# Largest square which can be formed using given rectangular blocks

Given an array arr[] of positive integers where each element of the array represents the length of the rectangular blocks. The task is to find the largest length of the square which can be formed using the rectangular blocks.

Examples:

Input: arr[] = {3, 2, 1, 5, 2, 4}
Output:
Explanation:
Using rectangular block of length 3, 5 and 4, square of side length 3 can be constructed as shown below:

Input: arr[] = {1, 2, 3}
Output:

Approach:

1. Sort the given array in decreasing order.
2. Initialise maximum sidelength(say maxLength) as 0.
3. Traverse the array arr[] and if arr[i] > maxLength then increment the maxLength and check this condition for next iteration.
4. If the above condition doesn’t satisfy then break the loop and print the maxLength.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to find maximum side``// length of square``void` `maxSide(``int` `a[], ``int` `n)``{``    ``int` `sideLength = 0;` `    ``// Sort array in asc order``    ``sort(a, a + n);` `    ``// Traverse array in desc order``    ``for` `(``int` `i = n - 1; i >= 0; i--) {` `        ``if` `(a[i] > sideLength) {``            ``sideLength++;``        ``}``        ``else` `{``            ``break``;``        ``}``    ``}``    ``cout << sideLength << endl;``}` `// Driver Code``int` `main()``{``    ``int` `N = 6;` `    ``// Given array arr[]``    ``int` `arr[] = { 3, 2, 1, 5, 2, 4 };` `    ``// Function Call``    ``maxSide(arr, N);``    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.util.Arrays;` `class` `GFG{ ``    ` `// Function to find maximum side``// length of square``static` `void` `maxSide(``int` `a[], ``int` `n)``{``    ``int` `sideLength = ``0``;` `    ``// Sort array in asc order``    ``Arrays.sort(a);` `    ``// Traverse array in desc order``    ``for``(``int` `i = n - ``1``; i >= ``0``; i--)``    ``{``       ``if` `(a[i] > sideLength)``       ``{``           ``sideLength++;``       ``}``       ``else``       ``{``           ``break``;``       ``}``    ``}``    ``System.out.println(sideLength);``}``    ` `// Driver code ``public` `static` `void` `main (String[] args) ``{ ``    ``int` `N = ``6``;` `    ``// Given array arr[]``    ``int` `arr[] = ``new` `int``[]{ ``3``, ``2``, ``1``,``                           ``5``, ``2``, ``4` `};``    ` `    ``// Function Call``    ``maxSide(arr, N);``} ``} ` `// This code is contributed by Pratima Pandey `

## Python3

 `# Python3 program for the above approach ` `# Function to find maximum side ``# length of square ``def` `maxSide(a, n): ` `    ``sideLength ``=` `0` `    ``# Sort array in asc order ``    ``a.sort ` `    ``# Traverse array in desc order ``    ``for` `i ``in` `range``(n ``-` `1``, ``-``1``, ``-``1``):``        ``if` `(a[i] > sideLength):``            ``sideLength ``+``=` `1``        ``else``: ``            ``break``            ` `    ``print``(sideLength)``    ` `# Driver code``N ``=` `6` `# Given array arr[] ``arr ``=` `[ ``3``, ``2``, ``1``, ``5``, ``2``, ``4` `] ` `# Function Call ``maxSide(arr, N) ` `# This code is contributed by divyeshrabadiya07`

## C#

 `// C# program for the above approach ``using` `System;``class` `GFG{ ``    ` `// Function to find maximum side``// length of square``static` `void` `maxSide(``int` `[]a, ``int` `n)``{``    ``int` `sideLength = 0;` `    ``// Sort array in asc order``    ``Array.Sort(a);` `    ``// Traverse array in desc order``    ``for``(``int` `i = n - 1; i >= 0; i--)``    ``{``        ``if` `(a[i] > sideLength)``        ``{``            ``sideLength++;``        ``}``        ``else``        ``{``            ``break``;``        ``}``    ``}``    ``Console.Write(sideLength);``}``    ` `// Driver code ``public` `static` `void` `Main() ``{ ``    ``int` `N = 6;` `    ``// Given array arr[]``    ``int` `[]arr = ``new` `int``[]{ 3, 2, 1,``                           ``5, 2, 4 };``    ` `    ``// Function Call``    ``maxSide(arr, N);``} ``} ` `// This code is contributed by Code_Mech`

## Javascript

 ``

Output
```3

```

Time Complexity: O(N*log N)
Auxiliary Space: O(1)

### Approach: Binary Search

Idea: Start by setting the low and high values as 1 and the sum of all elements in the array, respectively. Then, check if a square of mid-length can be formed using the given blocks. If yes, update the low value as mid+1. Otherwise, update the high value as mid-1.

