# 2013. 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:

Input
[[], [[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.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.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.

## Solutions

Solution 1: Hash Table

We can use a hash table $cnt$ to maintain all the information of the points, where $cnt[x][y]$ represents the count of point $(x, y)$.

When calling the $add(x, y)$ method, we increase the value of $cnt[x][y]$ by $1$.

When calling the $count(x_1, y_1)$ method, we need to get three other points to form an axis-aligned square. We can enumerate the point $(x_2, y_1)$ that is parallel to the $x$-axis and at a distance $d$ from $(x_1, y_1)$. If such a point exists, based on these two points, we can determine the other two points as $(x_1, y_1 + d)$ and $(x_2, y_1 + d)$, or $(x_1, y_1 - d)$ and $(x_2, y_1 - d)$. We can add up the number of schemes for these two situations.

In terms of time complexity, the time complexity of calling the $add(x, y)$ method is $O(1)$, and the time complexity of calling the $count(x_1, y_1)$ method is $O(n)$; the space complexity is $O(n)$. Here, $n$ is the number of points in the data stream.

• class DetectSquares {
private Map<Integer, Map<Integer, Integer>> cnt = new HashMap<>();

public DetectSquares() {
}

int x = point[0], y = point[1];
cnt.computeIfAbsent(x, k -> new HashMap<>()).merge(y, 1, Integer::sum);
}

public int count(int[] point) {
int x1 = point[0], y1 = point[1];
if (!cnt.containsKey(x1)) {
return 0;
}
int ans = 0;
for (var e : cnt.entrySet()) {
int x2 = e.getKey();
if (x2 != x1) {
int d = x2 - x1;
var cnt1 = cnt.get(x1);
var cnt2 = e.getValue();
ans += cnt2.getOrDefault(y1, 0) * cnt1.getOrDefault(y1 + d, 0)
* cnt2.getOrDefault(y1 + d, 0);
ans += cnt2.getOrDefault(y1, 0) * cnt1.getOrDefault(y1 - d, 0)
* cnt2.getOrDefault(y1 - d, 0);
}
}
return ans;
}
}

/**
* Your DetectSquares object will be instantiated and called as such:
* DetectSquares obj = new DetectSquares();
* int param_2 = obj.count(point);
*/

• class DetectSquares {
public:
DetectSquares() {
}

int x = point[0], y = point[1];
++cnt[x][y];
}

int count(vector<int> point) {
int x1 = point[0], y1 = point[1];
if (!cnt.count(x1)) {
return 0;
}
int ans = 0;
for (auto& [x2, cnt2] : cnt) {
if (x2 != x1) {
int d = x2 - x1;
auto& cnt1 = cnt[x1];
ans += cnt2[y1] * cnt1[y1 + d] * cnt2[y1 + d];
ans += cnt2[y1] * cnt1[y1 - d] * cnt2[y1 - d];
}
}
return ans;
}

private:
unordered_map<int, unordered_map<int, int>> cnt;
};

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

• class DetectSquares:
def __init__(self):
self.cnt = defaultdict(Counter)

def add(self, point: List[int]) -> None:
x, y = point
self.cnt[x][y] += 1

def count(self, point: List[int]) -> int:
x1, y1 = point
if x1 not in self.cnt:
return 0
ans = 0
for x2 in self.cnt.keys():
if x2 != x1:
d = x2 - x1
ans += self.cnt[x2][y1] * self.cnt[x1][y1 + d] * self.cnt[x2][y1 + d]
ans += self.cnt[x2][y1] * self.cnt[x1][y1 - d] * self.cnt[x2][y1 - d]
return ans

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


• type DetectSquares struct {
cnt map[int]map[int]int
}

func Constructor() DetectSquares {
return DetectSquares{map[int]map[int]int{} }
}

func (this *DetectSquares) Add(point []int) {
x, y := point[0], point[1]
if _, ok := this.cnt[x]; !ok {
this.cnt[x] = map[int]int{}
}
this.cnt[x][y]++
}

func (this *DetectSquares) Count(point []int) (ans int) {
x1, y1 := point[0], point[1]
if cnt1, ok := this.cnt[x1]; ok {
for x2, cnt2 := range this.cnt {
if x2 != x1 {
d := x2 - x1
ans += cnt2[y1] * cnt1[y1+d] * cnt2[y1+d]
ans += cnt2[y1] * cnt1[y1-d] * cnt2[y1-d]
}
}
}
return
}

/**
* Your DetectSquares object will be instantiated and called as such:
* obj := Constructor();