Naive Approach

The simplest approach is to iterate through each meeting, reschedule it to every possible valid time slot, and calculate the maximum free time for each rescheduling. We keep track of the overall maximum free time found so far.

1. Iterate through each meeting:
   • For each meeting i, store its original start and end times.
2. Iterate through possible new start times:
   • Check all possible start times from 0 up to eventTime.
   • Ensure the new start time allows the meeting to finish within the event time.
   • Check for overlaps with other meetings. If no overlap, reschedule the meeting.
3. Calculate Maximum Free Time: After rescheduling, sort the meetings by start time and calculate the longest continuous free time.
4. Update Maximum: Update the overall maximum free time if the current rescheduling results in a longer free period.
5. Revert: Put the meeting i back to it's original start and end times.

This approach is extremely inefficient due to its nested loops and overlap checks.

• Big(O) Runtime: O(n² * eventTime * n log n), where n is the number of meetings. The n² comes from the outer and inner loops, the eventTime comes from the possible reschedule locations, and the n log n comes from sorting to determine the longest free time.
• Big(O) Space Usage: O(n) due to storing meeting times.

Optimal Approach

The optimal approach involves these steps:

1. Calculate initial gaps: First, calculate the free time gaps between the given meetings without rescheduling. Sort the meetings by start time.
2. Consider rescheduling each meeting: Iterate through each meeting and consider rescheduling it to fill the largest gap found so far.
3. Find the largest gap: Identify the largest gap. Reschedule a meeting into the start of this gap.
4. Calculate new maximum free time: Find longest free time by merging the two longest free times if possible

• Big(O) Runtime: O(n log n), where n is the number of meetings. This is primarily due to sorting the meetings.
• Big(O) Space Usage: O(n) due to storing meeting times.

Edge Cases

• No meetings: If there are no meetings, the maximum free time is simply eventTime.
• Meetings cover the entire event time: If the meetings cover the entire event time with no gaps, the maximum free time is 0.
• Single meeting: With only one meeting, the free time is from 0 to start time and from end time to eventTime.
• Meetings already optimally scheduled: The initial arrangement might already provide the maximum possible free time.

Code (Python)

def solve():
    eventTime = 10
    startTime = [0, 3, 7, 9]
    endTime = [1, 4, 8, 10]

    n = len(startTime)
    meetings = sorted(zip(startTime, endTime))

    def calculate_max_free_time(meetings_list):
        meetings_list.sort()
        max_free = max(meetings_list[0][0], eventTime - meetings_list[-1][1])
        for i in range(len(meetings_list) - 1):
            max_free = max(max_free, meetings_list[i+1][0] - meetings_list[i][1])
        return max_free

    max_free_time = calculate_max_free_time(meetings)

    for i in range(n):
        original_start, original_end = meetings[i]

        # Try moving the meeting to the beginning
        new_meetings = meetings[:i] + meetings[i+1:]
        new_start = 0
        new_end = original_end - original_start
        valid = True
        for start, end in new_meetings:
            if not (new_end <= start or new_start >= end):
                valid = False
                break
        if valid:
            max_free_time = max(max_free_time, calculate_max_free_time(new_meetings + [(new_start, new_end)]))

        # Try moving the meeting to the end
        new_meetings = meetings[:i] + meetings[i+1:]
        new_start = eventTime - (original_end - original_start)
        new_end = eventTime
        valid = True
        for start, end in new_meetings:
            if not (new_end <= start or new_start >= end):
                valid = False
                break
        if valid:
            max_free_time = max(max_free_time, calculate_max_free_time(new_meetings + [(new_start, new_end)]))

        # Try moving the meeting to other spots
        for j in range(len(meetings)):
             if j != i:
                new_meetings = meetings[:i] + meetings[i+1:]
                new_start = meetings[j][1]
                new_end = new_start + (original_end - original_start)

                if new_end > eventTime:
                    continue

                valid = True
                for start, end in new_meetings:
                    if not (new_end <= start or new_start >= end):
                        valid = False
                        break

                if valid:
                    max_free_time = max(max_free_time, calculate_max_free_time(new_meetings + [(new_start, new_end)]))




    print(max_free_time)


solve()