Introduction

In this blog post, we will explore how to solve the LeetCode problem "3454" 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 covered by squares above the line equals the total area covered by squares below the line. Answers within 10-5 of the actual answer will be accepted. Note: Squares may overlap. Overlapping areas should be counted only once in this version. Example 1: Input: squares = [[0,0,1],[2,2,1]] Output: 1.00000 Explanation: Any horizontal line between y = 1 and y = 2 results in an equal split, with 1 square unit above and 1 square unit below. The minimum y-value is 1. Example 2: Input: squares = [[0,0,2],[1,1,1]] Output: 1.00000 Explanation: Since the blue square overlaps with the red square, it will not be counted again. Thus, the line y = 1 splits the squares into two equal parts. 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 1015.

Explanation

Here's the breakdown of the solution:

• Binary Search: Perform a binary search on the possible range of y-coordinates to find the optimal split. The lower bound is the minimum y-coordinate among all squares, and the upper bound is the maximum y-coordinate.

• Area Calculation: For a given y-coordinate (the mid-point in binary search), calculate the area of each square above and below the line. Since squares can overlap, we calculate the covered area by considering the intersections of the horizontal line with the squares.

• Comparison: Compare the total area above and below the line. Adjust the binary search range (lower or upper bound) based on whether the area above is greater or smaller than the area below.

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

Code

    def solve():
    squares = eval(input())

    def calculate_area(y):
        area_above = 0
        area_below = 0

        for x, y_bottom, l in squares:
            y_top = y_bottom + l

            if y >= y_top:
                area_below += l * l
            elif y <= y_bottom:
                area_above += l * l
            else:
                above_height = y_top - y
                below_height = y - y_bottom
                area_above += above_height * l
                area_below += below_height * l
        return area_above, area_below

    total_area = sum(l * l for _, _, l in squares)

    min_y = min(y for _, y, _ in squares)
    max_y = max(y + l for _, y, l in squares)

    low = float(min_y)
    high = float(max_y)

    for _ in range(100):
        mid = (low + high) / 2
        area_above, area_below = calculate_area(mid)

        if abs(area_above - area_below) < 1e-6:
            print(f"{mid:.5f}")
            return

        if area_above > area_below:
            low = mid
        else:
            high = mid

    print(f"{mid:.5f}")

solve()