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15 changes: 13 additions & 2 deletions Sprint-2/improve_with_caches/fibonacci/fibonacci.py
Original file line number Diff line number Diff line change
@@ -1,4 +1,15 @@
memo = {}

def fibonacci(n):

if n in memo:
return memo[n]

if n <= 1:
return n
return fibonacci(n - 1) + fibonacci(n - 2)
result = n
else:
# Recursive
result = fibonacci(n - 1) + fibonacci(n - 2)

memo[n] = result
return result
39 changes: 25 additions & 14 deletions Sprint-2/improve_with_caches/making_change/making_change.py
Original file line number Diff line number Diff line change
Expand Up @@ -7,26 +7,37 @@ def ways_to_make_change(total: int) -> int:

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])

coins = [200, 100, 50, 20, 10, 5, 2, 1]
memo = {}
return ways_to_make_change_helper(total, coins, 0, memo)


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, memo:dict) -> 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 == 0 :

return 1
if coin_index == len(coins):
return 0
Comment on lines +23 to 24

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This is unrelated to cache, but when there is only one coin left to consider, there is a much quicker way to find out whether there is 0 or 1 way to make change for total.


key = (total, coin_index)

if key in memo:
return memo[key]

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]
count_of_coin = 0

while count_of_coin * coin <= total:
remaining = total - count_of_coin * coin
ways += ways_to_make_change_helper(remaining, coins, coin_index + 1, memo)
count_of_coin += 1

memo[key] = ways

return ways
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