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Area of the largest rectangle formed by lines parallel to X and Y axis from given set of points
  • Last Updated : 04 Dec, 2020

Given an array arr[] consisting of N pair of integers representing coordinates of N points, the task is to find the area of the largest rectangle formed by straight lines drawn parallel to X and Y-axis rom given set of points.

Examples:

Input: arr[] = {{0, 0}, {1, 1}}
Output: 1
Explanation: The area of the largest rectangle is 1 formed by the coordinates (0, 0), (0, 1), (1, 0), (1, 1).

Input: arr[] = {{-2, 0}, {2, 0}, {4, 0}, {4, 2}}
Output: 8
Explanation: The area of the largest rectangle possible is 8 ( length = 4 and breadth = 2 ) by the coordinates (-2, 0), (2, 0), (2, 2), (-2, 2).

Approach: The problem can be solved using the sorting technique. Follow the steps below to solve the problem:



  1. Store X and Y coordinates in two different arrays, say x[] and y[].
  2. Sort the arrays x[] and y[].
  3. Find the maximum adjacent difference from both arrays and store in variables X_Max and Y_Max.
  4. The maximum rectangle area possible is the product of X_Max and Y_Max.

Below is the implementation of the above approach:

C++

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// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the area of the largest
// rectangle formed by lines parallel to X
// and Y axis from given set of points
int maxRectangle(vector<vector<int> > sequence,
                 int size)
{
    // Initialize two arrays
    long long int X_Cord[size], Y_Cord[size];
 
    // Store x and y coordinates
    for (int i = 0; i < size; i++) {
        X_Cord[i] = sequence[i][0];
        Y_Cord[i] = sequence[i][1];
    }
 
    // Sort arrays
    sort(X_Cord, X_Cord + size);
    sort(Y_Cord, Y_Cord + size);
 
    // Initialize max differences
    long long int X_Max = 0, Y_Max = 0;
 
    // Find max adjacent differences
    for (int i = 0; i < size - 1; i++) {
        X_Max = max(X_Max, X_Cord[i + 1]
                               - X_Cord[i]);
        Y_Max = max(Y_Max, Y_Cord[i + 1]
                               - Y_Cord[i]);
    }
 
    // Return answer
    return X_Max * Y_Max;
}
 
// Driver Code
int main()
{
    // Given points
    vector<vector<int> > point
        = { { -2, 0 }, { 2, 0 }, { 4, 0 }, { 4, 2 } };
 
    // Total points
    int n = point.size();
 
    // Function call
    cout << maxRectangle(point, n);
}

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Java

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// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to return the area of the largest
// rectangle formed by lines parallel to X
// and Y axis from given set of points
static int maxRectangle(int[][] sequence,
                        int size)
{
     
    // Initialize two arrays
    int[] X_Cord = new int[size];
    int[] Y_Cord = new int[size];
 
    // Store x and y coordinates
    for(int i = 0; i < size; i++)
    {
        X_Cord[i] = sequence[i][0];
        Y_Cord[i] = sequence[i][1];
    }
 
    // Sort arrays
    Arrays.sort(X_Cord);
    Arrays.sort(Y_Cord);
 
    // Initialize max differences
    int X_Max = 0, Y_Max = 0;
 
    // Find max adjacent differences
    for(int i = 0; i < size - 1; i++)
    {
        X_Max = Math.max(X_Max, X_Cord[i + 1] -
                                X_Cord[i]);
        Y_Max = Math.max(Y_Max, Y_Cord[i + 1] -
                                Y_Cord[i]);
    }
     
    // Return answer
    return X_Max * Y_Max;
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given points
    int[][] point = { { -2, 0 }, { 2, 0 },
                      { 4, 0 }, { 4, 2 } };
 
    // Total points
    int n = point.length;
 
    // Function call
    System.out.print(maxRectangle(point, n));
}
}
 
// This code is contributed by shikhasingrajput

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Python3

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# Python3 program for the
# above approach
 
# Function to return the
# area of the largest
# rectangle formed by lines
# parallel to X and Y axis
# from given set of points
def maxRectangle(sequence, size):
 
    # Initialize two arrays
    X_Cord = [0] * size
    Y_Cord = [0] * size
 
    # Store x and y coordinates
    for i in range(size):
        X_Cord[i] = sequence[i][0]
        Y_Cord[i] = sequence[i][1]
 
    # Sort arrays
    X_Cord.sort()
    Y_Cord.sort()
 
    # Initialize max differences
    X_Max = 0
    Y_Max = 0
 
    # Find max adjacent differences
    for i in range(size - 1):
        X_Max = max(X_Max,
                    X_Cord[i + 1] -
                    X_Cord[i])
        Y_Max = max(Y_Max,
                    Y_Cord[i + 1] -
                    Y_Cord[i])
 
    # Return answer
    return X_Max * Y_Max
 
# Driver Code
if __name__ == "__main__":
   
    # Given points
    point = [[-2, 0], [2, 0],
             [4, 0], [4, 2]]
 
    # Total points
    n = len(point)
 
    # Function call
    print(maxRectangle(point, n))
 
# This code is contributed by Chitranayal

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C#

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// C# program for the above approach
using System;
 
class GFG{
 
// Function to return the area of the largest
// rectangle formed by lines parallel to X
// and Y axis from given set of points
static int maxRectangle(int[,] sequence,
                        int size)
{
     
    // Initialize two arrays
    int[] X_Cord = new int[size];
    int[] Y_Cord = new int[size];
     
    // Store x and y coordinates
    for(int i = 0; i < size; i++)
    {
        X_Cord[i] = sequence[i, 0];
        Y_Cord[i] = sequence[i, 1];
    }
 
    // Sort arrays
    Array.Sort(X_Cord);
    Array.Sort(Y_Cord);
 
    // Initialize max differences
    int X_Max = 0, Y_Max = 0;
 
    // Find max adjacent differences
    for(int i = 0; i < size - 1; i++)
    {
        X_Max = Math.Max(X_Max, X_Cord[i + 1] -
                                X_Cord[i]);
        Y_Max = Math.Max(Y_Max, Y_Cord[i + 1] -
                                Y_Cord[i]);
    }
     
    // Return answer
    return X_Max * Y_Max;
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given points
    int[,] point = { { -2, 0 }, { 2, 0 },
                     { 4, 0 }, { 4, 2 } };
 
    // Total points
    int n = point.GetLength(0);
 
    // Function call
    Console.Write(maxRectangle(point, n));
}
}
 
// This code is contributed by shikhasingrajput

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Output: 

8

 

Time Complexity: O(N*log(N)) where N is the total number of points.
Auxiliary Space: O(N) 

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