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Area of the largest square that can be formed from the given length sticks using Hashing

  • Last Updated : 06 Apr, 2021

Given an array arr[] of N integers representing the heights of the sticks. The task is to find the area of the largest square that can be formed using these sticks and the count of such squares. Note that a single side of the square can only use a single stick.
Examples: 

Input: arr[] = {5, 3, 2, 3, 6, 3, 3} 
Output: 
Area = 9 
Count = 1 
Side of the square will be 3 and 
only one such square is possible.
Input: arr[] = {2, 2, 2, 9, 2, 2, 2, 2, 2} 
Output: 
Area = 4 
Count = 2 

Approach: Count the frequencies of all the elements of the array. Now, starting from the maximum (in order to maximize the area) find the first frequency which is at least 4 so that a square can be formed then the area can be calculated as freq[i] * freq[i] and the count of such squares will be freq[i] / 4.
Below is the implementation of the above approach: 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the area of the largest
// square that can be formed
// and the count of such squares
void findMaxSquare(int arr[], int n)
{
 
    // Maximum value from the array
    int maxVal = *max_element(arr, arr + n);
 
    // Update the frequencies of
    // the array elements
    int freq[maxVal + 1] = { 0 };
    for (int i = 0; i < n; i++)
        freq[arr[i]]++;
 
    // Starting from the maximum length sticks
    // in order to maximize the area
    for (int i = maxVal; i > 0; i--) {
 
        // The count of sticks with the current
        // length has to be at least 4
        // in order to form a square
        if (freq[i] >= 4) {
            cout << "Area = " << (pow(i, 2));
            cout << "\nCount = " << (freq[i] / 4);
            return;
        }
    }
 
    // Impossible to form a square
    cout << "-1";
}
 
// Driver code
int main()
{
    int arr[] = { 2, 2, 2, 9, 2, 2, 2, 2, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    findMaxSquare(arr, n);
 
    return 0;
}

Java




// Java implementation of the approach
import java.util.*;
 
class GFG
{
 
// Function to find the area of the largest
// square that can be formed
// and the count of such squares
static void findMaxSquare(int arr[], int n)
{
 
    // Maximum value from the array
    int maxVal = Arrays.stream(arr).max().getAsInt();
 
    // Update the frequencies of
    // the array elements
    int []freq = new int[maxVal + 1];
    for (int i = 0; i < n; i++)
        freq[arr[i]]++;
 
    // Starting from the maximum length sticks
    // in order to maximize the area
    for (int i = maxVal; i > 0; i--)
    {
 
        // The count of sticks with the current
        // length has to be at least 4
        // in order to form a square
        if (freq[i] >= 4)
        {
            System.out.print("Area = " +
                            (Math.pow(i, 2)));
            System.out.print("\nCount = " +
                            (freq[i] / 4));
            return;
        }
    }
 
    // Impossible to form a square
    System.out.print("-1");
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 2, 2, 2, 9, 2, 2, 2, 2, 2 };
    int n = arr.length;
 
    findMaxSquare(arr, n);
}
}
 
// This code is contributed by Princi Singh

Python3




# Python3 implementation of the approach
 
# Function to find the area of the largest
# square that can be formed
# and the count of such squares
def findMaxSquare(arr, n) :
 
    # Maximum value from the array
    maxVal = max(arr);
 
    # Update the frequencies of
    # the array elements
    freq = [0] * (maxVal + 1) ;
    for i in range(n) :
        freq[arr[i]] += 1;
 
    # Starting from the maximum length sticks
    # in order to maximize the area
    for i in range(maxVal, 0, -1) :
 
        # The count of sticks with the current
        # length has to be at least 4
        # in order to form a square
        if (freq[i] >= 4) :
            print("Area = ", pow(i, 2));
            print("Count =", freq[i] // 4);
            return;
 
    # Impossible to form a square
    print("-1");
 
# Driver code
if __name__ == "__main__" :
 
    arr = [ 2, 2, 2, 9, 2, 2, 2, 2, 2 ];
    n = len(arr);
 
    findMaxSquare(arr, n);
 
# This code is contributed by AnkitRai01

C#




// C# implementation of the approach
using System;
using System.Linq;
 
class GFG
{
 
// Function to find the area of the largest
// square that can be formed
// and the count of such squares
static void findMaxSquare(int []arr, int n)
{
 
    // Maximum value from the array
    int maxVal = arr.Max();
 
    // Update the frequencies of
    // the array elements
    int []freq = new int[maxVal + 1];
    for (int i = 0; i < n; i++)
        freq[arr[i]]++;
 
    // Starting from the maximum length sticks
    // in order to maximize the area
    for (int i = maxVal; i > 0; i--)
    {
 
        // The count of sticks with the current
        // length has to be at least 4
        // in order to form a square
        if (freq[i] >= 4)
        {
            Console.Write("Area = " +
                         (Math.Pow(i, 2)));
            Console.Write("\nCount = " +
                         (freq[i] / 4));
            return;
        }
    }
 
    // Impossible to form a square
    Console.Write("-1");
}
 
// Driver code
public static void Main(String[] args)
{
    int []arr = { 2, 2, 2, 9, 2, 2, 2, 2, 2 };
    int n = arr.Length;
 
    findMaxSquare(arr, n);
}
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
 
// Javascript implementation of the approach
 
// Function to find the area of the largest
// square that can be formed
// and the count of such squares
function findMaxSquare(arr, n)
{
 
    // Maximum value from the array
    var maxVal = Math.max(...arr);
 
    // Update the frequencies of
    // the array elements
    var freq = Array(maxVal + 1).fill(0);
    for (var i = 0; i < n; i++)
        freq[arr[i]]++;
 
    // Starting from the maximum length sticks
    // in order to maximize the area
    for (var i = maxVal; i > 0; i--) {
 
        // The count of sticks with the current
        // length has to be at least 4
        // in order to form a square
        if (freq[i] >= 4) {
            document.write("Area = " + (Math.pow(i, 2)));
            document.write("<br>Count = " + (freq[i] / 4));
            return;
        }
    }
 
    // Impossible to form a square
    document.write("-1");
}
 
// Driver code
var arr = [ 2, 2, 2, 9, 2, 2, 2, 2, 2 ];
var n = arr.length;
findMaxSquare(arr, n);
 
</script>
Output: 
Area = 4
Count = 2

 

Time Complexity: O(n)

Auxiliary Space: O(n)

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