# Largest area square in an array when elements can be shuffled

Given an array arr[] of N integers where arr[i] is the height of the ith chocolate and all the chocolated are 1 unit wide, the task is to find the maximum area for any square made from the chocolates when the chocolates can be arranged in any order.

Examples:

Input: arr[] = {1, 3, 4, 5, 5}
Output: 9
Square with side = 3 can be obtained
from either {3, 4, 5} or {4, 5, 5}.

Input: arr[] = {6, 1, 6, 6, 6}
Output: 16

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

Approach: A square of side a can be obtained if there exists atleast a element in the array which are either equal to or greater than a. Binary Search can be used to find the maximum side of the square that could be achieved within the range of 0 to N.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function that returns true if it ` `// is possible to make a square ` `// with side equal to l ` `bool` `isSquarePossible(``int` `arr[], ``int` `n, ``int` `l) ` `{ ` ` `  `    ``// To store the count of elements ` `    ``// greater than or equal to l ` `    ``int` `cnt = 0; ` `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``// Increment the count ` `        ``if` `(arr[i] >= l) ` `            ``cnt++; ` ` `  `        ``// If the count becomes greater ` `        ``// than or equal to l ` `        ``if` `(cnt >= l) ` `            ``return` `true``; ` `    ``} ` ` `  `    ``return` `false``; ` `} ` ` `  `// Function to return the ` `// maximum area of the square ` `// that can be obtained ` `int` `maxArea(``int` `arr[], ``int` `n) ` `{ ` `    ``int` `l = 0, r = n; ` `    ``int` `len = 0; ` `    ``while` `(l <= r) { ` `        ``int` `m = l + ((r - l) / 2); ` ` `  `        ``// If square is possible with ` `        ``// side length m ` `        ``if` `(isSquarePossible(arr, n, m)) { ` `            ``len = m; ` `            ``l = m + 1; ` `        ``} ` ` `  `        ``// Try to find a square with ` `        ``// smaller side length ` `        ``else` `            ``r = m - 1; ` `    ``} ` ` `  `    ``// Return the area ` `    ``return` `(len * len); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 3, 4, 5, 5 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(``int``); ` ` `  `    ``cout << maxArea(arr, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach  ` `class` `GFG ` `{ ` `     `  `    ``// Function that returns true if it  ` `    ``// is possible to make a square  ` `    ``// with side equal to l  ` `    ``static` `boolean` `isSquarePossible(``int` `arr[],  ` `                                    ``int` `n, ``int` `l)  ` `    ``{  ` `     `  `        ``// To store the count of elements  ` `        ``// greater than or equal to l  ` `        ``int` `cnt = ``0``;  ` `        ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``{  ` `     `  `            ``// Increment the count  ` `            ``if` `(arr[i] >= l)  ` `                ``cnt++;  ` `     `  `            ``// If the count becomes greater  ` `            ``// than or equal to l  ` `            ``if` `(cnt >= l)  ` `                ``return` `true``;  ` `        ``}  ` `        ``return` `false``;  ` `    ``}  ` `     `  `    ``// Function to return the  ` `    ``// maximum area of the square  ` `    ``// that can be obtained  ` `    ``static` `int` `maxArea(``int` `arr[], ``int` `n)  ` `    ``{  ` `        ``int` `l = ``0``, r = n;  ` `        ``int` `len = ``0``;  ` `        ``while` `(l <= r)  ` `        ``{  ` `            ``int` `m = l + ((r - l) / ``2``);  ` `     `  `            ``// If square is possible with  ` `            ``// side length m  ` `            ``if` `(isSquarePossible(arr, n, m)) ` `            ``{  ` `                ``len = m;  ` `                ``l = m + ``1``;  ` `            ``}  ` `     `  `            ``// Try to find a square with  ` `            ``// smaller side length  ` `            ``else` `                ``r = m - ``1``;  ` `        ``}  ` `     `  `        ``// Return the area  ` `        ``return` `(len * len);  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `main (String[] args) ` `    ``{  ` `        ``int` `arr[] = { ``1``, ``3``, ``4``, ``5``, ``5` `};  ` `        ``int` `n = arr.length;  ` `     `  `        ``System.out.println(maxArea(arr, n));  ` `    ``}  ` `} ` ` `  `// This code is contributed by kanugargng `

