Count Axis-Aligned Squares in a Point Stream

### Problem Implement the persistent `DetectSquares` class. `add(point)` stores a new point occurrence, with duplicate coordinates treated as separate choices. `count(point)` returns how many selections of three stored occurrences form a positive-area axis-aligned square together with the supplied query point. The query point is an additional corner and need not have been added; counting does not insert it. ```cpp class DetectSquares { public: DetectSquares(); void add(vector<int> point); int count(vector<int> point); }; ``` ### Examples **Example 1:** ```text Calls: add([0,0]), add([0,2]), add([2,0]), count([2,2]) Results: void, void, void, 1 Explanation: The three stored corners complete a side-two square. ``` **Example 2:** ```text Calls on a fresh object: add([1,1]), add([1,3]), add([3,1]), add([3,1]), count([3,3]), count([1,1]) Results: void, void, void, void, 2, 0 Explanation: Either occurrence at [3,1] completes the first query; [3,3] is absent for the second. ``` ### Constraints - Every point has two coordinates in `[0,1000]`. - At most 3000 calls to add and count in total. ### Clarifying Questions Rotated squares and zero-area shapes do not count. Counting does not insert the query point. ```hint Inspect the structure If you choose the diagonal corner opposite the query, are the other two corners forced? ``` ### Reasoning checkpoints - Count only positive-area squares with sides parallel to the coordinate axes. - Preserve stored-point multiplicity and count each valid selection of three occurrences once. - Maintain the original persistent constructor and add/count protocol. ### Optional extensions How could you reduce query work when many distinct points have been stored?