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

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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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



Output:

3

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