Maximizing Meetings: An Efficient Scheduling Approach

In the realm of competitive programming, the "Maximum Meetings in a Room" problem is a classic exercise in optimizing scheduling. Today's Problem of the Day (POTD) on GeeksForGeeks presents an intriguing challenge: Given a set of meetings with specified start and end times, determine the maximum number of non-overlapping meetings that can be scheduled in a single room. The solution involves a clever use of greedy algorithms, focusing on selecting the most meetings that can be attended without conflicts.

Problem Description

The task requires arranging a series of meetings, each with a specified start and end time, in such a way that the maximum number of meetings can be accommodated in a single room without any overlap. The challenge is to choose the meetings in such a way that the total number of non-overlapping meetings is maximized.

Approach and Solution

To solve this problem, we can use a greedy algorithm that prioritizes meetings based on their end times. This approach ensures that we can schedule as many meetings as possible by always choosing the next meeting that finishes the earliest, thus freeing up the room for subsequent meetings. Here's the implementation:

class Meeting {

public:

    int start;

    int end;

    int pos;

    Meeting() : start(0), end(0), pos(-1) {}

};

class Solution {

public:

    // Comparator function to sort meetings based on end time

    static bool compare(Meeting a, Meeting b) {

        return a.end < b.end;

    }

    int maxMeetings(int n, int start[], int end[]) {

        vector<Meeting> arrange(n);

        for (int i = 0; i < n; i++) {

            arrange[i].start = start[i];

            arrange[i].end = end[i];

            arrange[i].pos = i;

        }

        sort(arrange.begin(), arrange.end(), compare);

        int count = 1;  

        int freetime = arrange[0].end;

        for (int i = 1; i < n; i++) {

            if (arrange[i].start > freetime) {

                count++;

                freetime = arrange[i].end; 

            }

        }

        return count;

    }

};

```

        

BreakDown of the Solution :

1. Data Structure Setup:

- We define a Meeting class to store the start and end times of each meeting, along with its position in the input array.

2. Comparator Function:

- The compare function is a static method that sorts meetings based on their end times. This is crucial because, by finishing earlier, a meeting leaves the room free sooner for the next meeting.

3. Greedy Scheduling:

- After sorting the meetings, we initialize the count of meetings to 1 (the first meeting is always selected) and set freetime to the end time of the first meeting.

- We then iterate through the sorted meetings, and for each meeting, we check if its start time is greater than the freetime. If true, we can schedule this meeting, increment the count, and update freetime to the current meeting's end time.

4. Return the Result:

- The final count represents the maximum number of non-overlapping meetings that can be accommodated.

Conclusion

This problem exemplifies the power of greedy algorithms in optimization problems. By always selecting the meeting that finishes the earliest, we maximize the number of meetings that can be attended. The solution is efficient, with a time complexity dominated by the sorting step, i.e., \(O(n \log n)\). This approach is both intuitive and effective, making it a great practice problem for those looking to hone their algorithmic skills.

For anyone tackling scheduling problems, understanding and implementing greedy strategies like this one is essential. Not only does it provide a solution to a specific problem, but it also instills a way of thinking about optimizing resources and time management in a broader sense.

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Bonus Problem solutions (Stickler Theif)

First Approach :

Second Approach:

Third Approach:

Fourth and Best Approach:

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