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Area of the largest rectangle possible from given coordinates
  • Difficulty Level : Basic
  • Last Updated : 08 Nov, 2020

Given two arrays arr1[] and arr2[] of N denoting N coordinates of the form (arr1[i], 0) and M positive integers denotingM coordinates of the form (arr2[j], 1) where 1 ≤ i ≤ N and 1≤ j ≤ M. The task is to find the area of the largest rectangle that can be formed using these coordinates. Print 0 if no rectangle can be formed.

Examples:

Input: arr1[] = {1, 2, 4}, arr2[] = {1, 3, 4}
Output: 3
Explanation: The largest rectangle possible is {(1, 0), (1, 1), (4, 0), (4, 1)}.
Therefore, area of rectangle = length * breadth = (4 – 1)*(1 – 0) = 3

Input: arr1[] = {1, 3, 5}, arr2[] = {4, 2, 10}
Output: 0
Explanation: No rectangle can be formed. Therefore, the answer is zero.

Naive Approach: The simplest approach is to traverse the given array arr1[] and for each i, traverse the points in arr2[] using a variable j. If arr1[i] is equal to arr2[j], then store the value arr1[i] in the separate array, say ans[]. After finding all such values, sort the ans[] array and print the maximum area as ans[L] – ans[0] where L is the index of the last element of ans[] as the breadth of the rectangle will always be 1.



Time Complexity: O(N*M)
Auxiliary Space: O(M + N)

Efficient Approach: The idea is to use the Sorting Algorithm and Two-Pointer Technique. Observe that the breadth of the rectangle will always be 1. Therefore, the maximum area can be found by maximizing its length. Follow the below steps to solve the problem:

  • Sort the given arrays arr1[] and arr2[].
  • Initialize the variables, start and end with 0 to store the starting and ending point of the length. Also, initialize the variables i and j to traverse the array arr1[] and arr2[] respectively.
  • While i is smaller than N and j smaller than M, check the following conditions:
    • If arr1[i] is the same as arr2[j] and start is 0, update start as start = arr1[i]. Otherwise, update end as end = arr1[i] then increment i and j by 1.
    • If arr1[i] is greater than arr2[j], increment j by 1. Otherwise, increment i by 1.
  • If start or end is 0, print 0. Otherwise, print (end – start) as the maximum possible area.

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 find the maximum possible
// area of a rectangle
int largestArea(int arr1[], int n,
                int arr2[], int m)
{
    // Initialize variables
    int end = 0, start = 0, i = 0, j = 0;
 
    // Sort array arr1[]
    sort(arr1, arr1 + n);
 
    // Sort array arr2[]
    sort(arr2, arr2 + m);
 
    // Traverse arr1[] and arr2[]
    while (i < n and j < m) {
 
        // If arr1[i] is same as arr2[j]
        if (arr1[i] == arr2[j]) {
 
            // If no starting point
            // is found yet
            if (start == 0)
                start = arr1[i];
            else
                // Update maximum end
                end = arr1[i];
            i++;
            j++;
        }
 
        // If arr[i] > arr2[j]
        else if (arr1[i] > arr2[j])
            j++;
        else
            i++;
    }
 
    // If no rectangle is found
    if (end == 0 or start == 0)
        return 0;
    else
        // Return the area
        return (end - start);
}
 
// Driver Code
int main()
{
    // Given point
    int arr1[] = { 1, 2, 4 };
 
    // Given length
    int N = sizeof(arr1) / sizeof(arr1[0]);
 
    // Given points
    int arr2[] = { 1, 3, 4 };
 
    // Given length
    int M = sizeof(arr2) / sizeof(arr2[0]);
 
    // Function Call
    cout << largestArea(arr1, N, arr2, M);
 
    return 0;
}

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Java

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// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to find the maximum possible
// area of a rectangle
static int largestArea(int arr1[], int n,
                       int arr2[], int m)
{
     
    // Initialize variables
    int end = 0, start = 0, i = 0, j = 0;
 
    // Sort array arr1[]
    Arrays.sort(arr1);
 
    // Sort array arr2[]
    Arrays.sort(arr1);
 
