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Find Four points such that they form a square whose sides are parallel to x and y axes

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  • Difficulty Level : Medium
  • Last Updated : 12 Aug, 2022

Given ‘n’ pair of points, the task is to find four points such that they form a square whose sides are parallel to x and y axes or print “No such square” otherwise. If more than one square is possible then choose the one with the maximum area.

Examples:

Input : n = 6, points = (1, 1), (4, 4), (3, 4), (4, 3), (1, 4), (4, 1)
Output : side of the square is : 3, points of the square are 1, 1 4, 1 1, 4 4, 4
Explanation: The points 1, 1 4, 1 1, 4 4, 4 form a square of side 3

Input :n= 6, points= (1, 1), (4, 5), (3, 4), (4, 3), (7, 4), (3, 1)
Output :No such square

Simple approach: Choose all possible pairs of points with four nested loops and then see if the points form a square which is parallel to principal axes. If yes then check if it is the largest square so far in terms of the area and store the result, and then display the result at the end of the program. Time Complexity: O(N^4) Efficient Approach: Create a nested loop for the top right and bottom left corner of the square and form a square with those two points, then check if the other two points were assumed actually exist. To check if a point exists or not, create a map and store the points on the map to reduce the time to check whether the points exist. Also, keep in check the largest square by area so far and print it in the end.

Below is the implementation of the above approach: 

CPP




// C++ implemenataion of the above approach
#include <bits/stdc++.h>
using namespace std;
 
// find the largest square
void findLargestSquare(long long int points[][2], int n)
{
    // map to store which points exist
    map<pair<long long int, long long int>, int> m;
 
    // mark the available points
    for (int i = 0; i < n; i++) {
        m[make_pair(points[i][0], points[i][1])]++;
    }
    long long int side = -1, x = -1, y = -1;
 
    // a nested loop to choose the opposite corners of square
    for (int i = 0; i < n; i++) {
 
        // remove the chosen point
        m[make_pair(points[i][0], points[i][1])]--;
        for (int j = 0; j < n; j++) {
 
            // remove the chosen point
            m[make_pair(points[j][0], points[j][1])]--;
 
            // check if the other two points exist
            if (i != j
                  && (points[i][0]-points[j][0]) == (points[i][1]-points[j][1])){
                if (m[make_pair(points[i][0], points[j][1])] > 0
                     && m[make_pair(points[j][0], points[i][1])] > 0) {
 
                    // if the square is largest then store it
                    if (side < abs(points[i][0] - points[j][0])
                         || (side == abs(points[i][0] - points[j][0])
                            && ((points[i][0] * points[i][0]
                                   + points[i][1] * points[i][1])
                                      < (x * x + y * y)))) {
                        x = points[i][0];
                        y = points[i][1];
                        side = abs(points[i][0] - points[j][0]);
                    }
                }
            }
 
            // add the removed point
            m[make_pair(points[j][0], points[j][1])]++;
        }
 
        // add the removed point
        m[make_pair(points[i][0], points[i][1])]++;
    }
 
    // display the largest square
    if (side != -1)
        cout << "Side of the square is : " << side
             << ", \npoints of the square are " << x << ", " << y
             << " "
             << (x + side) << ", " << y
             << " "
             << (x) << ", " << (y + side)
             << " "
             << (x + side) << ", " << (y + side) << endl;
    else
        cout << "No such square" << endl;
}
 
//Driver code
int main()
{
    int n = 6;
 
    // given points
    long long int points[n][2]
      = { { 1, 1 }, { 4, 4 }, { 3, 4 }, { 4, 3 }, { 1, 4 }, { 4, 1 } };
 
