Detect Squares

You are given a stream of points on the X-Y plane. Design an algorithm that: - **Adds** new points from the stream into a data structure. **Duplicate** points are allowed and should be treated as different points. - Given a query point, **counts** the number of ways to choose three points from the data structure such that the three points and the query point form an **axis-aligned square** with **positive area**. An **axis-aligned square** is a square whose edges are all the same length and are either parallel or perpendicular to the x-axis and y-axis. Implement the `DetectSquares` class: - `DetectSquares()` Initializes the object with an empty data structure. - `void add(int[] point)` Adds a new point `point = [x, y]` to the data structure. - `int count(int[] point)` Counts the number of ways to form **axis-aligned squares** with point `point = [x, y]` as described above.   **Example 1:** **Input** ["DetectSquares", "add", "add", "add", "count", "count", "add", "count"] [[], [[3, 10]], [[11, 2]], [[3, 2]], [[11, 10]], [[14, 8]], [[11, 2]], [[11, 10]]] **Output** [null, null, null, null, 1, 0, null, 2] **Explanation** DetectSquares detectSquares = new DetectSquares(); detectSquares.add([3, 10]); detectSquares.add([11, 2]); detectSquares.add([3, 2]); detectSquares.count([11, 10]); // return 1. You can choose: // - The first, second, and third points detectSquares.count([14, 8]); // return 0. The query point cannot form a square with any points in the data structure. detectSquares.add([11, 2]); // Adding duplicate points is allowed. detectSquares.count([11, 10]); // return 2. You can choose: // - The first, second, and third points // - The first, third, and fourth points ```   **Constraints:** - `point.length == 2` - `0 <= x, y <= 1000` - At most `3000` calls **in total** will be made to `add` and `count`.