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Essentially, you want to add the number of ways for stairmaster( n -1) , stairmaster(n-2), and stairmaster(n-3), to get the value of stairmaster(n). Since it’s really easy to calculate stairmaster() for n = 1, 2, or 3, we can just include that at the beginning of the function to give us a starting point for bigger numbers.
Then we just run a for loop for 4 to n and on each iteration:
We set d = sum of the three previous values of a, b, and c
We update the values of a, b, c so that they are the values needed for the next value of i.
We are then counting up to stairmaster (n) eventually. Once i reaches n, d is the sum of the 3 previous values of d, which is the number of permutations of 1, 2, 3 that equal n number of stairs, so we terminate the loop and return d as our answer.
def stairmaster(n):
a = 1
b = 2
c = 4
d = 0 # fourth variable (placehold for when n > 3)
if (n == 0 or n == 1 or n == 2):
return n
if (n == 3):
return c
for i in range(4,n+1):
# starting from 4 because we already know the solution for, 1, 2, or 3 stairs
d = c + b + a # already counted for 3 stairs
a = b
b = c
c = d
return d
print(stairmaster(10))
def stairmasterDP(n):
memo = {0:1,1:1,2:2,3:4}
if n in memo:
return memo[n]
else:
for i in range(4,n+1):
memo[i] = memo[i-1] + memo[i-2] + memo[i-3]
return memo[n]
print(stairmasterDP(10))
This challenge is like trying to find Fibonacci numbers, but with base cases: 1, 2, 4.
I used loops instead of recursion, and I stored the relevant previous values in a list.
def stairmaster(n, k = 3):
if (n <= 0):
return 0;
elif (n <= k):
return 2 ** (n - 1)
recursive = [2 ** i for i in range(k)]
for i in range(k + 1, n + 1):
current = sum(recursive)
recursive.pop(0)
recursive.append(current)
return current
A list probably wasn’t the best data structure to use for this. Maybe a deque would be better.
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