Below is the implementation of the above approach:

## C++

 `// C++ program of the above approach` `#include ``#include ``using` `namespace` `std;` `// Function to check if a square of length k can be formed``// using the given blocks``bool` `isSquarePossible(vector<``int``>& arr, ``int` `k)``{``    ``int` `count = 0;``    ``for` `(``int` `i = 0; i < arr.size(); i++) {``        ``count += arr[i] / k;``        ``if` `(count >= k)``            ``return` `true``;``    ``}``    ``return` `false``;``}` `int` `largestSquareLength(vector<``int``>& arr)``{``    ``int` `n = arr.size();``    ``int` `low = 1;``    ``int` `high = 0;``    ``// Calculate the total sum of all block lengths in the``    ``// array``    ``for` `(``int` `i = 0; i < n; i++)``        ``high += arr[i];``    ``int` `result = 0;` `    ``// Binary search loop``    ``while` `(low <= high) {``        ``int` `mid = low + (high - low) / 2;``        ``if` `(isSquarePossible(arr, mid)) {``            ``result = mid;``            ``low = mid + 1;``        ``}``        ``else` `{``            ``high = mid - 1;``        ``}``    ``}``    ``// Return the largest square length found``    ``return` `result;``}` `// Driver Code``int` `main()``{``    ``vector<``int``> arr = { 1, 2, 3 };``    ``int` `largestSquare = largestSquareLength(arr);``    ``cout << largestSquare << endl;` `    ``return` `0;``}`

## Java

 `import` `java.util.ArrayList;``import` `java.util.List;` `class` `GFG {``    ``// Function to check if it's possible to form a square ``  ``// with side length 'k' using elements in 'arr'``    ``public` `static` `boolean` `isSquarePossible(List arr, ``int` `k) {``        ``int` `count = ``0``;``        ``for` `(``int` `i = ``0``; i < arr.size(); i++) {``            ``// Count the number of squares that can be formed ``          ``// with each element in 'arr'``            ``count += arr.get(i) / k;``            ``// If the count is greater than or equal to 'k', ``          ``// it means we can form a square with side length 'k'``            ``if` `(count >= k)``                ``return` `true``;``        ``}``        ``// If no square of side length 'k' can be formed, return false``        ``return` `false``;``    ``}` `    ``// Function to find the largest possible side length of a ``  ``// square that can be formed using elements in 'arr'``    ``public` `static` `int` `largestSquareLength(List arr) {``        ``int` `n = arr.size();``        ``int` `low = ``1``;``        ``int` `high = ``0``;``        ``for` `(``int` `i = ``0``; i < n; i++)``            ``high += arr.get(i);``        ``int` `result = ``0``;` `        ``// Binary search to find the largest possible side ``      ``// length of the square``        ``while` `(low <= high) {``            ``int` `mid = low + (high - low) / ``2``;``            ``// Check if it's possible to form a square with ``          ``// side length 'mid'``            ``if` `(isSquarePossible(arr, mid)) {``                ``result = mid;``                ``// If it's possible to form a square with side ``              ``// length 'mid', try larger side lengths``                ``low = mid + ``1``;``            ``} ``else` `{``                ``// If it's not possible to form a square with side``              ``// length 'mid', try smaller side lengths``                ``high = mid - ``1``;``            ``}``        ``}` `        ``// Return the largest possible side length of the square ``      ``// that can be formed using elements in 'arr'``        ``return` `result;``    ``}` `    ``public` `static` `void` `main(String[] args) {``        ``List arr = ``new` `ArrayList<>();``        ``arr.add(``1``);``        ``arr.add(``2``);``        ``arr.add(``3``);``        ``// Find the largest possible side length of a square that can be formed using elements in 'arr'``        ``int` `largestSquare = largestSquareLength(arr);``        ``System.out.println(largestSquare); ``// Print the result``    ``}``}`

