Possible number of Rectangle and Squares with the given set of elements

Given ‘N’ number of sticks of length a₁, a₂, a₃…a_n. The task is to count the number of squares and rectangles possible. 

Note: One stick should be used only once i.e. either in any of the squares or rectangles.

Examples:

Input: arr[] = {1, 2, 1, 2}
Output: 1
Rectangle with sides 1 1 2 2

Input: arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9}
Output: 0
No square or rectangle is possible

Approach: Below is the step by step algorithm to solve this problem :

1. Initialize the number of sticks.
2. Initialize all the sticks with it’s lengths in an array.
3. Sort the array in an increasing order.
4. Calculate the number of pairs of sticks with the same length.
5. Divide the total number of pairs by 2, which will be the total possible rectangle and square.

Below is the implementation of above approach: 

C++

// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the possible
// rectangles and squares
int rectangleSquare(int arr[], int n)
{

    // sort all the sticks
    sort(arr, arr + n);
    int count = 0;

    // calculate all the pair of
    // sticks with same length
    for (int i = 0; i < n - 1; i++) {
        if (arr[i] == arr[i + 1]) {
            count++;
            i++;
        }
    }

    // divide the total number of pair
    // which will be the number of possible
    // rectangle and square
    return count / 2;
}

// Driver code
int main()
{

    // initialize all the stick lengths
    int arr[] = { 2, 2, 4, 4, 4, 4, 6, 6, 6, 7, 7, 9, 9 };
    int n = sizeof(arr) / sizeof(arr[0]);

    cout << rectangleSquare(arr, n);

    return 0;
}

Java

// Java implementation of above approach
import java.util.Arrays;

class GFG
{

    // Function to find the possible
    // rectangles and squares
    static int rectangleSquare(int arr[], int n)
    {

        // sort all the sticks
        Arrays.sort(arr);
        int count = 0;

        // calculate all the pair of
        // sticks with same length
        for (int i = 0; i < n - 1; i++)
        {
            if (arr[i] == arr[i + 1])
            {
                count++;
                i++;
            }
        }

        // divide the total number of pair
        // which will be the number of possible
        // rectangle and square
        return count / 2;
    }

    // Driver code
    public static void main(String[] args)
    {
        // initialize all the stick lengths
        int arr[] = {2, 2, 4, 4, 4, 4, 6, 6, 6, 7, 7, 9, 9};
        int n = arr.length;
        System.out.println(rectangleSquare(arr, n));
    }
}

// This code is contributed
// by PrinciRaj1992

Python3

# Python3 implementation of above approach


# Function to find the possible
# rectangles and squares
def rectangleSquare( arr, n):


    # sort all the sticks
    arr.sort()
    count = 0
    #print(" xx",arr[6])
    # calculate all the pair of
    # sticks with same length
    k=0
    for i in range(n-1):
        if(k==1):
            k=0
            continue

        if (arr[i] == arr[i + 1]):

            count=count+1

            k=1



    # divide the total number of pair
    # which will be the number of possible
    # rectangle and square
    return count/2


# Driver code

if __name__=='__main__':

# initialize all the stick lengths
    arr = [2, 2, 4, 4, 4, 4, 6, 6, 6, 7, 7, 9, 9]
    n = len(arr)

    print(rectangleSquare(arr, n))

# this code is written by ash264

C#

// C# implementation of above approach
using System;

class GFG
{

    // Function to find the possible
    // rectangles and squares
    static int rectangleSquare(int []arr, int n)
    {

        // sort all the sticks
        Array.Sort(arr);
        int count = 0;

        // calculate all the pair of
        // sticks with same length
        for (int i = 0; i < n - 1; i++)
        {
            if (arr[i] == arr[i + 1])
            {
                count++;
                i++;
            }
        }

        // divide the total number of pair
        // which will be the number of possible
        // rectangle and square
        return count / 2;
    }

    // Driver code
    public static void Main(String[] args)
    {
        // initialize all the stick lengths
        int []arr = {2, 2, 4, 4, 4, 4, 6,
                        6, 6, 7, 7, 9, 9};
        int n = arr.Length;
        Console.WriteLine(rectangleSquare(arr, n));
    }
}

// This code has been contributed
// by Rajput-Ji

PHP

<?php
// PHP implementation of above approach

// Function to find the possible
// rectangles and squares
function rectangleSquare($arr, $n)
{

    // sort all the sticks
    sort($arr);
    $count = 0;

    // calculate all the pair of
    // sticks with same length
    for ($i = 0; $i < $n - 1; $i++)
    {
        if ($arr[$i] == $arr[$i + 1])
        {
            $count++;
            $i++;
        }
    }

    // divide the total number of pair
    // which will be the number of possible
    // rectangle and square
    return ($count / 2);
}

// Driver code

// initialize all the stick lengths
$arr = array(2, 2, 4, 4, 4, 4, 6,
                6, 6, 7, 7, 9, 9 );
$n = sizeof($arr);

echo rectangleSquare($arr, $n);

// This code is contributed by Sachin.
?>

Javascript

<script>

// javascript implementation of above approach

// Function to find the possible
// rectangles and squares
function rectangleSquare(arr , n)
{

    // sort all the sticks
    arr.sort();
    var count = 0;

    // calculate all the pair of
    // sticks with same length
    for (i = 0; i < n - 1; i++)
    {
        if (arr[i] == arr[i + 1])
        {
            count++;
            i++;
        }
    }

    // divide the total number of pair
    // which will be the number of possible
    // rectangle and square
    return count / 2;
}

// Driver code
// initialize all the stick lengths
var arr = [2, 2, 4, 4, 4, 4, 6, 6, 6, 7, 7, 9, 9];
var n = arr.length;
document.write(rectangleSquare(arr, n));

// This code is contributed by 29AjayKumar

</script>

Output

3

Complexity Analysis:

• Time Complexity: O(n*log n) where n is the size of the array.
• Auxiliary Space: O(1)