Given an array of n positive integers that represent lengths. Find out the maximum possible area whose four sides are picked from the given array. Note that a rectangle can only be formed if there are two pairs of equal values in the 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 possibleMethod 1 (Sorting): The task basically reduces to finding two pairs of equal values in the array. If there are more than two pairs, then pick the two pairs with maximum values. A simple solution is to do the following.
- Sort the given array.
- Traverse array from largest to smallest value and return two pairs with maximum values.
Implementation:
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. ?> |
Javascript
<script> // Javascript program for finding maximum area possible // of a rectangle // function for finding max area function findArea(arr, n)
{ // sort array in non-increasing order
arr.sort((a,b)=>{return b-a;})
// Initialize two sides of rectangle
var dimension = [0,0];
// traverse through array
for (var 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 var arr = [ 4, 2, 1, 4, 6, 6, 2, 5 ];
var n = arr.length;
document.write( findArea(arr, n)); </script> |
Output
24
Time Complexity: O(n Log n) as one traversal of array takes O(n) time and sorting the array takes n logn extra time , hence overall time taken by the algorithm is O(n Log n)
Auxiliary Space: O(1) since no extra array is used so it takes constant space
Method 2 (Hashing): The idea is to insert all first occurrences of elements in a hash set. For second occurrences, keep track of the maximum of two values.
Below is the implementation of the above approach:
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 |
Python3
C#
Javascript
Output
24
Time Complexity: O(n) as only one traversal of array is required to complete all operations hence overall complexity turns out be O(n)
Auxiliary Space: O(n) since an unordered set is used, in worst case all elements will be inserted into the set thus occupying O(n) space.