Codility - FrogJmp

https://codility.com/programmers/task/frog_jmp

My solution

def solution(x, y, d)
  ((y - x).to_f / d).ceil
end

Learning points

• Paper and pencil (and algebra) helps. The equation is: x + dn >= y After that it was just a matter of finding n.
• Integer division can be tricky. Convert to float to retain precision, then convert back to integer afterwards (if the situation requires it).
• #to_i will chop off the decimal portion of the float; this may or may not be what you want.
• O(1) will usually mean there should be no loop used – at all. It is a straightforward computation.