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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: int, fib_cache: dict[int, int] = {0: 0, 1: 1}) -> int:
if n <= 1:
return n
return fibonacci(n - 1) + fibonacci(n - 2)
#check if already calc this num.
if n in fib_cache:
return fib_cache[n]
# if not in cache, calculate it, store it, and return
fib_cache[n] = fibonacci(n - 1) + fibonacci(n - 2)
return fib_cache[n]
43 changes: 26 additions & 17 deletions Sprint-2/improve_with_caches/making_change/making_change.py
Original file line number Diff line number Diff line change
@@ -1,32 +1,41 @@
from typing import List

from typing import List, Tuple, Dict

change_cache: Dict[Tuple[int, int], int] = {}
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, [200, 100, 50, 20, 10, 5, 2, 1],0)


def ways_to_make_change_helper(total: int, coins: List[int]) -> int:
def ways_to_make_change_helper(total: int, coins: List[int], coin_index: int) -> int:
"""
Helper function for ways_to_make_change to avoid exposing the coins parameter to callers.
"""
if total == 0 or len(coins) == 0:

#If total is 0, we found exactly 1 valid way to make change
if total == 0:
return 1

#If total becomes negative or we run out of coins, this way failed
if total < 0 or 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
# Cache check
cache_key = (total, coin_index)
if cache_key in change_cache:
return change_cache[cache_key]

current_coin = coins[coin_index]

# 1 use the current coin and stay on the same coin index
# 2 skip the current coin and move to the next coin index
ways = (
ways_to_make_change_helper(total - current_coin, coins, coin_index) +
ways_to_make_change_helper(total, coins, coin_index + 1)
Comment on lines +36 to +37

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The original "loop" version and your "without loop" version differ in terms of how deep the recursion can go, which affects the amount of stack space required at runtime. For a large total, the program may exceed the available stack space and fail.

Could you update your code to match the original approach as much as possible?

Notes:

  • Supposedly, one level of loop is enough.
  • When there is only one coin left to consider, there is a much quicker way to figure out if there is 0 or 1 way to make the change.

)
# store the result in the cache before the return
change_cache[cache_key] = ways
return ways
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