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


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;
}

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

Output:

3


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Improved By : ash264