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author | William Carroll <wpcarro@gmail.com> | 2020-03-30T13·14+0100 |
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committer | William Carroll <wpcarro@gmail.com> | 2020-03-30T13·14+0100 |
commit | 8d36c6d00fc2abba9ac04e9e2d4cc620e375d770 (patch) | |
tree | 46652ab67644affd902d15ee4154109eee95e75a /scratch/deepmind/part_two/todo.org | |
parent | 7e41aba8b7bf3b6ef450797739d8069a835a2af6 (diff) |
Solve InterviewCake's compute nth Fibonacci
While the "Dynamic programming and recursion" section hosts this problem, the optimal solution does not use recursion. Many cite the Fibonacci problem as a quintessential dynamic programming question. I assume these people expect an answer like: ```python def fib(n): cache = {0: 0, 1: 1} def do_fib(n): if n in cache: return cache[n] else: cache[n - 1] = do_fib(n - 1) cache[n - 2] = do_fib(n - 2) return cache[n - 1] + cache[n - 2] return do_fib(n) ``` The cache turns the runtime of the classic Fibonacci solution... ```python def fib(n): if n in {0, 1}: return n return fib(n - 1) + fib(n - 2) ``` ... from O(2^n) to a O(n). But both the cache itself and the additional stacks that the runtime allocates for each recursive call create an O(n) space complexity. InterviewCake wants the answer to be solved in O(n) time with O(1) space. To achieve this, instead of solving fib(n) from the top-down, we solve it from the bottom-up. I found this problem to be satisfying to solve.
Diffstat (limited to 'scratch/deepmind/part_two/todo.org')
-rw-r--r-- | scratch/deepmind/part_two/todo.org | 2 |
1 files changed, 1 insertions, 1 deletions
diff --git a/scratch/deepmind/part_two/todo.org b/scratch/deepmind/part_two/todo.org index ba128bb82d43..9c76da754167 100644 --- a/scratch/deepmind/part_two/todo.org +++ b/scratch/deepmind/part_two/todo.org @@ -29,7 +29,7 @@ ** DONE Find Repeat, Space Edition BEAST MODE * Dynamic programming and recursion ** DONE Recursive String Permutations -** TODO Compute nth Fibonacci Number +** DONE Compute nth Fibonacci Number ** TODO Making Change ** TODO The Cake Thief ** DONE Balanced Binary Tree |