Maximum area rectangle created by selecting four sides from an array

Given a set of n positive integers representing lengths. Determine the maximum possible area whose four sides are chosen from the given array. It should be noted that a rectangle can only be formed if the given array contains two pairs of equal values.

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

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.

Sort the given array.
Traverse array from largest to smallest value and return two pairs with maximum values.

Program in 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;
}

Output

Time Complexity : O (n Log n)

Method 2 (Hashing)

In a hash set, all first occurrences of elements are inserted. Keep track of no more than two values for second occurrences.

Program in 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;
}

Output

Time Complexity : O (n)

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