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)
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