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
Time complexity : O(1)
Auxiliary Space : O(1)
This article is contributed by Harshit Agrawal. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to email@example.com. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.