Meeting Rooms LeetCode

In this Educative Answer, we’ll work on the meeting rooms problem, which is a classical example of scheduling algorithms and time management skills. We can understand the meeting rooms scheduling problem by imagining that we’ve multiple meetings scheduled throughout the day, each represented by a specific time interval. Our task is to determine if a person can attend all these meetings without overlaps. For instance, if we have meetings from 9:00 a.m to 10:00 a.m. and another from 10:30 a.m to 11:30 a.m., we could attend both as there’s a gap. This problem is crucial for time management and scheduling optimization in various scenarios, from business meetings to event planning.

Problem statement

We’re given an array of meeting intervals, intervals where intervals[i] = [start[i], end[i]]. We have to determine if a person can attend all meetings without any overlap. Return TRUE if possible, otherwise return FALSE.

Constraints:

$0 \leq$ intervals.length $\leq 10^3$
intervals[i].length $= 2$ (because each cell consists of only two values—the start time and the end time)
$0 \leq$ start[i] $<$ end[i] $\leq 10^6$

Let’s look at some examples to better understand this problem:

Algorithm

We now have a clearer idea of the task at hand. To solve the meeting room scheduling problem, we can begin by sorting the meetings based on their start times. Then, we iterate through the sorted meetings, checking if a current meeting starts before the previous one ends. If this happens, it indicates a conflict, and attending all meetings becomes impossible. This is a common technique used in the greedy algorithm approach for interval scheduling problems.

Note: If one meeting ends exactly when the next one starts (i.e., end[i] == start[i+1]), it is not considered a conflict in this problem. A person can attend back-to-back meetings without any issue if the start time of the next meeting is exactly the end time of the previous one.

Now, let’s look at the following illustration to get a better understanding of the algorithm:

def can_attend_meetings(intervals):
    # Sort meetings by start time
    intervals.sort(key=lambda x: x[0])
    for i in range(len(intervals) - 1):
        # Check for overlaps
        if intervals[i][1] > intervals[i + 1][0]:
            return False
    return True
def main():
    test_cases = [
        [[0, 1], [2, 4], [1, 3], [5, 6]],  
        [[7, 10], [2, 4], [1, 3], [5, 6]], 
        [[0, 1], [1, 2], [2, 3], [3, 4]],  
        [[1, 2], [2, 3], [1, 3], [3, 4]],   
        [[0, 5], [1, 2], [3, 4], [6, 7]],  
        [[7, 10], [2, 4]] 
    ]
  
    for i, meetings in enumerate(test_cases):
        print(f"{i+1}. \tSchedule of meetings:",test_cases[i])
        if can_attend_meetings(meetings):
            print("\tWe can attend all meetings without conflicts.")
        else:
            print("\tThere are conflicts in the meeting schedule.")
        print("-" * 100)
if __name__ == "__main__":
    main()

Meeting Rooms in Python

Frequently asked questions

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Can the meeting rooms problem be solved using other algorithms?

Yes, while the greedy algorithm is commonly used, other approaches may be applicable depending on the constraints, such as dynamic programming for more complex variations.

What are the common challenges in solving the meeting rooms problem?

The main challenge is efficiently checking for overlapping intervals, which can become computationally expensive if the meetings aren’t sorted.

What real-life applications can the meeting rooms problem algorithm be used for?

This algorithm can be applied to event planning, class scheduling, business meetings, and any scenario involving resource allocation or time management.

How do you handle edge cases in the meeting rooms problem?

Edge cases include scenarios with no meetings (intervals.length = 0), meetings that start and end at the same time, or meetings where all intervals are non-overlapping. The algorithm is designed to handle these efficiently.

How does this meeting rooms problem's solution scale with an increasing number of meetings?

The solution scales well for up to a thousand meetings $(n ≤ 1000)$ , as the time complexity is $O(n \log n)$ due to sorting, making it efficient even for large datasets.

Meeting Rooms LeetCode

Problem statement

Knowledge test