Introduction

In this blog post, we will explore how to solve the LeetCode problem "3161" 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 exists an infinite number line, with its origin at 0 and extending towards the positive x-axis. You are given a 2D array queries, which contains two types of queries: For a query of type 1, queries[i] = [1, x]. Build an obstacle at distance x from the origin. It is guaranteed that there is no obstacle at distance x when the query is asked. For a query of type 2, queries[i] = [2, x, sz]. Check if it is possible to place a block of size sz anywhere in the range [0, x] on the line, such that the block entirely lies in the range [0, x]. A block cannot be placed if it intersects with any obstacle, but it may touch it. Note that you do not actually place the block. Queries are separate. Return a boolean array results, where results[i] is true if you can place the block specified in the ith query of type 2, and false otherwise. Example 1: Input: queries = [[1,2],[2,3,3],[2,3,1],[2,2,2]] Output: [false,true,true] Explanation: For query 0, place an obstacle at x = 2. A block of size at most 2 can be placed before x = 3. Example 2: Input: queries = [[1,7],[2,7,6],[1,2],[2,7,5],[2,7,6]] Output: [true,true,false] Explanation: Place an obstacle at x = 7 for query 0. A block of size at most 7 can be placed before x = 7. Place an obstacle at x = 2 for query 2. Now, a block of size at most 5 can be placed before x = 7, and a block of size at most 2 before x = 2. Constraints: 1 <= queries.length <= 15 104 2 <= queries[i].length <= 3 1 <= queries[i][0] <= 2 1 <= x, sz <= min(5 104, 3 * queries.length) The input is generated such that for queries of type 1, no obstacle exists at distance x when the query is asked. The input is generated such that there is at least one query of type 2.

Explanation

Here's the breakdown of the problem and an efficient solution:

• Key Idea: Store the obstacles in a sorted manner (e.g., using a set). For each type 2 query, iterate through the obstacles within the range [0, x] to find the largest gap. Check if this gap is greater than or equal to the block size.

• Optimization: Use a set to maintain the obstacles. The set allows for efficient insertion and ordered retrieval of obstacles, which is crucial for finding the largest gap within the specified range.

• Complexity:

  • Runtime: O(N log N) where N is the number of queries. This comes from the possible logN insertion cost into the set of obstacles. Type 2 queries are answered in O(M log M) worst case where M is the number of obstacles, but each obstacle is added only once, making the cumulative time complexity O(N log N).
  • Storage: O(N) to store the obstacles in the set and the results array.

Code

    def solve():
    queries = eval(input())
    obstacles = set()
    results = []

    for query in queries:
        if query[0] == 1:
            obstacles.add(query[1])
        else:
            x = query[1]
            sz = query[2]

            obstacles_in_range = sorted([obs for obs in obstacles if obs <= x])

            if not obstacles_in_range:
                if x >= sz:
                    results.append(True)
                else:
                    results.append(False)
                continue

            max_gap = obstacles_in_range[0]

            for i in range(1, len(obstacles_in_range)):
                max_gap = max(max_gap, obstacles_in_range[i] - obstacles_in_range[i - 1])

            max_gap = max(max_gap, x - obstacles_in_range[-1])

            if max_gap >= sz:
                results.append(True)
            else:
                results.append(False)

    print(results)

solve()