space complexity of the solution.

Meeting scheduling is a common problem, and this algorithm can be valuable in various real-world scenarios, from conference room booking systems to time management applications.

By implementing this solution, you can optimize meeting room usage and ensure efficient scheduling.

If you have any questions, comments, or suggestions, please feel free to share them in the comments section below.

We encourage you to engage with the content, ask questions, and provide feedback.

Sharing knowledge and fostering discussion is what makes the programming community thrive.

To put your newfound knowledge to the test, you can practice this problem on the LeetCode platform by following this link.

Happy coding!