Introduction

In this blog post, we will explore how to solve the LeetCode problem "1942" 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

There is a party where n friends numbered from 0 to n - 1 are attending. There is an infinite number of chairs in this party that are numbered from 0 to infinity. When a friend arrives at the party, they sit on the unoccupied chair with the smallest number. For example, if chairs 0, 1, and 5 are occupied when a friend comes, they will sit on chair number 2. When a friend leaves the party, their chair becomes unoccupied at the moment they leave. If another friend arrives at that same moment, they can sit in that chair. You are given a 0-indexed 2D integer array times where times[i] = [arrivali, leavingi], indicating the arrival and leaving times of the ith friend respectively, and an integer targetFriend. All arrival times are distinct. Return the chair number that the friend numbered targetFriend will sit on. Example 1: Input: times = [[1,4],[2,3],[4,6]], targetFriend = 1 Output: 1 Explanation: - Friend 0 arrives at time 1 and sits on chair 0. - Friend 1 arrives at time 2 and sits on chair 1. - Friend 1 leaves at time 3 and chair 1 becomes empty. - Friend 0 leaves at time 4 and chair 0 becomes empty. - Friend 2 arrives at time 4 and sits on chair 0. Since friend 1 sat on chair 1, we return 1. Example 2: Input: times = [[3,10],[1,5],[2,6]], targetFriend = 0 Output: 2 Explanation: - Friend 1 arrives at time 1 and sits on chair 0. - Friend 2 arrives at time 2 and sits on chair 1. - Friend 0 arrives at time 3 and sits on chair 2. - Friend 1 leaves at time 5 and chair 0 becomes empty. - Friend 2 leaves at time 6 and chair 1 becomes empty. - Friend 0 leaves at time 10 and chair 2 becomes empty. Since friend 0 sat on chair 2, we return 2. Constraints: n == times.length 2 <= n <= 104 times[i].length == 2 1 <= arrivali < leavingi <= 105 0 <= targetFriend <= n - 1 Each arrivali time is distinct.

Explanation

Here's the breakdown of the solution:

• Sort by Arrival Time: Sort the friends based on their arrival times to simulate the party chronologically.
• Track Available Chairs: Use a min-heap (priority queue) to efficiently find the smallest available chair number. Initialize the heap with chair numbers 0 to n-1. Also, keep track of occupied chairs and their leaving times.

• Simulate and Assign: Iterate through the sorted friends. Before assigning a chair, check if any occupied chairs have become available (leaving time has passed). If so, add those chair numbers back to the min-heap. Assign the smallest available chair from the min-heap to the current friend. If the current friend is the target friend, store the assigned chair number and return it at the end.

• Runtime Complexity: O(n log n) due to sorting and heap operations. Storage Complexity: O(n) to store the sorted arrival times, occupied chairs, and the min-heap.

Code

    import heapq

def smallestChair(times, targetFriend):
    n = len(times)
    arrivals = sorted([(times[i][0], i) for i in range(n)])  # (arrival_time, friend_index)
    available_chairs = list(range(n))
    heapq.heapify(available_chairs)
    occupied_chairs = []  # (leaving_time, chair_number)
    friend_to_chair = {}

    for arrival_time, friend_index in arrivals:
        # Release occupied chairs
        while occupied_chairs and occupied_chairs[0][0] <= arrival_time:
            leaving_time, chair_number = heapq.heappop(occupied_chairs)
            heapq.heappush(available_chairs, chair_number)

        # Assign a chair
        chair_number = heapq.heappop(available_chairs)
        friend_to_chair[friend_index] = chair_number
        heapq.heappush(occupied_chairs, (times[friend_index][1], chair_number))

        if friend_index == targetFriend:
            return chair_number