Find minimum area of rectangle with given set of coordinates

• Difficulty Level : Hard
• Last Updated : 24 May, 2022

Given an array of set of points in the X-Y plane. The task is to find the minimum area of a rectangle that can be formed from these points. The sides of the rectangle should be parallel to the X and Y axes. If a rectangle cannot be formed with the given points then print .
Examples:

Input: arr[][] = [[1, 1], [1, 3], [3, 1], [3, 3], [2, 2]]
Output:
The only rectangle possible will be formed with the points (1, 1), (1, 3), (3, 1) and (3, 3)
Input: arr[][] = [[1, 1], [1, 3], [3, 1], [3, 3], [4, 1], [4, 3]]
Output:

Approach: Group the points by coordinates, so that points on straight vertical lines are grouped together. Then, for every pair of points in a group, for eg. coordinates (X, Y1) and (X, Y2), we check for the smallest rectangle with this pair of points as the rightmost edge of the rectangle to be formed. We can do this by keeping track of all other pairs of points we’ve visited before. Finally return the minimum possible area of the rectangle obtained.
Below is the implementation of the above approach:

CPP

 `// C++ Implementation of above approach``#include ``using` `namespace` `std;` `// function to find minimum area of Rectangle``int` `minAreaRect(vector> A){` `    ``// creating empty columns``    ``map<``int``,vector<``int``>> columns;` `    ``// fill columns with coordinates``    ``for``(``auto` `i:A)``        ``columns[i[0]].push_back(i[1]);` `    ``map,``int` `> lastx;` `    ``int` `ans = INT_MAX;` `    ``for` `(``auto` `x:columns)``    ``{``        ``vector<``int``> column = x.second;``        ``sort(column.begin(), column.end());``        ``for` `(``int` `j = 0; j < column.size(); j++)``        ``{``            ``for` `(``int` `i = 0; i < j; i++)``            ``{``                ``int` `y1 = column[i];` `                ``// check if rectangle can be formed``                ``if` `(lastx.find({y1, column[j]}) != lastx.end())``                ``{``                    ``ans = min(ans, (x.first - lastx[{y1, column[j]}]) *``                            ``(column[j] - column[i]));``                ``}``                ``lastx[{y1, column[j]}] = x.first;``            ``}``            ``}``        ``}` `    ``if` `(ans < INT_MAX)``        ``return` `ans;``    ``else``        ``return` `0;``}` `// Driver code``int` `main()``{``    ``vector> A = {{1, 1}, {1, 3}, {3, 1}, {3, 3}, {2, 2}};``    ``cout << (minAreaRect(A));``    ``return` `0;``}` `// This code is contributed by mohit kumar 29`

Java

 `/*package whatever //do not write package name here */``// Java Implementation of above approach``import` `java.io.*;``import` `java.util.*;` `class` `GFG {` `  ``//# function to find minimum area of Rectangle``  ``public` `static` `int` `minAreaRect(``int``[][] points)``  ``{` `    ``// creating empty columns``    ``@SuppressWarnings``(``"unchecked"``)``    ``Set columns = ``new` `HashSet();` `    ``// fill columns with coordinates``    ``for` `(``int``[] point : points)``      ``columns.add(``40001` `* point[``0``] + point[``1``]);` `    ``int` `ans = Integer.MAX_VALUE;``    ``for` `(``int` `i = ``0``; i < points.length; ++i)``      ``for` `(``int` `j = i + ``1``; j < points.length; ++j) {``        ``if` `(points[i][``0``] != points[j][``0``]``            ``&& points[i][``1``] != points[j][``1``]) {``          ``if` `(columns.contains(``40001``                               ``* points[i][``0``]``                               ``+ points[j][``1``])``              ``&& columns.contains(``                ``40001` `* points[j][``0``]``                ``+ points[i][``1``])) {``            ``ans = Math.min( ans, Math.abs(points[j][``0``]``                            ``- points[i][``0``])``              ``* Math.abs(points[j][``1``]``                         ``- points[i][``1``]));``          ``}``        ``}``      ``}` `    ``return` `ans < Integer.MAX_VALUE ? ans : ``0``;``  ``}``  ` `  ``// Driver code``  ``public` `static` `void` `main(String[] args)``  ``{` `    ``int``[][] A = {{``1``, ``1``}, {``1``, ``3``}, {``3``, ``1``}, {``3``, ``3``}, {``2``, ``2``}};``    ``System.out.println(minAreaRect(A));``  ``}``}` `// This code is contributed by maheshwaripiyush9`

Python

 `# Python Implementation of above approach``import` `collections` `# function to find minimum area of Rectangle``def` `minAreaRect(A):` `    ``# creating empty columns``    ``columns ``=` `collections.defaultdict(``list``)` `    ``# fill columns with coordinates``    ``for` `x, y ``in` `A:``        ``columns[x].append(y)` `    ``lastx ``=` `{}``    ``ans ``=` `float``(``'inf'``)` `    ``for` `x ``in` `sorted``(columns):``        ``column ``=` `columns[x]``        ``column.sort()``        ``for` `j, y2 ``in` `enumerate``(column):``            ``for` `i ``in` `range``(j):``                ``y1 ``=` `column[i]` `                ``# check if rectangle can be formed``                ``if` `(y1, y2) ``in` `lastx:``                    ``ans ``=` `min``(ans, (x ``-` `lastx[y1, y2]) ``*` `(y2 ``-` `y1))``                ``lastx[y1, y2] ``=` `x` `    ``if` `ans < ``float``(``'inf'``):``        ``return` `ans``    ``else``:``        ``return` `0` `# Driver code``A ``=` `[[``1``, ``1``], [``1``, ``3``], [``3``, ``1``], [``3``, ``3``], [``2``, ``2``]]``print``(minAreaRect(A))`

Javascript

 ``

Output:

`4`

Time Complexity: O(N*N), as we are using nested loops.

Auxiliary Space: O(N*N), as we are using extra space.

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