# Detect Squares Solution leetcode – 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.

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:Detect Squares Solution leetcode 2021

`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]]]OutputDetect Squares Solution leetcode 2021[null, null, null, null, 1, 0, null, 2]ExplanationDetectSquares 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
`5000`

calls**in total**will be made to`add`

and`count`

.