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10 changes: 8 additions & 2 deletions Sprint-2/improve_with_caches/fibonacci/fibonacci.py
Original file line number Diff line number Diff line change
@@ -1,4 +1,10 @@
def fibonacci(n):
def fibonacci(n, cache={}):
if n in cache:
return cache[n]

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

result = fibonacci(n - 1) + fibonacci(n - 2)
cache[n] = result
return result
39 changes: 21 additions & 18 deletions Sprint-2/improve_with_caches/making_change/making_change.py
Original file line number Diff line number Diff line change
@@ -1,32 +1,35 @@
from typing import List


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.
"""
return ways_to_make_change_helper(total, [200, 100, 50, 20, 10, 5, 2, 1])
return ways_to_make_change_helper(total, 0)


def ways_to_make_change_helper(total: int, coins: List[int]) -> int:
def ways_to_make_change_helper(total: int, coin_index: int, cache={}) -> int:
"""
Helper function for ways_to_make_change to avoid exposing the coins parameter to callers.
"""
if total == 0 or len(coins) == 0:
coins = (200, 100, 50, 20, 10, 5, 2, 1)

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Declaring coin as a local variable means a different tuple (with identical items) is created in every recursive function call.

Why not pass the coin list through the parameter so that the function can be reused for different coin list?
When we pass a list to a function, we are only passing its reference -- the cost is negligible.

However, if the function is expected to accept different coin lists, it would not be appropriate to declare the cache as a parameter with a default value.

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I have refactored and made COINS a global variable

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Why not pass the coin list through the parameter?


key = (total, coin_index)

if key in cache:
return cache[key]

if total == 0:
return 1

if coin_index == len(coins):
return 0

ways = 0
for coin_index in range(len(coins)):
coin = coins[coin_index]
count_of_coin = 1
while coin * count_of_coin <= total:
total_from_coins = coin * count_of_coin
if total_from_coins == total:
ways += 1
else:
intermediate = ways_to_make_change_helper(total - total_from_coins, coins=coins[coin_index+1:])
ways += intermediate
count_of_coin += 1
coin = coins[coin_index]

if coin > total:
ways = ways_to_make_change_helper(total, coin_index + 1)
else:
ways = (ways_to_make_change_helper(total - coin, coin_index) + ways_to_make_change_helper(total, coin_index + 1))

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It seems the code on lines 30-32 could be shorten.

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The if/else structure, once flattened, breaks the flow for large inputs like the (ways_to_make_change(9176))

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One of the function calls will always happen. That would suggest it does not need to reside in an if-else block.


cache[key] = ways
return ways
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