Meeting Rooms

class Solution {
public:
    bool isPossibleToAttend(vector<vector<int>>& timetable) {
        if (timetable.empty()) {
            return true;
        }

        // Sorting the timetable by start times
        sort(timetable.begin(), timetable.end());
        for (size_t index = 0; index < timetable.size() - 1; index++) {
            // Check for overlap
            if (timetable[index][1] > timetable[index + 1][0]) {
                return false;
            }
        }
        return true;
    }
};

The "Meeting Rooms" problem requires determining if a person can attend all meetings without any overlaps in their schedule. The solution involves checking the start and end times of the meetings.

• Start by checking if the timetable vector is empty. If it is, return true indicating that all meetings can be attended when no meetings are scheduled.

• Sort the timetable based on the start times of meetings using the sort() function. This ensures that the meetings are checked in the order they begin.

• Using a for loop, iterate through the sorted timetable and compare the end time of the current meeting to the start time of the next meeting.

• If the end time of a meeting is greater than the start time of the subsequent meeting, return false, indicating an overlap, hence it's not possible to attend all meetings.

• If the loop completes without finding any overlaps, return true, confirming that all meetings can be attended without any conflicts.

This approach effectively determines the possibility of attending all scheduled meetings in a given timetable by leveraging sorting and sequential comparison.

class Solution {
    public boolean canParticipate(int[][] times) {
        Arrays.sort(times, (first, second) -> Integer.compare(first[0], second[0]));
        for (int index = 0; index < times.length - 1; index++) {
            if (times[index][1] > times[index + 1][0]) {
                return false;
            }
        }
        return true;
    }
}

The Java solution provided addresses the problem of determining if a person can attend all meetings without any overlaps. The canParticipate method takes an array times where each sub-array represents the start and end time of a single meeting.

Steps of the solution:

1. Sort the times array based on the start time of the meetings. This helps in comparing each meeting with the next one in chronological order.
2. Iterate through the sorted array.
3. During each iteration, check if the current meeting's end time overlaps with the next meeting's start time.
4. If any overlap is found, return false. This indicates that attending all meetings is not possible.
5. If no overlaps are detected after checking all pairs of consecutive meetings, return true.

This approach ensures that the function quickly determines the feasibility of attending all the meetings, driven by efficient sorting and a single pass comparison.

class MeetingScheduler:
    def canSchedule(self, timeslots: List[List[int]]) -> bool:
        timeslots.sort()
        for j in range(len(timeslots) - 1):
            if timeslots[j][1] > timeslots[j + 1][0]:
                return False
        return True

The "Meeting Rooms" problem involves determining if a person can attend all meetings without any overlaps. The solution is implemented in Python and revolves around sorting the meeting intervals by their start times, then checking for overlaps.

Here's the summarized breakdown of how the solution works:

• First, the meeting time slots are sorted based on their starting times.
• Loop through the sorted list and check each meeting against the next one in the sequence.
• If the end time of the current meeting overlaps with the start time of the next meeting, the function returns False indicating a scheduling conflict.
• If no overlaps are found after checking all meeting pairs, the function returns True, confirming that all meetings can be attended without conflict.

This solution efficiently checks for possible meeting overlaps using a simplified for-loop and basic conditional statements, ensuring a straightforward approach to solving this common scheduling problem.