Introduction

In this blog post, we will explore how to solve the LeetCode problem "1353" 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 array of events where events[i] = [startDayi, endDayi]. Every event i starts at startDayi and ends at endDayi. You can attend an event i at any day d where startTimei <= d <= endTimei. You can only attend one event at any time d. Return the maximum number of events you can attend. Example 1: Input: events = [[1,2],[2,3],[3,4]] Output: 3 Explanation: You can attend all the three events. One way to attend them all is as shown. Attend the first event on day 1. Attend the second event on day 2. Attend the third event on day 3. Example 2: Input: events= [[1,2],[2,3],[3,4],[1,2]] Output: 4 Constraints: 1 <= events.length <= 105 events[i].length == 2 1 <= startDayi <= endDayi <= 105

Explanation

Here's an efficient solution to the maximum events attended problem:

• Greedy Approach with Priority Queue: Sort events by their start day. Iterate through each day, adding events that have started on or before that day to a priority queue (min-heap) ordered by their end day.
• Attend Events with Earliest Deadlines: For each day, attend the event in the priority queue with the earliest end day. This maximizes the remaining time for other events.

• Clean Up Expired Events: Before attending an event each day, remove expired events from the priority queue.

• Runtime Complexity: O(n log n), where n is the number of events (due to sorting and heap operations). Storage Complexity: O(n), for storing the priority queue.

Code

    import heapq

def maxEvents(events):
    """
    Calculates the maximum number of events that can be attended.

    Args:
        events: A list of events, where each event is a list [startDay, endDay].

    Returns:
        The maximum number of events that can be attended.
    """

    events.sort()  # Sort events by start day
    pq = []  # Min-heap to store events, ordered by end day
    event_idx = 0
    day = 1
    count = 0

    while event_idx < len(events) or pq:
        # Add events that have started on or before the current day
        while event_idx < len(events) and events[event_idx][0] <= day:
            heapq.heappush(pq, events[event_idx][1])  # Push end day to heap
            event_idx += 1

        # Remove events that have already ended
        while pq and pq[0] < day:
            heapq.heappop(pq)

        # Attend the event with the earliest end day
        if pq:
            heapq.heappop(pq)
            count += 1
            day += 1
        else:
            # If no events to attend today, advance to the next event's start day to save time
            if event_idx < len(events):
                day = events[event_idx][0]
            else:
                break # No more events to process

    return count