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Maximum Area Rectangle With Point Constraints II

Hard
100%
Updated 6/1/2025

Maximum Area Rectangle With Point Constraints II

What is this problem about?

This is the scaled-up Hard version of Part I. You are again asked to find the maximum area of an empty axis-aligned rectangle formed by points from the array. However, the number of points NN is massive (e.g., 10510^5). The O(N2)O(N^2) or O(N3)O(N^3) diagonal-scanning algorithms from Part I will result in an immediate Time Limit Exceeded. You must find an O(NlogN)O(N \log N) approach.

Why is this asked in interviews?

This is a highly advanced computational geometry problem. It is designed to test a candidate's mastery of the Sweep Line Algorithm and Segment Trees / Binary Indexed Trees. It is primarily seen in specialized algorithmic roles. It evaluates if you can process spatial data by moving an imaginary line across an axis, maintaining an active state of valid geometric bounds.

Algorithmic pattern used

This problem strictly requires a Sweep Line paired with a Segment Tree / Balanced BST.

  1. Group points by their X-coordinates and sort the unique X-coordinates to simulate a vertical line sweeping from left to right.
  2. For a specific X-coordinate, look at all Y-coordinates of points sitting on this line.
  3. To form an "empty" rectangle extending to the left, we look at the pairs of adjacent Y-coordinates on our current line. We query our active data structure (which stores historical Y-states) to find the closest previous X-coordinate that had points at these exact two Y-heights.
  4. Because the rectangle must be empty, we must use a Segment Tree (or Monotonic Stack logic depending on constraints) to verify that no points broke the boundary between the previous X and the current X.

Example explanation

Due to the immense complexity of Sweep Line, a simplified abstraction: Imagine moving a vertical scanner left to right. At X=1X = 1, we see points at Y=2Y = 2 and Y=5Y = 5. We record in a dictionary: the edge (Y: 2 to 5) was seen at X=1X = 1. At X=3X = 3, we see a point at Y=4Y = 4. It doesn't complete a rectangle, but it "poisons" the space for future rectangles spanning across X=3X=3. We record its presence in a segment tree. At X=5X = 5, we see points at Y=2Y = 2 and Y=5Y = 5. We check our dictionary: (Y: 2 to 5) was last seen at X=1X = 1. This forms a rectangle! We query the segment tree: "Were there any points between Y=2Y=2 and Y=5Y=5 inserted between X=1X=1 and X=5X=5?" The tree flags the point at (3,4)(3, 4). The rectangle is poisoned. We update the dictionary to state (Y: 2 to 5) is now seen at X=5X = 5 and keep sweeping.

Common mistakes candidates make

Candidates almost universally attempt to optimize the O(N2)O(N^2) solution using Hash Maps, failing to realize that worst-case inputs (like all points scattered randomly) still result in quadratic checks. Not recognizing that a massive NN in a geometry problem strictly dictates a Sweep Line algorithm is the primary failure point. Implementing the Segment Tree incorrectly, allowing points to "leak" through the query bounds, is the second most common error.

Interview preparation tip

If you face Maximum Area Rectangle With Point Constraints II, you are in a top-tier interview. Focus heavily on articulating the Sweep Line strategy verbally before coding. Explain how sorting by X reduces the 2D problem to 1D states, and how a Segment Tree acts as a fast query mechanism for "boundary contamination." Even if you don't finish the Segment Tree code, accurately designing the Sweep Line architecture is a massive pass signal.

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