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 3Input :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 squarevoid 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 codeint 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;} |
Java
// Java implemenataion of the above approachimport java.util.*;import java.util.Arrays.*;// Defining a class to implement equals and hashCode// methods for int[] as keyclass Array { public Integer[] arrayInstance; public void setArray(Integer[] a) { this.arrayInstance = a; } @Override public int hashCode() { final int prime = 31; int result = 1; result = prime * result + Arrays.deepHashCode(arrayInstance); return result; } @Override public boolean equals(Object obj) { if (this == obj) return true; if (obj == null) return false; if (getClass() != obj.getClass()) return false; Array other = (Array)obj; if (!Arrays.deepEquals(arrayInstance, other.arrayInstance)) return false; return true; }}class GFG { // find the largest square static void findLargestSquare(int[][] points, int n) { // map to store which points exist Map<Array, Integer> m = new HashMap<Array, Integer>(); // mark the available points for (int i = 0; i < n; i++) { Integer[] l1 = { points[i][0], points[i][1] }; Array a1 = new Array(); a1.setArray(l1); if (!m.containsKey(a1)) m.put(a1, 0); m.put(a1, m.get(a1) + 1); } 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 Integer[] l1 = { points[i][0], points[i][1] }; Array a1 = new Array(); a1.setArray(l1); m.put(a1, m.get(a1) - 1); for (int j = 0; j < n; j++) { Integer[] l2 = { points[j][0], points[j][1] }; Array a2 = new Array(); a2.setArray(l2); // remove the chosen point m.put(a2, m.get(a2) - 1); Integer[] l3 = { points[i][0], points[j][1] }; Integer[] l4 = { points[j][0], points[i][1] }; Array a3 = new Array(); a3.setArray(l3); Array a4 = new Array(); a4.setArray(l4); // 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(a3) && m.containsKey(a4)) { // 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.put(a2, m.get(a2) + 1); } // add the removed point m.put(a1, m.get(a1) + 1); } // display the largest square if (side != -1) System.out.println( "Side of the square is : " + side + ", \npoints of the square are " + x + ", " + y + " " + (x + side) + ", " + y + " " + (x) + ", " + (y + side) + " " + (x + side) + ", " + (y + side)); else System.out.println("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 |
Python3
# Python3 implemenataion of the above approach# find the largest squaredef 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 coden = 6# given pointspoints=[[ 1, 1 ],[ 4, 4 ],[ 3, 4 ],[ 4, 3 ],[ 1, 4 ],[ 4, 1 ] ]# find the largest squarefindLargestSquare(points, n)# This code is contributed by mohit kumar 29 |
C#
// C# implemenataion of the above approachusing System;using System.Collections.Generic;// Defining a class to implement Equals and GetHashCode// methods for int[] as keypublic 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 squarefunction 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 codelet n = 6;// given pointslet points = [[1, 1 ], [4, 4], [3, 4], [4, 3], [1, 4], [4, 1]];// find the largest squarefindLargestSquare(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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