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Maximum number of 2×2 squares that can be fit inside a right isosceles triangle

What is the maximum number of squares of size 2×2 units that can be fit in a right angled isosceles triangle of a given base (in units).
A side of the square must be parallel to the base of the triangle.
Examples:

Input  : 8
Output : 6
Please refer below diagram for explanation.
squares-in-triangle

Input  : 7
Output : 3

Since the triangle is isosceles, the given base would also be equal to the height. Now in the diagonal part, we would always need an extra length of 2 units in both height and base of the triangle to accommodate a triangle. (The CF and AM segment of the triangle in the image. The part that does not contribute to any square). In the remaining length of base, we can construct length / 2 squares. Since each square is of 2 units, same would be the case of height, there is no need to calculate that again.

So, for each level of given length we can construct “(length-2)/2” squares. This gives us a base of “(length-2)” above it. Continuing this process to get the no of squares for all available “length-2” height, we can calculate the squares.

while length > 2
    answer += (length - 2 )/2
    length = length - 2

For more effective way, we can use the formula of sum of AP n * ( n + 1 ) / 2, where n = length – 2

C++

// C++ program to count number of 2 x 2
// squares in a right isosceles triangle
#include<bits/stdc++.h>
using namespace std;
  
int numberOfSquares(int base)
{
   // removing the extra part we would
   // always need
   base = (base - 2);
  
   // Since each square has base of
   // length of 2
   base = base / 2;
  
   return base * (base + 1)/2;
}
  
// Driver code
int main()
{
   int base = 8;
   cout << numberOfSquares(base);
   return 0;
}

Java

// Java program to count number of 2 x 2
// squares in a right isosceles triangle
  
class Squares
{
  public static  int numberOfSquares(int base)
   {
      // removing the extra part 
      // we would always need
      base = (base - 2);
   
      // Since each square has 
      // base of length of 2
      base = base / 2;
   
      return base * (base + 1)/2;
   }
   
   // Driver code
   public static void main(String args[])
   {
        
      int base = 8;
      System.out.println(numberOfSquares(base));
   }
}
  
// This code is contributed by Anshika Goyal.

/div>

Python3

# Python3 program to count number
# of 2 x 2 squares in a right 
# isosceles triangle
def numberOfSquares(base):
      
    # removing the extra part we would
    # always need
    base = (base - 2)
      
    # Since each square has base of
    # length of 2
    base = base / 2
      
    return base * (base + 1) / 2
      
# Driver code
base = 8
  
print(numberOfSquares(base))
  
# This code is contributed by Anant Agarwal. 

C#

// C# program to count number of 2 x 2
// squares in a right isosceles triangle
using System;
  
class GFG {
      
    public static int numberOfSquares(int _base)
    {
          
        // removing the extra part 
        // we would always need
        _base = (_base - 2);
      
        // Since each square has 
        // base of length of 2
        _base = _base / 2;
      
        return _base * (_base + 1)/2;
    }
      
    // Driver code
    public static void Main()
    {
          
        int _base = 8;
        Console.WriteLine(numberOfSquares(_base));
    }
}
  
// This code is contributed by anuj_67.

PHP

<?php
// PHP program to count number of 2 x 2
// squares in a right isosceles triangle
  
    function numberOfSquares( $base)
    {
          
        // removing the extra 
        // part we would
        // always need
        $base = ($base - 2);
          
        // Since each square
        // has base of
        // length of 2
        $base = $base / 2;
      
        return $base * ($base + 1)/2;
    }
  
// Driver code
$base = 8;
echo numberOfSquares($base);
  
// This code is contributed by anuj_67.
?>


Output:

6

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



This article is attributed to GeeksforGeeks.org

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