    // find the largest square
    findLargestSquare(points, n);
 
    return 0;
}

Python3




# Python3 implemenataion of the above approach
 
# find the largest square
def findLargestSquare(points,n):
     
    # map to store which points exist
    m = dict()
 
    # mark the available points
    for i in range(n):
        m[(points[i][0], points[i][1])] = \
        m.get((points[i][0], points[i][1]), 0) + 1
 
    side = -1
    x = -1
    y = -1
 
    # a nested loop to choose the opposite corners of square
    for i in range(n):
 
        # remove the chosen point
        m[(points[i][0], points[i][1])]-=1
        for j in range(n):
 
            # remove the chosen point
            m[(points[j][0], points[j][1])]-=1
 
            # check if the other two points exist
            if (i != j and (points[i][0]-points[j][0]) == \
                (points[i][1]-points[j][1])):
                if (m[(points[i][0], points[j][1])] > 0 and
                    m[(points[j][0], points[i][1])] > 0):
 
                    # if the square is largest then store it
                    if (side < abs(points[i][0] - points[j][0])
                        or (side == abs(points[i][0] - points[j][0])
                            and ((points[i][0] * points[i][0]
                                + points[i][1] * points[i][1])
                                    < (x * x + y * y)))):
                        x = points[i][0]
                        y = points[i][1]
                        side = abs(points[i][0] - points[j][0])
 
            # add the removed point
            m[(points[j][0], points[j][1])] += 1
 
        # add the removed point
        m[(points[i][0], points[i][1])] += 1
 
    # display the largest square
    if (side != -1):
        print("Side of the square is : ",side
            ,", \npoints of the square are ",x,", ",y
            ," "
            ,(x + side),", ",y
            ," "
            ,(x),", ",(y + side)
            ," "
            ,(x + side),", ",(y + side))
    else:
        print("No such square")
 
# Driver code
n = 6
 
# given points
points=[[ 1, 1 ],[ 4, 4 ],[ 3, 4 ],[ 4, 3 ],[ 1, 4 ],[ 4, 1 ] ]
 
# find the largest square
findLargestSquare(points, n)
 
# This code is contributed by mohit kumar 29

C#




// C# implemenataion of the above approach
using System;
using System.Collections.Generic;
 
// Defining a class to implement Equals and GetHashCode
// methods for int[] as key
public class MyEqualityComparer : IEqualityComparer<int[]>
{
    // This method compares two int[] arrays
    public bool Equals(int[] x, int[] y)
    {
        if (x.Length != y.Length)
        {
            return false;
        }
        for (int i = 0; i < x.Length; i++)
        {
            if (x[i] != y[i])
            {
                return false;
            }
        }
        return true;
    }
 
    // This method generates a hashcode for an int[] key
    public int GetHashCode(int[] obj)
    {
        int result = 17;
        for (int i = 0; i < obj.Length; i++)
        {
            unchecked
            {
                result = result * 23 + obj[i];
            }
        }
        return result;
    }
}
 
class GFG
{
  // find the largest square
  static void findLargestSquare(int[,] points, int n)
  {
    // map to store which points exist
    Dictionary<int[], int> m = new Dictionary<int[], int>(new MyEqualityComparer());
 
    // mark the available points
    for (int i = 0; i < n; i++) {
      int[] l1 = {points[i, 0], points[i, 1]};
      if (!m.ContainsKey(l1))
        m[l1] = 0;
      m[l1]++;
    }
    int side = -1, x = -1, y = -1;
 
    // a nested loop to choose the opposite corners of square
    for (int i = 0; i < n; i++) {
 
      // remove the chosen point
      int[] l1= {points[i, 0], points[i, 1]};
      m[l1]--;
      for (int j = 0; j < n; j++) {
 
        int[] l2 = {points[j,0], points[j,1]};
        // remove the chosen point
        m[l2]--;
 
        int[] l3 = {points[i,0], points[j,1]};
        int[] l4 = {points[j,0], points[i,1]};
 
        // check if the other two points exist
        if (i != j
            && (points[i,0]-points[j,0]) == (points[i,1]-points[j,1])){
          if (m.ContainsKey(l3)
              && m.ContainsKey(l4)) {
 