## Python3

 `# Python3 implementation of the approach  ` ` `  `# Function that returns true if it  ` `# is possible to make a square  ` `# with side equal to l  ` `def` `isSquarePossible(arr, n, l) :  ` ` `  `    ``# To store the count of elements  ` `    ``# greater than or equal to l  ` `    ``cnt ``=` `0` `    ``for` `i ``in` `range``(n) : ` ` `  `        ``# Increment the count  ` `        ``if` `arr[i] >``=` `l :  ` `            ``cnt ``+``=` `1` ` `  `        ``# If the count becomes greater  ` `        ``# than or equal to l  ` `        ``if` `cnt >``=` `l :  ` `            ``return` `True` ` `  `    ``return` `False` ` `  `# Function to return the  ` `# maximum area of the square  ` `# that can be obtained  ` `def` `maxArea(arr, n) :  ` ` `  `    ``l , r ``=` `0``, n  ` `    ``len` `=` `0` `    ``while` `l <``=` `r :  ` `        ``m ``=` `l ``+` `((r ``-` `l) ``/``/` `2``)  ` ` `  `        ``# If square is possible with  ` `        ``# side length m  ` `        ``if` `isSquarePossible(arr, n, m) :  ` `            ``len` `=` `m  ` `            ``l ``=` `m ``+` `1` ` `  `        ``# Try to find a square with  ` `        ``# smaller side length  ` `        ``else` `: ` `            ``r ``=` `m ``-` `1` ` `  `    ``# Return the area  ` `    ``return` `(``len` `*` `len``) ` ` `  `# Driver code  ` `arr ``=` `[ ``1``, ``3``, ``4``, ``5``, ``5` `]  ` `n ``=` `len``(arr) ` ` `  `print``(maxArea(arr, n))  ` ` `  `# This code is contributed by divyamohan `

## C#

 `// C# implementation of the approach  ` `using` `System; ` ` `  `class` `GFG  ` `{  ` `     `  `    ``// Function that returns true if it  ` `    ``// is possible to make a square  ` `    ``// with side equal to l  ` `    ``static` `bool` `isSquarePossible(``int` `[]arr,  ` `                                 ``int` `n, ``int` `l)  ` `    ``{  ` `     `  `        ``// To store the count of elements  ` `        ``// greater than or equal to l  ` `        ``int` `cnt = 0;  ` `        ``for` `(``int` `i = 0; i < n; i++)  ` `        ``{  ` `     `  `            ``// Increment the count  ` `            ``if` `(arr[i] >= l)  ` `                ``cnt++;  ` `     `  `            ``// If the count becomes greater  ` `            ``// than or equal to l  ` `            ``if` `(cnt >= l)  ` `                ``return` `true``;  ` `        ``}  ` `        ``return` `false``;  ` `    ``}  ` `     `  `    ``// Function to return the  ` `    ``// maximum area of the square  ` `    ``// that can be obtained  ` `    ``static` `int` `maxArea(``int` `[]arr, ``int` `n)  ` `    ``{  ` `        ``int` `l = 0, r = n;  ` `        ``int` `len = 0;  ` `        ``while` `(l <= r)  ` `        ``{  ` `            ``int` `m = l + ((r - l) / 2);  ` `     `  `            ``// If square is possible with  ` `            ``// side length m  ` `            ``if` `(isSquarePossible(arr, n, m))  ` `            ``{  ` `                ``len = m;  ` `                ``l = m + 1;  ` `            ``}  ` `     `  `            ``// Try to find a square with  ` `            ``// smaller side length  ` `            ``else` `                ``r = m - 1;  ` `        ``}  ` `     `  `        ``// Return the area  ` `        ``return` `(len * len);  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `Main()  ` `    ``{  ` `        ``int` `[]arr = { 1, 3, 4, 5, 5 };  ` `        ``int` `n = arr.Length;  ` `     `  `        ``Console.WriteLine(maxArea(arr, n));  ` `    ``}  ` `} ` ` `  `// This code is contributed by AnkitRai01 `

Output:

```9
```

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