How do you solve Reschedule Meetings for Maximum Free Time II on LeetCode?

Reschedule Meetings for Maximum Free Time II (LeetCode #3440) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Gaps between meetings can be maximized by considering moving one meeting.

What is the time complexity of Reschedule Meetings for Maximum Free Time II?

The optimal solution for Reschedule Meetings for Maximum Free Time II runs in O(n log n) time and uses O(n) space. The sorting step takes O(n log n) time, and we use O(n) space to store the meetings and gaps.

What is the brute force solution for Reschedule Meetings for Maximum Free Time II?

The brute force approach for Reschedule Meetings for Maximum Free Time II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach tries to reschedule each meeting one by one and checks the maximum free time possible after each rescheduling. This method is straightforward but inefficient as it checks all possible combinations. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Reschedule Meetings for Maximum Free Time II?

Reschedule Meetings for Maximum Free Time II uses the following concepts: Array, Greedy, Enumeration. The recommended patterns to study are: Greedy, Sorting, Enumeration.

What are the interview tips for Reschedule Meetings for Maximum Free Time II?

Clarify the constraints and edge cases before diving into coding. Think about how sorting can simplify the problem. Discuss your thought process with the interviewer to showcase your understanding.

What are common mistakes when solving Reschedule Meetings for Maximum Free Time II?

Not considering edge cases where moving a meeting might overlap with the event boundaries. Forgetting to check both earlier and later movements of meetings.

#3440

Reschedule Meetings for Maximum Free Time II

Medium

ArrayGreedyEnumerationGreedySortingEnumeration

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

💡

Intuition

Time O(n log n)Space O(n)

The optimal solution sorts the meetings and then calculates the maximum free time by considering the gaps between meetings. It efficiently checks if moving one meeting can create a larger gap, maximizing free time.

⚙️

Algorithm

4 steps

1Step 1: Create a list of meetings as pairs of (startTime[i], endTime[i]).
2Step 2: Sort the meetings based on their end times.
3Step 3: Calculate the free time between consecutive meetings and find the maximum free time.
4Step 4: For each meeting, check if moving it to the previous or next free gap can increase the maximum free time.

solution.py30 lines

1# Full working Python code
2
3def maxFreeTime(eventTime, startTime, endTime):
4    meetings = sorted(zip(startTime, endTime))
5    max_free_time = 0
6    prev_end = 0
7    gaps = []
8
9    for start, end in meetings:
10        max_free_time = max(max_free_time, start - prev_end)
11        gaps.append((prev_end, start))
12        prev_end = end
13
14    # Check moving each meeting
15    for i in range(len(meetings)):
16        start, end = meetings[i]
17        duration = end - start
18        # Check moving earlier
19        if i > 0:
20            new_start = meetings[i - 1][1]
21            if new_start + duration <= eventTime:
22                max_free_time = max(max_free_time, new_start - gaps[i - 1][0])
23        # Check moving later
24        if i < len(meetings) - 1:
25            new_end = meetings[i + 1][0]
26            if new_end - duration >= 0:
27                max_free_time = max(max_free_time, gaps[i][1] - new_end)
28
29    return max_free_time
30

ℹ

Complexity note: The sorting step takes O(n log n) time, and we use O(n) space to store the meetings and gaps.

1Gaps between meetings can be maximized by considering moving one meeting.
2Sorting meetings helps in efficiently calculating free time.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.