Largest square which can be formed using given rectangular blocks

Last Updated : 24 Mar, 2021

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: 3
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: 2

Approach:

Sort the given array in decreasing order.
Initialise maximum sidelength(say maxLength) as 0.
Traverse the array arr[] and if arr[i] > maxLength then increment the maxLength and check this condition for next iteration.
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 <bits/stdc++.h>
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

<script>
// Javascript program for the above approach
    // Function to find maximum side
    // length of square
    function maxSide( a, n) {
        let sideLength = 0;
 
        // Sort array in asc order
        a.sort();
 
        // Traverse array in desc order
        for ( i = n - 1; i >= 0; i--) {
            if (a[i] > sideLength) {
                sideLength++;
            } else {
                break;
            }
        }
        document.write(sideLength);
    }
 
    // Driver code
      
        let N = 6;
 
        // Given array arr
        let arr = [3, 2, 1, 5, 2, 4 ];
 
        // Function Call
        maxSide(arr, N);
     
 
// This code contributed by aashish1995
</script>

Output:

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

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