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

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 top right and bottom left corner of the square and form a square with those two points, then check if the other two points which were assumed actually exist. To check if a point exists or not, create a map and store the points in 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; ` `} ` |

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**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)

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