tomatocream
1dec88aaf2
Add populate-notes.mjs that fetches problem descriptions and Python/C++ code stubs from LeetCode's GraphQL API. Populated all 197 NeetCode 150 note files with: - Problem description (examples, constraints) - Python code stub (function signature) - C++ code stub (function signature + includes) API responses cached in leetcode/.cache/leetcode/ for instant re-runs.
2.8 KiB
2.8 KiB
TODO 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 pointpoint = [x, y]to the data structure.int count(int[] point)Counts the number of ways to form axis-aligned squares with pointpoint = [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]]]
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 pointsConstraints:
point.length == 20 <= x, y <= 1000- At most
3000calls in total will be made toaddandcount.
TODO Approach
Write your approach here.
TODO 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)TODO C++
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);
*/