# Solution For Detect Squares

Problem statement: You are given a stream of points in the plane where points[i] = [xi, yi]. Each point [xi, yi] represents a point on the 2D plane. We want to know how many squares can be formed by these points.

A square is a quadrilateral with four equal sides and four equal angles (90-degree angles). All four sides and four angles are equal.

Solution:

Approach 1: Brute force

This approach involves finding all possible pairs of points and check if they form a square. To do this check, we calculate all possible distances between the points, and check if there exist two pairs of points that have equal distance and intersect at right angles. This approach is inefficient as the time complexity is O(n^4).

Approach 2: Hash Map

We can use a hash map to store the frequency of all pairs of points. Two pairs of points can form a square if they have equal distance and intersect at right angles. Therefore, we can calculate the distance and angle between each pair of points, and store them in the hash map using a tuple with distance and angle as key. We can then iterate through the hash map and check if there exist four points that can form a square.

For a pair of points with coordinates (x1, y1) and (x2, y2), the distance is the square root of the sum of the squares of the differences between the coordinates. The angle between the two points is the arctangent of the difference in y-coordinates over the difference in x-coordinates.

The time complexity of this approach is O(n^2 log n) since we have to calculate the distance and angle for each pair of points, and sort the hash maps.

Approach 3: Count occurrences of sides and diagonals

We can count the frequency of sides and diagonals with a hash map. For each pair of points, we calculate the distance between them. We add this distance to the hash map with the frequency of this distance. For a square, there should be two pairs of sides with equal distance and a pair of diagonals with equal distance.

We can iterate through the hash map and count the number of pairs of sides and diagonals that form a square. The time complexity of this approach is O(n^2 log n) since we have to calculate the distance for each pair of points, and sort the hash maps.

Approach 4: Count occurrences of diagonal endpoints

We can count the frequency of diagonal endpoints with a hash map. For each pair of points, we calculate the distance between them. We add this distance to the hash map with the frequency of this distance and the two endpoints as a tuple. For a square, there should be at least one pair of diagonal endpoints with equal distance.

We can iterate through the hash map and count the number of pairs of diagonal endpoints that form a square. The time complexity of this approach is O(n^2 log n) since we have to calculate the distance for each pair of points, and sort the hash maps.

Approach 5: Count occurrences of midpoints

We can count the frequency of midpoints with a hash map. For each pair of points, we calculate the midpoint. We add this midpoint to the hash map with the frequency of this midpoint. For a square, there should be four points with equal distance from the midpoint.

We can iterate through the hash map and count the number of quadruples of midpoints that form a square. The time complexity of this approach is O(n^2) since we only have to calculate the midpoint for each pair of points.

Approach 6: Count occurrences of slopes and intercepts of sides

We can count the frequency of slopes and intercepts of sides with a hash map. For each pair of points, we calculate the slope and intercept of the line passing through them. We add this slope-intercept pair to the hash map with the frequency of this pair. For a square, there should be two pairs of sides with the same slope and equal intercept.

We can iterate through the hash map and count the number of pairs of slope-intercept pairs that form a square. The time complexity of this approach is O(n^2) since we only have to calculate the slope and intercept for each pair of points.

Approach 7: Count occurrences of slope and midpoint

We can count the frequency of slopes and midpoint with a hash map. For each pair of points, we calculate the slope and midpoint of the line passing through them. We add this slope-midpoint pair to the hash map with the frequency of this pair. For a square, there should be one slope-intercept pair that forms the diagonal of the square and is perpendicular to two other sides.

We can iterate through the hash map and count the number of slope-midpoint pairs that form a square. The time complexity of this approach is O(n^2) since we only have to calculate the slope and midpoint for each pair of points.

Among these approaches, approach 5 and approach 6 are the most efficient with a time complexity of O(n^2) and space complexity of O(n), making them ideal for large datasets.

## Step by Step Implementation For Detect Squares

There are many ways to solve this problem. One way is to use a brute force approach, which is to check every possible square in the matrix. However, this is not very efficient. A better way to solve this problem is to use a more sophisticated algorithm, such as the one described in this paper: https://www.cs.cmu.edu/~ckingsf/bioinfo-lectures/linalg.pdf This algorithm is more efficient because it only needs to check a small number of potential squares.

def detectSquares(matrix): # Your code here return

var detectSquares = function(arr) { let squares = []; for (let i = 0; i < arr.length; i++) { for (let j = 0; j < arr[0].length; j++) { if (arr[i][j] === 1) { let k = 1; while (i + k < arr.length && j + k < arr[0].length && arr[i + k][j + k] === 1) { k++; } if (k > 1) { squares.push([i, j, k]); } } } } return squares; };

We can use a brute force approach to detect squares in a given image. For each pixel, we can check if there is a square of side length k that contains that pixel as its top-left corner. If so, we increment the count of squares by 1. The time complexity of this approach is O(nmk), where n and m are the dimensions of the image and k is the maximum side length of a square. The space complexity is O(nm). #include#include using namespace std; int detectSquares(vector >& image, int k) { int count = 0; for (int i = 0; i < image.size(); i++) { for (int j = 0; j < image[0].size(); j++) { if (image[i][j] == 1) { // check if there is a square of side length k that contains this pixel as its top-left corner int side = 1; while (side < k && i + side < image.size() && j + side < image[0].size()) { // check if all the pixels in this square are 1 bool isSquare = true; for (int x = 0; x < side; x++) { for (int y = 0; y < side; y++) { if (image[i + x][j + y] != 1) { isSquare = false; break; } } if (!isSquare) { break; } } if (isSquare) { count++; } side++; } } } } return count; } int main() { vector > image = {{1, 1, 0, 0, 0}, {1, 1, 0, 0, 0}, {0, 0, 1, 1, 1}, {0, 0, 1, 1, 1}, {0, 0, 1, 1, 1}}; int k = 2; cout << detectSquares(image, k) << endl; return 0; }

public class Solution { public IList> DetectSquares(int[][] matrix) { // check for empty matrix if (matrix == null || matrix.Length == 0) { return new List >(); } // get number of rows and columns int rows = matrix.Length; int cols = matrix[0].Length; // create a list to store the results List > result = new List >(); // iterate over the matrix for (int i = 0; i < rows; i++) { for (int j = 0; j < cols; j++) { // check if current element is 1 if (matrix[i][j] == 1) { // check if the element is part of a square if (isSquare(matrix, i, j)) { // if element is part of a square, add it to the result result.Add(new List () { i, j }); } } } } return result; } // helper function to check if an element is part of a square private bool isSquare(int[][] matrix, int i, int j) { // get number of rows and columns int rows = matrix.Length; int cols = matrix[0].Length; // variable to store the size of the square int size = 1; // check if element is part of a square of size 2 if (i + 1 < rows && j + 1 < cols && matrix[i+1][j] == 1 && matrix[i][j+1] == 1) { size = 2; } // check if element is part of a square of size 3 or more while (i + size < rows && j + size < cols) { // variable to keep track of whether all elements are 1 bool allOnes = true; // check all elements in the square for (int x = i; x <= i + size; x++) { for (int y = j; y <= j + size; y++) { // if any element is not 1, set allOnes to false and break if (matrix[x][y] != 1) { allOnes = false; break; } } // if allOnes is false, break if (!allOnes) { break; } } // if allOnes is still true, the element is part of a square if (allOnes) { return true; } // otherwise, try the next size size++; } // if we get to this point, the element is not part of a square return false; } }