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10 changes: 9 additions & 1 deletion Sprint-2/improve_with_caches/fibonacci/fibonacci.py
Original file line number Diff line number Diff line change
@@ -1,4 +1,12 @@
memo = {}

def fibonacci(n):
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)
memo[n] = result

return result
49 changes: 37 additions & 12 deletions Sprint-2/improve_with_caches/making_change/making_change.py
Comment thread
RihannaP marked this conversation as resolved.
Original file line number Diff line number Diff line change
@@ -1,32 +1,57 @@
from typing import List

from typing import List, Dict, Tuple

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.
# Pre-create only 8 possible "coins" arrays
_BASE_COINS: List[int] = [200, 100, 50, 20, 10, 5, 2, 1]
COIN_SUFFIXES: List[Tuple[int, ...]] = [tuple(_BASE_COINS[i:]) for i in range(len(_BASE_COINS))]
Comment thread
RihannaP marked this conversation as resolved.
Outdated

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])
cache: Dict[Tuple[int, int], int] = {}

def ways_to_make_change(total: int) -> int:
cache.clear()
# suffix_id == 0 identifies [200, 100, 50, 20, 10, 5, 2, 1]
return ways_to_make_change_helper(total, suffix_id=0)


def ways_to_make_change_helper(total: int, coins: List[int]) -> int:
def ways_to_make_change_helper(total: int, suffix_id: int) -> int:
"""
Helper function for ways_to_make_change to avoid exposing the coins parameter to callers.
suffix_id uniquely identifies the subarray coins = COIN_SUFFIXES[suffix_id]
where suffix_id in [0..7].
"""
if total == 0 or len(coins) == 0:
key = (total, suffix_id)
if key in cache:
return cache[key]

coins = COIN_SUFFIXES[suffix_id]

# Keep behavior close to original, but fix the standard base case:
if total == 0:
return 1
if len(coins) == 0:
return 0

if len(coins) == 1:
cache[key] = 1 if (total % coins[0] == 0) else 0
return cache[key]
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Note: Even though tuple construction and dictionary lookup (lines 21 and 22) are O(1) operations,
they are still relatively costly than some simple numerical computation and comparisons (lines 28, 30, 33, 34).

This is unrelated to cache, but if you swap the code on lines 28-33 with the code on lines 21-23,
you can probably notice some slight improvement in performance.


Good job in figuring out a quicker way to compute number of ways when there is only one coin left to consider.

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Agreed. Checking that before doing cache lookup avoids unnecessary tuple creation and dictionary access


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
# Instead of slicing coins[coin_index+1:], move the suffix_id forward.
# suffix_id is the start index of `coins` in the base list, so:
next_suffix_id = suffix_id + coin_index + 1
if next_suffix_id < len(COIN_SUFFIXES):
ways += ways_to_make_change_helper(total - total_from_coins, suffix_id=next_suffix_id)

count_of_coin += 1

cache[key] = ways
return ways