# Solving Leetcode Interviews in Seconds with AI: Block Placement Queries


	# 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](https://www.chatmagic.app), 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
	```python
	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()
	```