Introduction

In this blog post, we will explore how to solve the LeetCode problem "699" 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 are several squares being dropped onto the X-axis of a 2D plane. You are given a 2D integer array positions where positions[i] = [lefti, sideLengthi] represents the ith square with a side length of sideLengthi that is dropped with its left edge aligned with X-coordinate lefti. Each square is dropped one at a time from a height above any landed squares. It then falls downward (negative Y direction) until it either lands on the top side of another square or on the X-axis. A square brushing the left/right side of another square does not count as landing on it. Once it lands, it freezes in place and cannot be moved. After each square is dropped, you must record the height of the current tallest stack of squares. Return an integer array ans where ans[i] represents the height described above after dropping the ith square. Example 1: Input: positions = [[1,2],[2,3],[6,1]] Output: [2,5,5] Explanation: After the first drop, the tallest stack is square 1 with a height of 2. After the second drop, the tallest stack is squares 1 and 2 with a height of 5. After the third drop, the tallest stack is still squares 1 and 2 with a height of 5. Thus, we return an answer of [2, 5, 5]. Example 2: Input: positions = [[100,100],[200,100]] Output: [100,100] Explanation: After the first drop, the tallest stack is square 1 with a height of 100. After the second drop, the tallest stack is either square 1 or square 2, both with heights of 100. Thus, we return an answer of [100, 100]. Note that square 2 only brushes the right side of square 1, which does not count as landing on it. Constraints: 1 <= positions.length <= 1000 1 <= lefti <= 108 1 <= sideLengthi <= 106

Explanation

Here's the breakdown of the solution:

• Represent Squares: Each square is represented by its left edge, right edge, and height.
• Find Landing Height: For each new square, iterate through the existing landed squares to find the maximum landing height. This involves checking for overlap in the x-axis.

• Update Max Height: After each square is dropped, update the overall maximum height.

• Runtime Complexity: O(n^2), where n is the number of squares. Storage Complexity: O(n).

Code

    def fallingSquares(positions):
    """
    Calculates the height of the tallest stack of squares after each square is dropped.

    Args:
        positions: A list of lists, where each inner list represents a square with its left edge and side length.

    Returns:
        A list of integers representing the height of the tallest stack after each square is dropped.
    """

    landed = []  # List to store landed squares (left, right, height)
    ans = []
    max_height = 0

    for left, side in positions:
        right = left + side
        height = side  # Initial height of the current square

        # Find the maximum height the current square can land on
        for l, r, h in landed:
            if left < r and right > l:  # Check for overlap
                height = max(height, h + side)

        # Add the current square to the landed squares
        landed.append((left, right, height))

        # Update the maximum height
        max_height = max(max_height, height)
        ans.append(max_height)

    return ans