Maximum number of 2x2 squares that can be fit inside a right isosceles triangle - GeeksforGeeks

Maximum number of 2x2 squares that can be fit inside a right isosceles triangle - GeeksforGeeks
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.

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

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;

}

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