> For the complete documentation index, see [llms.txt](https://jaywin.gitbook.io/leetcode/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://jaywin.gitbook.io/leetcode/solutions/2013-detect-squares.md).

# 2013. Detect Squares

<https://leetcode.com/problems/detect-squares>

## Description

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`.

## ac

```java
```


---

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