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Java-based LeetCode algorithm problem solutions, regularly updated

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2013\. Detect Squares

Medium

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:**

![](https://assets.leetcode.com/uploads/2021/09/01/image.png)

**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`.