Considering the classical staircase problem as "Davis has a number of staircases in his house and he likes to climb each staircase 1, 2, or 3 steps at a time. Being a very precocious child, he wonders how many ways there are to reach the top of the staircase."
My approach is to use memoization with recursion as
# TimeO(N), SpaceO(N), DP Bottom Up + Memoization
def stepPerms(n, memo = {}):
if n < 3:
return n
elif n == 3:
return 4
if n in memo:
return memo[n]
else:
memo[n] = stepPerms(n - 1, memo) + stepPerms(n - 2 ,memo) + stepPerms(n - 3 ,memo)
return memo[n]
The question that comes to my mind is that, is this solution bottom-up or top-down. My way of approaching it is that since we go all the way down to calculate the upper N values (imagine the recursion tree). I consider this bottom-up. Is this correct?