Introduction

In this blog post, we will explore how to solve the LeetCode problem "3453" 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

You are given a 2D integer array squares. Each squares[i] = [xi, yi, li] represents the coordinates of the bottom-left point and the side length of a square parallel to the x-axis. Find the minimum y-coordinate value of a horizontal line such that the total area of the squares above the line equals the total area of the squares below the line. Answers within 10-5 of the actual answer will be accepted. Note: Squares may overlap. Overlapping areas should be counted multiple times. Example 1: Input: squares = [[0,0,1],[2,2,1]] Output: 1.00000 Explanation: Any horizontal line between y = 1 and y = 2 will have 1 square unit above it and 1 square unit below it. The lowest option is 1. Example 2: Input: squares = [[0,0,2],[1,1,1]] Output: 1.16667 Explanation: The areas are: Below the line: 7/6 2 (Red) + 1/6 (Blue) = 15/6 = 2.5. Above the line: 5/6 2 (Red) + 5/6 (Blue) = 15/6 = 2.5. Since the areas above and below the line are equal, the output is 7/6 = 1.16667. Constraints: 1 <= squares.length <= 5 * 104 squares[i] = [xi, yi, li] squares[i].length == 3 0 <= xi, yi <= 109 1 <= li <= 109 The total area of all the squares will not exceed 1012.

Explanation

Here's a breakdown of the solution:

• Binary Search: The core idea is to use binary search on the possible range of y-coordinate values to find the horizontal line that balances the areas above and below.
• Area Calculation: For a given y-coordinate (mid in the binary search), calculate the total area of squares above the line and the total area of squares below the line. This involves iterating through each square and calculating the overlapping area based on the square's y-coordinate and side length.

• Adjustment: Based on whether the area above is greater or smaller than the area below, adjust the search range (high or low) accordingly.

• Runtime Complexity: O(N log(Y)), where N is the number of squares, and Y is the range of possible y-coordinates. Storage Complexity: O(1).

Code

    def find_minimum_y(squares):
    """
    Finds the minimum y-coordinate of a horizontal line such that the total area
    of the squares above the line equals the total area of the squares below the line.

    Args:
        squares: A 2D integer array representing the squares.

    Returns:
        The minimum y-coordinate value of the horizontal line.
    """

    def calculate_area_difference(y):
        """
        Calculates the difference between the total area above the line y and
        the total area below the line y.
        """
        area_above = 0
        area_below = 0
        for x, bottom_y, side in squares:
            if bottom_y + side <= y:
                area_below += side * side
            elif bottom_y >= y:
                area_above += side * side
            else:
                above_height = bottom_y + side - y
                below_height = y - bottom_y
                area_above += above_height * side
                area_below += below_height * side
        return area_above - area_below

    low = min(square[1] for square in squares)
    high = max(square[1] + square[2] for square in squares)

    while high - low > 1e-6:
        mid = (low + high) / 2
        difference = calculate_area_difference(mid)
        if difference > 0:
            low = mid  # Too much area above, lower the line
        else:
            high = mid  # Too much area below, raise the line

    return low