3453. Separate Squares I - Detailed Explanation

Problem Statement

You are given a 2D integer array squares where each squares[i] = [xᵢ, yᵢ, lᵢ] describes a square of side length lᵢ whose bottom‑left corner is at (xᵢ, yᵢ). Find the minimum y‑coordinate Y of a horizontal line such that the total area of all squares strictly above that line equals the total area of all squares strictly below that line. Answers within 10⁻⁵ of the true value are accepted. Squares may overlap, and overlapping regions are counted for each square independently.

Examples

Example 1

Input: squares = [[0,0,1],[2,2,1]]
Output: 1.00000
Explanation: Any line Y with 1 ≤ Y ≤ 2 splits one unit of area above and one unit below. The minimum is Y = 1.

Example 2

Input: squares = [[0,0,2],[1,1,1]]
Output: 1.16667
Explanation: Total area = 4 + 1 = 5. Half is 2.5.  
For Y = 7/6 ≈ 1.16667:
  • Square 1 at (0,0) of side 2 contributes 
      below: (7/6−0)*2 = 14/6 ≈ 2.33333  
      above: (2−7/6)*2 = 10/6 ≈ 1.66667  
  • Square 2 at (1,1) of side 1 contributes 
      below: (7/6−1)*1 = 1/6 ≈ 0.16667  
      above: (2−7/6−?)*1 = 5/6 ≈ 0.83333  
  Totals both 2.5 exactly.

Example 3

Input: squares = [[0,0,4]]
Output: 2.00000
Explanation: A single square of area 16 is split in half when Y = 2.

Constraints

• 1 ≤ squares.length ≤ 5·10⁴
• squares[i].length == 3
• 0 ≤ xᵢ, yᵢ ≤ 10⁹
• 1 ≤ lᵢ ≤ 10⁹
• Sum of all lᵢ·lᵢ (total area) ≤ 10¹²

Hints

1. For a fixed line at Y, you can compute in O(n) the total area above and below that line by iterating through all squares.
2. As you raise Y, the area above the line decreases and the area below increases monotonically. That suggests a binary‑search on Y.
3. Alternatively, view each square as contributing a “weight” of lᵢ to the vertical span [yᵢ, yᵢ+lᵢ]. The area‐below function is then a piecewise‐linear integral of that weight; you can solve for the exact break‐point in one sweep.

Approach 1 Binary Search on Y

Explanation

Define for any candidate Y:

areaAbove(Y) = ∑ᵢ   { lᵢ²,                    if Y ≤ yᵢ
                       0,                     if Y ≥ yᵢ+lᵢ
                       (yᵢ+lᵢ − Y)·lᵢ,        otherwise }
areaBelow(Y) = ∑ᵢ   { lᵢ²,                    if Y ≥ yᵢ+lᵢ
                       0,                     if Y ≤ yᵢ
                       (Y − yᵢ)·lᵢ,           otherwise }

We want areaAbove(Y) == areaBelow(Y). Let

diff(Y) = areaAbove(Y) − areaBelow(Y).

As Y increases, areaAbove(Y) decreases and areaBelow(Y) increases, so diff(Y) is strictly decreasing. We can binary‐search in the range

low = minᵢ yᵢ,   high = maxᵢ (yᵢ + lᵢ)

to find the unique Y where diff(Y) = 0, within error ε = 10⁻⁶.

Step‑by‑step Walkthrough

1. Compute totalArea = ∑ lᵢ².
2. Initialize low = min(yᵢ), high = max(yᵢ + lᵢ).
3. Repeat ~60 times (enough for 10⁻⁵ accuracy):
   • mid = (low + high) / 2
   • Evaluate d = diff(mid)
   • If d > 0 (too much area above), move the line up: low = mid
   • Else move down: high = mid
4. Return low (or high).

Python Code