Maximum Squares possible parallel to both axes from N distinct points

• Last Updated : 26 Nov, 2021

Given Xi and Yi co-ordinate of N distinct points in the 2-D Plane, the task is to count the number of squares that can be formed from these points that are parallel to both X and Y-axis.
Examples:

Input : X[] = { 0, 2, 0, 2 }, Y[] = { 0, 2, 2, 0 }
Output :
Explanation: Only one Square can be formed using these points –
(0, 2)(2, 2)
(0, 0)(2, 0)
Input : X[] = { 3, 2, 0, 6 }, Y[] = { 0, 2, 0, 6 }
Output :
Explanation: No Square can be formed using these points –
(3,0),(2,0), (0,0), (6,6)

Naive Approach: Iterate over all possible combinations of four points and check if a square can be formed that is parallel to both X and Y-axis. The time complexity of this approach would be O(N4)
Efficient Approach: We can observe that for any four points to make a required square, the following conditions must hold true –

• The points lying in the same horizontal line must have the same Y co-ordinates.

• The points lying in the same vertical line must have the same X co-ordinates.

• The distance between these points should be same.

Thus the four points of any square can be written as P1(X1, Y1), P2(X2, Y1), P3(X2, Y2) and P4(X1, Y2) in clockwise direction.
Consider any two points on the same horizontal or vertical line from the given ones. Calculate the distance between them. Based on that, form the other two points. Now check if both of the given points are present or not. If so, it ensures that a square is present. Hence, increase the count and proceed further.
Below is the implementation of the above approach:

C++

 // C++ implementation of the approach#include using namespace std; // Function that returns the count of// squares parallel to both X and Y-axis// from a given set of pointsint countSquares(int* X, int* Y, int N){    // Initialize result    int count = 0;     // Initialize a set to store points    set > points;     // Initialize a map to store the    // points in the same vertical line    map > vertical;     // Store the points in a set    for (int i = 0; i < N; i++) {        points.insert({ X[i], Y[i] });    }     // Store the points in the same vertical line    // i.e. with same X co-ordinates    for (int i = 0; i < N; i++) {        vertical[X[i]].push_back(Y[i]);    }     // Check for every two points    // in the same vertical line    for (auto line : vertical) {        int X1 = line.first;        vector yList = line.second;         for (int i = 0; i < yList.size(); i++) {            int Y1 = yList[i];            for (int j = i + 1; j < yList.size(); j++) {                int Y2 = yList[j];                int side = abs(Y1 - Y2);                int X2 = X1 + side;                 // Check if other two point are present or not                if (points.find({ X2, Y1 }) != points.end()                and points.find({ X2, Y2 }) != points.end())                    count++;            }        }    }     return count;} // Driver Codeint main(){    int X[] = { 0, 2, 0, 2 }, Y[] = { 0, 2, 2, 0 };     int N = sizeof(X) / sizeof(X);     cout << countSquares(X, Y, N);     return 0;}

Python3

 # Python3 implementation of the approach # Function that returns the count of# squares parallel to both X and Y-axis# from a given set of pointsdef countSquares(X,  Y, N) :     # Initialize result    count = 0;     # Initialize a set to store points    points = [];     # Initialize a map to store the    # points in the same vertical line    vertical = dict.fromkeys(X, None);     # Store the points in a set    for i in range(N) :        points.append((X[i], Y[i]));     # Store the points in the same vertical line    # i.e. with same X co-ordinates    for i in range(N) :        if vertical[X[i]] is None :            vertical[X[i]] = [Y[i]];        else :            vertical[X[i]].append(Y[i]);     # Check for every two points    # in the same vertical line    for line in vertical :        X1 = line;        yList = vertical[line];         for i in range(len(yList)) :            Y1 = yList[i];            for j in range(i + 1, len(yList)) :                Y2 = yList[j];                side = abs(Y1 - Y2);                X2 = X1 + side;                 # Check if other two point are present or not                if ( X2, Y1 ) in points and ( X2, Y2 ) in points :                    count += 1;     return count; # Driver Codeif __name__ == "__main__" :     X = [ 0, 2, 0, 2 ]; Y = [ 0, 2, 2, 0 ];     N = len(X);     print(countSquares(X, Y, N)); # This code is contributed by AnkitRai01
Output:
1

Time Complexity: O(N2).

Auxiliary Space: O(N)

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