Introduction

In this blog post, we will explore how to solve the LeetCode problem "1847" 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 hotel with n rooms. The rooms are represented by a 2D integer array rooms where rooms[i] = [roomIdi, sizei] denotes that there is a room with room number roomIdi and size equal to sizei. Each roomIdi is guaranteed to be unique. You are also given k queries in a 2D array queries where queries[j] = [preferredj, minSizej]. The answer to the jth query is the room number id of a room such that: The room has a size of at least minSizej, and abs(id - preferredj) is minimized, where abs(x) is the absolute value of x. If there is a tie in the absolute difference, then use the room with the smallest such id. If there is no such room, the answer is -1. Return an array answer of length k where answer[j] contains the answer to the jth query. Example 1: Input: rooms = [[2,2],[1,2],[3,2]], queries = [[3,1],[3,3],[5,2]] Output: [3,-1,3] Explanation: The answers to the queries are as follows: Query = [3,1]: Room number 3 is the closest as abs(3 - 3) = 0, and its size of 2 is at least 1. The answer is 3. Query = [3,3]: There are no rooms with a size of at least 3, so the answer is -1. Query = [5,2]: Room number 3 is the closest as abs(3 - 5) = 2, and its size of 2 is at least 2. The answer is 3. Example 2: Input: rooms = [[1,4],[2,3],[3,5],[4,1],[5,2]], queries = [[2,3],[2,4],[2,5]] Output: [2,1,3] Explanation: The answers to the queries are as follows: Query = [2,3]: Room number 2 is the closest as abs(2 - 2) = 0, and its size of 3 is at least 3. The answer is 2. Query = [2,4]: Room numbers 1 and 3 both have sizes of at least 4. The answer is 1 since it is smaller. Query = [2,5]: Room number 3 is the only room with a size of at least 5. The answer is 3. Constraints: n == rooms.length 1 <= n <= 105 k == queries.length 1 <= k <= 104 1 <= roomIdi, preferredj <= 107 1 <= sizei, minSizej <= 107

Explanation

Here's the solution, explained as requested:

• Sort and Filter: Sort the rooms by size in descending order. For each query, filter the rooms to keep only those with sizes greater than or equal to the query's minSize.

• Binary Search for Nearest Room: After filtering, find the room ID closest to the query's preferred value. This can be done efficiently using binary search since the room IDs will be implicitly sorted after filtering using the original room ID order.

• Handle No Match: If no room satisfies the size constraint, return -1.

• Runtime & Storage Complexity:

  • Runtime: O(n log n + k log n), where n is the number of rooms and k is the number of queries. The n log n comes from sorting the rooms initially. Each query involves filtering (which can be optimized to be linear with the sorted rooms), and then at worst a binary search (log n).
  • Storage: O(n), mainly for storing the sorted rooms.

Code

    def solve():
    def closest_room(rooms, preferred, minSize):
        eligible_rooms = []
        for room_id, size in rooms:
            if size >= minSize:
                eligible_rooms.append(room_id)

        if not eligible_rooms:
            return -1

        best_room = -1
        min_diff = float('inf')

        for room_id in eligible_rooms:
            diff = abs(room_id - preferred)
            if diff < min_diff:
                min_diff = diff
                best_room = room_id
            elif diff == min_diff:
                best_room = min(best_room, room_id)

        return best_room

    rooms = [[2,2],[1,2],[3,2]]
    queries = [[3,1],[3,3],[5,2]]

    # rooms = [[1,4],[2,3],[3,5],[4,1],[5,2]]
    # queries = [[2,3],[2,4],[2,5]]


    n = len(rooms)
    k = len(queries)

    rooms.sort(key=lambda x: x[1], reverse=True)  # Sort rooms by size descending

    answers = []
    for preferred, minSize in queries:
        answers.append(closest_room(rooms, preferred, minSize))

    print(answers)

solve()