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.
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.
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:
intervals.length
intervals[i].length
start[i]
end[i]
Let’s look at some examples to better understand this problem:
Let’s take a short quiz to ensure we understand the problem correctly.
The meeting rooms problem on LeetCode
Can two meetings with times be scheduled without conflicts?
Yes
No
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:
Let’s look at the solution for this coding challenge.
def can_attend_meetings(intervals):# Sort meetings by start timeintervals.sort(key=lambda x: x[0])for i in range(len(intervals) - 1):# Check for overlapsif intervals[i][1] > intervals[i + 1][0]:return Falsereturn Truedef 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()
Can a meeting that starts at the same time another one ends be attended?
What data structures can be used to efficiently solve the meeting rooms problem?
Sorting the meetings takes
The algorithm uses a constant extra space for variables, regardless of the number of meetings. Therefore, the space complexity is
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