Is Separate Squares II easy or hard?

Separate Squares II is classified as a Hard problem because it combines computational geometry, coordinate compression, sweep line processing, and segment trees. Implementing the data structure correctly and handling union area calculations requires strong algorithmic experience.

How to solve Separate Squares II in O(n log n)?

Convert each square into vertical sweep events and compress all x-coordinates. While sweeping along the y-axis, update the segment tree with interval add/remove operations. The root node stores the total covered x-length. Multiply that width by the vertical distance between events to accumulate union area and detect where half the area is reached.

What is the best approach for Separate Squares II?

The most efficient approach uses a line sweep combined with a segment tree to compute the union area of the squares. As the sweep moves vertically, active x-intervals are maintained in the segment tree to track the covered width. Area is accumulated between sweep events until half of the total area is reached. This runs in O(n log n) time.

Is Separate Squares II asked at Google/Amazon/Meta?

Problems involving line sweep, interval union, and segment trees frequently appear in interviews at companies like Google, Amazon, and Meta. Separate Squares II combines computational geometry with advanced data structures, making it a strong representation of hard interview questions.

What data structure is used in Separate Squares II?

A segment tree is used to maintain the total length of covered x-intervals during the sweep. Each node tracks how many intervals cover a range and the resulting union length. This allows fast range updates and real-time computation of the merged coverage width.

What is the time complexity of Separate Squares II?

The optimal solution runs in O(n log n) time and O(n) space. Sorting sweep events takes O(n log n), and each event updates a segment tree in O(log n). This structure efficiently maintains the union width of active intervals during the sweep.

Separate Squares II Python or Java solution approach

Both Python and Java implementations follow the same strategy: generate sweep events, compress x-coordinates, and maintain active coverage using a segment tree. Events are processed in sorted order of y. After each update, the tree provides the current covered width used to compute incremental union area.

Description Hints 2 Solution Video 51:32

3454. Separate Squares II

Hard59.2% acceptance2 companies

You are given a 2D integer array squares. Each squares[i] = [x_i, y_i, l_i] 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] = [x_i, y_i, l_i]
squares[i].length == 3
0 <= x_i, y_i <= 10⁹
1 <= l_i <= 10⁹
The total area of all the squares will not exceed 10¹⁵.

Asked at 2 companies

Google Amazon

Topics

Hints open one at a time

Hint 1

Use a line sweep and a segment tree.

Hint 2

The line must lie in one of the squares.

51:32

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