Maximum area rectangle by picking four sides from array

Given an array of n positive integers that represent lengths. Find out the maximum possible area whose four sides are picked from given array. Note that a rectangle can only be formed if there are two pairs of equal values in given array.

Examples:

Input : arr[] = {2, 1, 2, 5, 4, 4}
Output : 8
Explanation : Dimension will be 4 * 2

Input : arr[] = {2, 1, 3, 5, 4, 4}
Output : 0
Explanation : No rectangle possible

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Method 1 (Sorting)
The task basically reduces to finding two pairs of equal values in array. If there are more than two pairs, then pick the two pairs with maximum values. A simple solution is to do following.
1) Sort the given array.
2) Traverse array from largest to smallest value and return two pairs with maximum values.

C++

// CPP program for finding maximum area possible 
// of a rectangle 
#include <bits/stdc++.h> 
using namespace std; 
  
// function for finding max area 
int findArea(int arr[], int n) 
{ 
    // sort array in non-increasing order 
    sort(arr, arr + n, greater<int>()); 
  
    // Initialize two sides of rectangle 
    int dimension[2] = { 0, 0 }; 
  
    // traverse through array 
    for (int i = 0, j = 0; i < n - 1 && j < 2; i++) 
  
        // if any element occurs twice 
        // store that as dimension 
        if (arr[i] == arr[i + 1]) 
            dimension[j++] = arr[i++]; 
  
    // return the product of dimensions 
    return (dimension[0] * dimension[1]); 
} 
  
// driver function 
int main() 
{ 
    int arr[] = { 4, 2, 1, 4, 6, 6, 2, 5 }; 
    int n = sizeof(arr) / sizeof(arr[0]); 
    cout << findArea(arr, n); 
    return 0; 
} 

Java

// Java program for finding maximum area  
// possible of a rectangle 
import java.util.Arrays; 
import java.util.Collections; 
  
public class GFG  
{      
    // function for finding max area 
    static int findArea(Integer arr[], int n) 
    { 
        // sort array in non-increasing order 
        Arrays.sort(arr, Collections.reverseOrder()); 
       
        // Initialize two sides of rectangle 
        int[] dimension = { 0, 0 }; 
       
        // traverse through array 
        for (int i = 0, j = 0; i < n - 1 && j < 2;  
                                           i++) 
       
            // if any element occurs twice 
            // store that as dimension 
            if (arr[i] == arr[i + 1]) 
                dimension[j++] = arr[i++]; 
       
        // return the product of dimensions 
        return (dimension[0] * dimension[1]); 
    } 
       
    // driver function 
    public static void main(String args[]) 
    { 
        Integer arr[] = { 4, 2, 1, 4, 6, 6, 2, 5 }; 
        int n = arr.length; 
        System.out.println(findArea(arr, n)); 
    } 
} 
// This code is contributed by Sumit Ghosh 

Python3

# Python3 program for finding  
# maximum area possible of  
# a rectangle 
  
# function for finding 
# max area 
def findArea(arr, n): 
  
    # sort array in 
    # non-increasing order 
    arr.sort(reverse = True) 
  
    # Initialize two  
    # sides of rectangle 
    dimension = [0, 0] 
  
    # traverse through array 
    i = 0
    j = 0
    while(i < n - 1 and j < 2): 
  
        # if any element occurs twice 
        # store that as dimension 
        if (arr[i] == arr[i + 1]): 
            dimension[j] = arr[i] 
            j += 1
            i += 1
        i += 1
          
    # return the product  
    # of dimensions 
    return (dimension[0] * 
            dimension[1]) 
  
# Driver code 
arr = [4, 2, 1, 4, 6, 6, 2, 5] 
n = len(arr) 
print(findArea(arr, n)) 
  
# This code is contributed  
# by Smitha 

C#

// C# program for finding maximum area  
// possible of a rectangle  
using System; 
using System.Collections; 
  
class GFG  
{  
// function for finding max area  
static int findArea(int []arr, int n)  
{  
    // sort array in non-increasing order  
    Array.Sort(arr);  
    Array.Reverse(arr); 
  
    // Initialize two sides of rectangle  
    int[] dimension = { 0, 0 };  
  
    // traverse through array  
    for (int i = 0, j = 0;  
             i < n - 1 && j < 2; i++)  
  
        // if any element occurs twice  
        // store that as dimension  
        if (arr[i] == arr[i + 1])  
            dimension[j++] = arr[i++];  
  
    // return the product of dimensions  
    return (dimension[0] * dimension[1]);  
}  
  
// Driver Code 
public static void Main(String []args)  
{  
    int []arr = { 4, 2, 1, 4, 6, 6, 2, 5 };  
    int n = arr.Length;  
    Console.Write(findArea(arr, n));  
}  
}  
  
// This code is contributed  
// by PrinciRaj1992 

PHP

<?php 
// PHP program for finding maximum area possible 
// of a rectangle 
  
// function for finding max area 
function findArea($arr, $n) 
{ 
      
    // sort array in non- 
    // increasing order 
    rsort($arr); 
  
    // Initialize two sides  
    // of rectangle 
    $dimension = array( 0, 0 ); 
  
    // traverse through array 
    for( $i = 0, $j = 0; $i < $n - 1 &&  
                           $j < 2; $i++) 
  
        // if any element occurs twice 
        // store that as dimension 
        if ($arr[$i] == $arr[$i + 1]) 
            $dimension[$j++] = $arr[$i++]; 
  
    // return the product 
    // of dimensions 
    return ($dimension[0] *  
            $dimension[1]); 
} 
  
    // Driver Code 
    $arr = array(4, 2, 1, 4, 6, 6, 2, 5); 
    $n =count($arr); 
    echo findArea($arr, $n); 
      
// This code is contributed by anuj_67. 
?> 

Output:

Time Complexity : O(n Log n)

Method 2 (Hashing)
The idea is to insert all first occurrences of elements in a hash set. For second occurrences, keep track of maximum two values.

