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15 changes: 14 additions & 1 deletion Sprint-2/improve_with_caches/fibonacci/fibonacci.py
Original file line number Diff line number Diff line change
@@ -1,4 +1,17 @@
# Create a dictionary to use as our "Notepad" (Cache)
memo = {}

def fibonacci(n):
# Check if we already calculated this number
if n in memo:
return memo[n]

if n <= 1:
return n
return fibonacci(n - 1) + fibonacci(n - 2)

result = fibonacci(n - 1) + fibonacci(n - 2)

# Save the result in our Notepad before returning
memo[n] = result

return result
19 changes: 19 additions & 0 deletions Sprint-2/improve_with_caches/making_change/making_change.py
Original file line number Diff line number Diff line change
@@ -1,22 +1,38 @@
from typing import List

cache = {}
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Better to make cache local inside ways_to_make_change(), and define
ways_to_make_change_helper() as an inner function inside ways_to_make_change().

Since that's not the purpose of this exercise, change is optional.


def ways_to_make_change(total: int) -> int:
"""
Given access to coins with the values 1, 2, 5, 10, 20, 50, 100, 200, returns a count of all of the ways to make the passed total value.

For instance, there are two ways to make a value of 3: with 3x 1 coins, or with 1x 1 coin and 1x 2 coin.
"""
# Clear cache before starting new calculation
cache.clear()
return ways_to_make_change_helper(total, [200, 100, 50, 20, 10, 5, 2, 1])


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.
"""
# We use tuple(coins) because lists cannot be keys.
key = (total, tuple(coins))
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@cjyuan cjyuan Jan 23, 2026

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From line 45, we know coins can only be one of the following 9 lists:

[200, 100, 50, 20, 10, 5, 2, 1]
[100, 50, 20, 10, 5, 2, 1]
[50, 20, 10, 5, 2, 1]
...
[1]
[]

We could further improve the performance if we can

  • avoid repeatedly creating the same sub-lists at line 45, and
  • create key as (total, a_value_identifiying_one_of_the_sublists) instead of as (total, tuple(coins))

I don't think this exercise expects trainees to optimize the code to this level. So change is optional.


if key in cache:
return cache[key]

if total == 0 or len(coins) == 0:
return 0

# Check if only one coin type remains
if len(coins) == 1:
if total % coins[0] == 0:
return 1
else:
return 0

ways = 0
for coin_index in range(len(coins)):
coin = coins[coin_index]
Expand All @@ -29,4 +45,7 @@ def ways_to_make_change_helper(total: int, coins: List[int]) -> int:
intermediate = ways_to_make_change_helper(total - total_from_coins, coins=coins[coin_index+1:])
ways += intermediate
count_of_coin += 1

# Store result in cache
cache[key] = ways
return ways