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7 changes: 6 additions & 1 deletion Sprint-2/improve_with_caches/fibonacci/fibonacci.py
Original file line number Diff line number Diff line change
@@ -1,4 +1,9 @@
cache = {}


def fibonacci(n):
if n <= 1:
return n
return fibonacci(n - 1) + fibonacci(n - 2)
if n not in cache:
cache[n] = fibonacci(n - 1) + fibonacci(n - 2)
return cache[n]
12 changes: 11 additions & 1 deletion Sprint-2/improve_with_caches/making_change/making_change.py
Original file line number Diff line number Diff line change
Expand Up @@ -10,6 +10,9 @@ def ways_to_make_change(total: int) -> int:
return ways_to_make_change_helper(total, [200, 100, 50, 20, 10, 5, 2, 1])


cache = {}


def ways_to_make_change_helper(total: int, coins: List[int]) -> int:
"""
Helper function for ways_to_make_change to avoid exposing the coins parameter to callers.
Expand All @@ -26,7 +29,14 @@ def ways_to_make_change_helper(total: int, coins: List[int]) -> int:
if total_from_coins == total:
ways += 1
else:
intermediate = ways_to_make_change_helper(total - total_from_coins, coins=coins[coin_index+1:])
key = (total - total_from_coins, tuple(coins[coin_index + 1 :]))
if key not in cache:

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converting the list to a tuple every time is going to be expensive. Also at each stage you're creating a new list, which will also be expensive. The new lists are always the same list, but starting later. Can you think of a way you could refactor the code so that it always uses the same tuple, and the recursion is achieved by passing a parameter indicating where the first usable entry in the tuple is? In that situation, how much can you simplify the key to the cache as well?

cache[key] = ways_to_make_change_helper(
total - total_from_coins, coins=coins[coin_index + 1 :]
)
intermediate = cache[key]
else:
intermediate = cache[key]
ways += intermediate
count_of_coin += 1
return ways