Introduction

In this blog post, we will explore how to solve the LeetCode problem "3439" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given an integer eventTime denoting the duration of an event, where the event occurs from time t = 0 to time t = eventTime. You are also given two integer arrays startTime and endTime, each of length n. These represent the start and end time of n non-overlapping meetings, where the ith meeting occurs during the time [startTime[i], endTime[i]]. You can reschedule at most k meetings by moving their start time while maintaining the same duration, to maximize the longest continuous period of free time during the event. The relative order of all the meetings should stay the same and they should remain non-overlapping. Return the maximum amount of free time possible after rearranging the meetings. Note that the meetings can not be rescheduled to a time outside the event. Example 1: Input: eventTime = 5, k = 1, startTime = [1,3], endTime = [2,5] Output: 2 Explanation: Reschedule the meeting at [1, 2] to [2, 3], leaving no meetings during the time [0, 2]. Example 2: Input: eventTime = 10, k = 1, startTime = [0,2,9], endTime = [1,4,10] Output: 6 Explanation: Reschedule the meeting at [2, 4] to [1, 3], leaving no meetings during the time [3, 9]. Example 3: Input: eventTime = 5, k = 2, startTime = [0,1,2,3,4], endTime = [1,2,3,4,5] Output: 0 Explanation: There is no time during the event not occupied by meetings. Constraints: 1 <= eventTime <= 109 n == startTime.length == endTime.length 2 <= n <= 105 1 <= k <= n 0 <= startTime[i] < endTime[i] <= eventTime endTime[i] <= startTime[i + 1] where i lies in the range [0, n - 2].

Explanation

Here's the breakdown of the problem and the optimal solution:

• High-Level Approach:

  • Calculate initial gaps between meetings. These gaps represent potential free time.
  • Iterate through all possible subarrays of gaps. For each subarray, determine the minimum number of meetings that need to be rescheduled to merge the gaps into a single free time period.
  • Maximize the length of the merged free time period, ensuring that the number of rescheduled meetings does not exceed k.

• Complexity:

  • Runtime Complexity: O(n²)
  • Storage Complexity: O(n)

Code

    def max_free_time(eventTime: int, k: int, startTime: list[int], endTime: list[int]) -> int:
    """
    Calculates the maximum amount of free time possible after rescheduling at most k meetings.

    Args:
        eventTime: The duration of the event.
        k: The maximum number of meetings that can be rescheduled.
        startTime: A list of the start times of the meetings.
        endTime: A list of the end times of the meetings.

    Returns:
        The maximum amount of free time possible after rearranging the meetings.
    """

    n = len(startTime)
    gaps = []
    gaps.append(startTime[0])  # Gap before the first meeting

    for i in range(n - 1):
        gaps.append(startTime[i + 1] - endTime[i])

    gaps.append(eventTime - endTime[n - 1])  # Gap after the last meeting

    max_free = 0
    for i in range(len(gaps)):
        for j in range(i, len(gaps)):
            reschedules_needed = 0
            total_free_time = 0

            # Count reschedules and accumulate free time in the range [i, j]
            for l in range(i, j):
                if gaps[l] == 0:
                    reschedules_needed +=1
                else:
                    total_free_time += gaps[l]

            if reschedules_needed <= k:
                total_free_time += gaps[j]
                max_free = max(max_free, total_free_time)

    return max_free