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. 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.

## 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.

Output:

6

This article is contributed by **Harshit Agrawal**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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