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.

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++ program to count number of 2 x 2
// squares in a right isosceles triangle
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;



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