            // if the square is largest then store it
            if (side < Math.Abs(points[i,0] - points[j,0])
                || (side == Math.Abs(points[i,0] - points[j,0])
                    && ((points[i,0] * points[i,0]
                         + points[i,1] * points[i,1])
                        < (x * x + y * y)))) {
              x = points[i,0];
              y = points[i,1];
              side = Math.Abs(points[i,0] - points[j,0]);
            }
          }
        }
 
        // add the removed point
        m[l2]++;
      }
 
      // add the removed point
      m[l1]++;
    }
 
    // display the largest square
    if (side != -1)
      Console.WriteLine("Side of the square is : " + side +
                        ", \npoints of the square are " + x + ", " + y
                        + " "
                        + (x + side) + ", " + y
                        + " "
                        + (x) + ", " + (y + side)
                        + " "
                        + (x + side) + ", " + (y + side));
    else
      Console.WriteLine("No such square");
  }
 
  //Driver code
  public static void Main(string[] args)
  {
    int n = 6;
 
    // given points
    int[,] points = { { 1, 1 }, { 4, 4 }, { 3, 4 }, { 4, 3 }, { 1, 4 }, { 4, 1 } };
 
    // find the largest square
    findLargestSquare(points, n);
 
  }
}
 
 
// This code is contributed by phasing17

Javascript




// JavaScript implemenataion of the above approach
 
// find the largest square
function findLargestSquare(points, n)
{
    // map to store which points exist
    let m = new Map();
 
    // mark the available points
    for (let i = 0; i < n; i++) {
        if(m.has([points[i][0], points[i][1]].join())){
            m.set([points[i][0],  points[i][1]].join(), m.get([points[i][0], points[i][1]].join() + 1));
        }
        else{
            m.set([points[i][0], points[i][1]].join(), 1);
        }
    }
    let side = -1, x = -1, y = -1;
 
    // a nested loop to choose the opposite corners of square
    for (let i = 0; i < n; i++) {
 
        // remove the chosen point
        m.set([points[i][0], points[i][1]].join(), m.get([points[i][0], points[i][1]].join()) - 1);
 
        for (let j = 0; j < n; j++) {
 
            // remove the chosen point
            m.set([points[j][0], points[j][1]].join(), m.get([points[j][0], points[j][1]].join()) - 1);
 
            // check if the other two points exist
            if (i != j && (points[i][0]-points[j][0]) == (points[i][1]-points[j][1])){
                if (m.get([points[i][0], points[j][j]].join()) > 0 && m.get([points[j][0], points[i][1]].join()) > 0) {
                    // if the square is largest then store it
                    if (side < Math.abs(points[i][0] - points[j][0])
                         || (side == Math.abs(points[i][0] - points[j][0]) && ((points[i][0] * points[i][0] + points[i][1] * points[i][1]) < (x * x + y * y)))) {
                        x = points[i][0];
                        y = points[i][1];
                        side = Math.abs(points[i][0] - points[j][0]);
                    }
                }
            }
 
            // add the removed point
            m.set([points[j][0], points[j][1]].join(), m.get([points[j][0], points[j][1]].join()) + 1);
        }
 
        // add the removed point
        m.set([points[i][0], points[i][1]].join(), m.get([points[i][0], points[i][1]].join()) + 1);
    }
 
    // display the largest square
    if (side != -1){
        console.log("Side of the square is :", side, ", ");
        console.log("points of the square are", x, ",", y, x + side, ",", y, x, ",", y + side, x + side, ",", y + side);  
    }
    else
        console.log("No such square");
}
 
//Driver code
let n = 6;
 
// given points
let points = [[1, 1 ],  [4, 4], [3, 4], [4, 3], [1, 4], [4, 1]];
 
// find the largest square
findLargestSquare(points, n);
 
// The code is contributed by Nidhi goel

Output:

Side of the square is : 3, 
points of the square are 1, 1 4, 1 1, 4 4, 4

Time Complexity: O(N^2)
Auxiliary Space: O(N)


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