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Find minimum area of rectangle with given set of coordinates

  • Difficulty Level : Medium
  • Last Updated : 25 May, 2021

Given an array arr    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 0    .
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 X    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 <bits/stdc++.h>
using namespace std;
 
// function to find minimum area of Rectangle
int minAreaRect(vector<vector<int>> 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<pair<int,int>,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<vector<int>> 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<Integer> 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




<script>
/*package whatever //do not write package name here */
// Javascript Implementation of above approach
     
    //# function to find minimum area of Rectangle
    function minAreaRect(points)
    {
        // creating empty columns
    let columns = new Set();
  
    // fill columns with coordinates
    for (let point=0;point<points.length;point++)
      columns.add(40001 * points[point][0] + points[point][1]);
  
    let ans = Number.MAX_VALUE;
    for (let i = 0; i < points.length; ++i)
      for (let j = i + 1; j < points.length; ++j) {
        if (points[i][0] != points[j][0]
            && points[i][1] != points[j][1]) {
          if (columns.has(40001
                               * points[i][0]
                               + points[j][1])
              && columns.has(
                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 < Number.MAX_VALUE ? ans : 0;
    }
     
    // Driver code
    let A = [[1, 1], [1, 3], [3, 1], [3, 3], [2, 2]];
    document.write(minAreaRect(A));
     
     
// This code is contributed by patel2127
</script>
Output: 
4

 

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