## Python3

 `# Python code of the above mentioned approach` `# Function to check if a square of length k is possible``# by using the given blocks``def` `is_square_possible(arr, k):``    ``count ``=` `0``    ``for` `i ``in` `range``(``len``(arr)):``        ``count ``+``=` `arr[i] ``/``/` `k``        ``if` `count >``=` `k:``            ``return` `True``    ``return` `False` `# Function to calculate the length of the``# largest square``def` `largest_square_length(arr):``    ``n ``=` `len``(arr)``    ``low ``=` `1``    ``high ``=` `0``    ` `    ``# Finding the total sum of ``    ``# the all blocks of the arryas``    ``for` `i ``in` `range``(n):``        ``high ``+``=` `arr[i]``    ``result ``=` `0``    ` `    ``# Doing Binary Search``    ``while` `low <``=` `high:``        ``mid ``=` `low ``+` `(high ``-` `low) ``/``/` `2``        ``if` `is_square_possible(arr, mid):``            ``result ``=` `mid``            ``low ``=` `mid ``+` `1``        ``else``:``            ``high ``=` `mid ``-` `1``    ` `    ``# Returning the final Result``    ``return` `result` `# Driver Code``arr ``=` `[``1``, ``2``, ``3``]``largest_square ``=` `largest_square_length(arr)``print``(largest_square)`

## C#

 `using` `System;``using` `System.Collections.Generic;` `public` `class` `GFG``{``    ``// Function to check if a square of length k can be formed``    ``// using the given blocks``    ``public` `static` `bool` `IsSquarePossible(List<``int``> arr, ``int` `k)``    ``{``        ``int` `count = 0;``        ``foreach` `(``int` `block ``in` `arr)``        ``{``            ``count += block / k;``            ``if` `(count >= k)``                ``return` `true``;``        ``}``        ``return` `false``;``    ``}` `    ``public` `static` `int` `LargestSquareLength(List<``int``> arr)``    ``{``        ``int` `n = arr.Count;``        ``int` `low = 1;``        ``int` `high = 0;``        ``// Calculate the total sum of all block lengths in the array``        ``foreach` `(``int` `block ``in` `arr)``            ``high += block;``        ``int` `result = 0;` `        ``// Binary search loop``        ``while` `(low <= high)``        ``{``            ``int` `mid = low + (high - low) / 2;``            ``if` `(IsSquarePossible(arr, mid))``            ``{``                ``result = mid;``                ``low = mid + 1;``            ``}``            ``else``            ``{``                ``high = mid - 1;``            ``}``        ``}``        ``// Return the largest square length found``        ``return` `result;``    ``}` `    ``public` `static` `void` `Main(``string``[] args)``    ``{``        ``List<``int``> arr = ``new` `List<``int``> { 1, 2, 3 };``        ``int` `largestSquare = LargestSquareLength(arr);``        ``Console.WriteLine(largestSquare);``    ``}``}`

## Javascript

 `// Function to check if a square of length k can be formed``// using the given blocks``function` `isSquarePossible(arr, k) {``    ``let count = 0;``    ``// Iterate through each block length in the array``    ``for` `(let block of arr) {``        ``// Calculate how many blocks of size k can be formed from the current block``        ``count += Math.floor(block / k);``        ``// If we have enough blocks to form a square of size k, return true``        ``if` `(count >= k) {``            ``return` `true``;``        ``}``    ``}``    ``// If we couldn't form a square of size k using the given blocks, return false``    ``return` `false``;``}` `// Function to find the largest square length that can be formed``function` `largestSquareLength(arr) {``    ``const n = arr.length;``    ``let low = 1;``    ``let high = 0;``    ``// Calculate the total sum of all block lengths in the array``    ``for` `(let block of arr) {``        ``high += block;``    ``}``    ``let result = 0;` `    ``// Binary search loop to find the largest possible square length``    ``while` `(low <= high) {``        ``const mid = Math.floor(low + (high - low) / 2);``        ``if` `(isSquarePossible(arr, mid)) {``            ``// If a square of size mid is possible, update result and search higher``            ``result = mid;``            ``low = mid + 1;``        ``} ``else` `{``            ``// If a square of size mid is not possible, search lower``            ``high = mid - 1;``        ``}``    ``}``    ``// Return the largest square length found``    ``return` `result;``}` `// Example array of block lengths``const arr = [1, 2, 3];``// Find and print the largest square length that can be formed``const largestSquare = largestSquareLength(arr);``console.log(largestSquare);`

Output
```2

```

Time Complexity: O(n log M), where n is the number of elements in the input array and M is the sum of all elements in the array.
Auxiliary Space: O(1)

Previous
Next