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

• Difficulty Level : Basic
• Last Updated : 01 Sep, 2022

Given ‘N’ number of sticks of length a1, a2, a3…an. 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 ``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);` `    ``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)``    ``# 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

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## Javascript

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Output

`3`

Complexity Analysis:

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

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