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:
// 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;
} |
// Java implemenataion of the above approach import java.util.*;
import java.util.Arrays.*;
// Defining a class to implement equals and hashCode // methods for int[] as key class 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 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# 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 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 |
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)