C++

// CPP program for finding maximum area possible 
// of a rectangle 
#include <bits/stdc++.h> 
using namespace std; 
  
// function for finding max area 
int findArea(int arr[], int n) 
{ 
    unordered_set<int> s; 
  
    // traverse through array 
    int first = 0, second = 0; 
    for (int i = 0; i < n; i++) { 
  
        // If this is first occurrence of arr[i], 
        // simply insert and continue 
        if (s.find(arr[i]) == s.end()) { 
            s.insert(arr[i]); 
            continue; 
        } 
  
        // If this is second (or more) occurrence, 
        // update first and second maximum values. 
        if (arr[i] > first) { 
            second = first; 
            first = arr[i]; 
        } else if (arr[i] > second) 
            second = arr[i]; 
    } 
  
    // return the product of dimensions 
    return (first * second); 
} 
  
// driver function 
int main() 
{ 
    int arr[] = { 4, 2, 1, 4, 6, 6, 2, 5 }; 
    int n = sizeof(arr) / sizeof(arr[0]); 
    cout << findArea(arr, n); 
    return 0; 
} 

Java

// Java program for finding maximum  
// area possible of a rectangle 
import java.util.HashSet; 
import java.util.Set; 
  
public class GFG  
{      
    // function for finding max area 
    static int findArea(int arr[], int n) 
    { 
        //unordered_set<int> s; 
          
        Set<Integer> s = new HashSet<>();  
       
        // traverse through array 
        int first = 0, second = 0; 
        for (int i = 0; i < n; i++) { 
       
            // If this is first occurrence of  
            // arr[i], simply insert and continue 
            if (!s.contains(arr[i])) { 
                s.add(arr[i]); 
                continue; 
            } 
       
            // If this is second (or more)  
            // occurrence, update first and  
            // second maximum values. 
            if (arr[i] > first) { 
                second = first; 
                first = arr[i]; 
            } else if (arr[i] > second) 
                second = arr[i]; 
        } 
       
        // return the product of dimensions 
        return (first * second); 
    } 
       
    // driver function 
    public static void main(String args[]) 
    { 
        int arr[] = { 4, 2, 1, 4, 6, 6, 2, 5 }; 
        int n = arr.length; 
        System.out.println(findArea(arr, n)); 
    } 
} 
// This code is contributed by Sumit Ghosh 

Python 3

# Python 3 program for finding maximum  
# area possible of a rectangle 
  
# function for finding max area 
def findArea(arr, n): 
  
    s = [] 
  
    # traverse through array 
    first = 0
    second = 0
    for i in range(n) : 
  
        # If this is first occurrence of  
        # arr[i], simply insert and continue 
        if arr[i] not in s: 
            s.append(arr[i]) 
            continue
  
        # If this is second (or more) occurrence, 
        # update first and second maximum values. 
        if (arr[i] > first) : 
            second = first 
            first = arr[i] 
        elif (arr[i] > second): 
            second = arr[i] 
  
    # return the product of dimensions 
    return (first * second) 
  
# Driver Code 
if __name__ == "__main__": 
      
    arr = [ 4, 2, 1, 4, 6, 6, 2, 5 ] 
    n = len(arr) 
    print(findArea(arr, n)) 
  
# This code is contributed by ita_c 

C#

using System; 
using System.Collections.Generic; 
  
// c# program for finding maximum   
// area possible of a rectangle  
  
public class GFG 
{ 
    // function for finding max area  
    public static int findArea(int[] arr, int n) 
    { 
        //unordered_set<int> s;  
  
        ISet<int> s = new HashSet<int>(); 
  
        // traverse through array  
        int first = 0, second = 0; 
        for (int i = 0; i < n; i++) 
        { 
  
            // If this is first occurrence of   
            // arr[i], simply insert and continue  
            if (!s.Contains(arr[i])) 
            { 
                s.Add(arr[i]); 
                continue; 
            } 
  
            // If this is second (or more)   
            // occurrence, update first and   
            // second maximum values.  
            if (arr[i] > first) 
            { 
                second = first; 
                first = arr[i]; 
            } 
            else if (arr[i] > second) 
            { 
                second = arr[i]; 
            } 
        } 
  
        // return the product of dimensions  
        return (first * second); 
    } 
  
    // driver function  
    public static void Main(string[] args) 
    { 
        int[] arr = new int[] {4, 2, 1, 4, 6, 6, 2, 5}; 
        int n = arr.Length; 
        Console.WriteLine(findArea(arr, n)); 
    } 
} 
  
// This code is contributed by Shrikant13 

Output:

Time Complexity: O(n)

This article is contributed by Shivam Pradhan (anuj_charm). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

C++

Java

Python 3

C#

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