    // Traverse arr1[] and arr2[]
    while (i < n && j < m)
    {
         
        // If arr1[i] is same as arr2[j]
        if (arr1[i] == arr2[j])
        {
             
            // If no starting point
            // is found yet
            if (start == 0)
                start = arr1[i];
            else
             
                // Update maximum end
                end = arr1[i];
                 
            i++;
            j++;
        }
 
        // If arr[i] > arr2[j]
        else if (arr1[i] > arr2[j])
            j++;
        else
            i++;
    }
 
    // If no rectangle is found
    if (end == 0 || start == 0)
        return 0;
    else
     
        // Return the area
        return (end - start);
}
 
// Driver Code
public static void main(String args[])
{
     
    // Given point
    int arr1[] = { 1, 2, 4 };
 
    // Given length
    int N = arr1.length;
 
    // Given points
    int arr2[] = { 1, 3, 4 };
 
    // Given length
    int M = arr2.length;
 
    // Function Call
    System.out.println(largestArea(arr1, N,
                                   arr2, M));
}
}
   
// This code is contributed by bolliranadheer

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Python3

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# Python3 program for the above approach
 
# Function to find the maximum possible
# area of a rectangle
def largestArea(arr1, n, arr2, m):
     
    # Initialize variables
    end = 0
    start = 0
    i = 0
    j = 0
 
    # Sort array arr1[]
    arr1.sort(reverse = False)
 
    # Sort array arr2[]
    arr2.sort(reverse = False)
 
    # Traverse arr1[] and arr2[]
    while (i < n and j < m):
         
        # If arr1[i] is same as arr2[j]
        if (arr1[i] == arr2[j]):
             
            # If no starting point
            # is found yet
            if (start == 0):
                start = arr1[i]
            else:
                 
                # Update maximum end
                end = arr1[i]
                 
            i += 1
            j += 1
 
        # If arr[i] > arr2[j]
        elif (arr1[i] > arr2[j]):
            j += 1
        else:
            i += 1
 
    # If no rectangle is found
    if (end == 0 or start == 0):
        return 0
    else:
         
        # Return the area
        return (end - start)
 
# Driver Code
if __name__ == '__main__':
     
    # Given point
    arr1 = [ 1, 2, 4 ]
 
    # Given length
    N = len(arr1)
 
    # Given points
    arr2 = [ 1, 3, 4 ]
 
    # Given length
    M =  len(arr2)
 
    # Function Call
    print(largestArea(arr1, N, arr2, M))
 
# This code is contributed by ipg2016107

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C#

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// C# program for the above approach
using System;
 
class GFG{
     
// Function to find the maximum possible
// area of a rectangle
static int largestArea(int[] arr1, int n,
                       int[] arr2, int m)
{
     
    // Initialize variables
    int end = 0, start = 0, i = 0, j = 0;
  
    // Sort array arr1[]
    Array.Sort(arr1);
     
    // Sort array arr2[]
    Array.Sort(arr2);
  
    // Traverse arr1[] and arr2[]
    while (i < n && j < m)
    {
         
        // If arr1[i] is same as arr2[j]
        if (arr1[i] == arr2[j])
        {
             
            // If no starting point
            // is found yet
            if (start == 0)
                start = arr1[i];
            else
             
                // Update maximum end
                end = arr1[i];
                  
            i++;
            j++;
        }
  
        // If arr[i] > arr2[j]
        else if (arr1[i] > arr2[j])
            j++;
        else
            i++;
    }
  
    // If no rectangle is found
    if (end == 0 || start == 0)
        return 0;
    else
      
        // Return the area
        return (end - start);
}
 
// Driver code
static void Main()
{
     
    // Given point
    int[] arr1 = { 1, 2, 4 };
  
    // Given length
    int N = arr1.Length;
  
    // Given points
    int[] arr2 = { 1, 3, 4 };
  
    // Given length
    int M = arr2.Length;
  
    // Function Call
    Console.WriteLine(largestArea(arr1, N,
                                  arr2, M));
}
}
 
// This code is contributed by divyeshrabadiya07

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Output: 

3










 

Time Complexity: O(N*log N + M*log M)
Auxiliary Space: O(M+N)

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