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London | 25-SDC-July | Andrei Filippov | Sprint 2 | Improve code with caches #30

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London | 25-SDC-July | Andrei Filippov | Sprint 2 | Improve code with caches #30

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4fb5c72
improved fibonacci
Droid-An Oct 13, 2025
d66ee9b
improved making change with cache
Droid-An Oct 13, 2025
9561f64
actually done making change improvement
Droid-An Oct 13, 2025
bcc9ee4
refactor fibonacci function to simplify return statements
Droid-An Jan 26, 2026
c98d800
refactor ways_to_make_change function to improve caching
Droid-An Jan 28, 2026
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9 changes: 8 additions & 1 deletion Sprint-2/improve_with_caches/fibonacci/fibonacci.py
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Original file line number Diff line number Diff line change
@@ -1,4 +1,11 @@
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]
else:
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@OracPrime OracPrime Jan 26, 2026

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It's personal taste as to which way you do it, but the code would be more concise without lines 9 and 10, and with 11 outside the if.

return cache[n]
12 changes: 11 additions & 1 deletion Sprint-2/improve_with_caches/making_change/making_change.py
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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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@OracPrime OracPrime Jan 26, 2026

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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?

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@Droid-An Droid-An Jan 28, 2026

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yes, I can do this. In that case I only need to store the amount and index where to start as key

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