Maximum Squares possible parallel to both axes from N distinct points

Last Updated : 12 Sep, 2022

Given X_i and Y_i 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 : 1
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 : 0
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 <bits/stdc++.h>
using namespace std;
 
// Function that returns the count of
// squares parallel to both X and Y-axis
// from a given set of points
int countSquares(int* X, int* Y, int N)
{
    // Initialize result
    int count = 0;
 
    // Initialize a set to store points
    set<pair<int, int> > points;
 
    // Initialize a map to store the
    // points in the same vertical line
    map<int, vector<int> > 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<int> 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 Code
int main()
{
    int X[] = { 0, 2, 0, 2 }, Y[] = { 0, 2, 2, 0 };
 
    int N = sizeof(X) / sizeof(X[0]);
 
    cout << countSquares(X, Y, N);
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
public class GFG
{
    // Function that returns the count of
    // squares parallel to both X and Y-axis
    // from a given set of points
    static int countSquares(int[] X, int[] Y, int N)
    {
        // Initialize result
        int count = 0;
     
        // Initialize a set to store points
        HashSet<String> points = new HashSet<String>();
     
        // Initialize a map to store the
        // points in the same vertical line
        HashMap<Integer, ArrayList<Integer> > vertical = new HashMap<Integer, ArrayList<Integer>>();
     
        // Store the points in a set 
        for (int i = 0; i < N; i++) {
            points.add( String.valueOf(X[i]) + "#" + String.valueOf(Y[i]));
        }
     
        // Store the points in the same vertical line
        // i.e. with same X co-ordinates
        for (int i = 0; i < N; i++) {
            if (!vertical.containsKey(X[i]))
                vertical.put(X[i], new ArrayList<Integer>());
            ArrayList<Integer> l1 = vertical.get(X[i]);
            l1.add(Y[i]);
            vertical.put(X[i], l1);
        }
     
        // Check for every two points
        // in the same vertical line
        for (var line : vertical.entrySet()) {
            int X1 = line.getKey();
            ArrayList<Integer> yList = line.getValue();
     
            for (int i = 0; i < yList.size(); i++) {
                int Y1 = yList.get(i);
                for (int j = i + 1; j < yList.size(); j++) {
                    int Y2 = yList.get(j);
                    int side = Math.abs(Y1 - Y2);
                    int X2 = X1 + side;
     
                    // Check if other two point are present or not
                    if (points.contains( String.valueOf(X2) + "#" + String.valueOf(Y1)) && points.contains( String.valueOf(X2) + "#" + String.valueOf(Y2))) 
                        count++;
                }
            }
        }
     
        return count;
    }
     
    // Driver Code
    public static void main(String[] args)
    {
        int[] X = { 0, 2, 0, 2 };
        int[] Y = { 0, 2, 2, 0 };
     
        int N = X.length;
     
       System.out.println(countSquares(X, Y, N));
     
    }
}
 
 
// This code is contributed by phasing17

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 points 
def 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 Code 
if __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

C#

// C# implementation of the approach
 
using System;
using System.Collections.Generic;
 
 
class GFG
{
    // Function that returns the count of
    // squares parallel to both X and Y-axis
    // from a given set of points
    static int countSquares(int[] X, int[] Y, int N)
    {
        // Initialize result
        int count = 0;
     
        // Initialize a set to store points
        HashSet<string> points = new HashSet<string>();
     
        // Initialize a map to store the
        // points in the same vertical line
        Dictionary<int, List<int> > vertical = new Dictionary<int, List<int>>();
     
        // Store the points in a set 
        for (int i = 0; i < N; i++) {
            points.Add( Convert.ToString(X[i]) + "#" + Convert.ToString(Y[i]));
        }
     
        // Store the points in the same vertical line
        // i.e. with same X co-ordinates
        for (int i = 0; i < N; i++) {
            if (!vertical.ContainsKey(X[i]))
                vertical[X[i]] = new List<int>();
            vertical[X[i]].Add(Y[i]);
        }
     
        // Check for every two points
        // in the same vertical line
        foreach (var line in vertical) {
            int X1 = line.Key;
            List<int> yList = line.Value;
     
            for (int i = 0; i < yList.Count; i++) {
                int Y1 = yList[i];
                for (int j = i + 1; j < yList.Count; j++) {
                    int Y2 = yList[j];
                    int side = Math.Abs(Y1 - Y2);
                    int X2 = X1 + side;
     
                    // Check if other two point are present or not
                    if (points.Contains( Convert.ToString(X2) + "#" + Convert.ToString(Y1)) && points.Contains( Convert.ToString(X2) + "#" + Convert.ToString(Y2))) 
                        count++;
                }
            }
        }
     
        return count;
    }
     
    // Driver Code
    public static void Main(string[] args)
    {
        int[] X = { 0, 2, 0, 2 };
        int[] Y = { 0, 2, 2, 0 };
     
        int N = X.Length;
     
       Console.WriteLine(countSquares(X, Y, N));
     
    }
}

Javascript

// JS implementation of the approach
 
 
// Function that returns the count of
// squares parallel to both X and Y-axis
// from a given set of points
function countSquares(X, Y, N)
{
    // Initialize result
    let count = 0;
 
    // Initialize a set to store points
    let points = new Set();
 
    // Initialize a map to store the
    // points in the same vertical line
    let vertical = {};
 
    // Store the points in a set 
    for (var i = 0; i < N; i++) {
        points.add(X[i] + "#" + Y[i]);
    }
 
    // Store the points in the same vertical line
    // i.e. with same X co-ordinates
    for (var i = 0; i < N; i++) {
        if (!vertical.hasOwnProperty(X[i]))
            vertical[X[i]] = [];
        vertical[X[i]].push(Y[i]);
    }
     
    // Check for every two points
    // in the same vertical line
    for (var [X1, yList] of Object.entries(vertical)) {
        X1 = parseInt(X1)
        for (var i = 0; i < yList.length; i++) {
            let Y1 = yList[i];
            for (let j = i + 1; j < yList.length; j++) {
                let Y2 = yList[j];
                let side = Math.abs(Y1 - Y2);
                let X2 = X1 + side;
                 
                let p1 = X2 + "#" + Y1
                let p2 = X2 + "#" + Y2
 
                // Check if other two point are present or not
                if (points.has(p1) && points.has(p2))
                    count++;
            }
        }
    }
     
 
 
    return count;
}
 
 
// Driver Code
let X = [ 0, 2, 0, 2 ]
let Y = [ 0, 2, 2, 0 ];
 
let N = X.length;
 
console.log(countSquares(X, Y, N));
 
 
 
// This code is contributed by phasing17