cpp-flashcards/org/study_deck_02/dsa/math-geometry/2013-detect-squares.org

#+PROPERTY: STUDY_DECK_02
* TODO 2013. Detect Squares :medium:
:PROPERTIES:
:NEETCODE: [[file:../../roadmap.org::*2013. Detect Squares][2013. Detect Squares]]
:END:

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:*


#+begin_src
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
#+end_src


*Constraints:*

 - ~point.length == 2~

 - ~0 <= x, y <= 1000~

 - At most ~3000~ calls *in total* will be made to ~add~ and ~count~.

** TODO Approach
Write your approach here.

** TODO Python
#+begin_src python
class DetectSquares:

    def __init__(self):
        

    def add(self, point: List[int]) -> None:
        

    def count(self, point: List[int]) -> int:
        


# Your DetectSquares object will be instantiated and called as such:
# obj = DetectSquares()
# obj.add(point)
# param_2 = obj.count(point)
#+end_src

** TODO C++
#+begin_src cpp
class DetectSquares {
public:
    DetectSquares() {
        
    }
    
    void add(vector<int> point) {
        
    }
    
    int count(vector<int> point) {
        
    }
};

/**
 * Your DetectSquares object will be instantiated and called as such:
 * DetectSquares* obj = new DetectSquares();
 * obj->add(point);
 * int param_2 = obj->count(point);
 */
#+end_src