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London | 26-SDC-March | Jamal Laqdiem| Sprint 2 | Improve code with caches #214
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51b891b
feat: optomise fibonacci using manual dictionary memoization
jamallaqdiem 66cfc34
feat: optimize change helper with index-based recursion and caching
jamallaqdiem 88578a2
fix: remove the globale scope and use a default argument in the funct…
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -1,4 +1,10 @@ | ||
| def fibonacci(n): | ||
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| 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
43
Sprint-2/improve_with_caches/making_change/making_change.py
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -1,32 +1,41 @@ | ||
| from typing import List | ||
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| from typing import List, Tuple, Dict | ||
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| 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. | ||
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| 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) | ||
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| 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: | ||
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| #If total is 0, we found exactly 1 valid way to make change | ||
| if total == 0: | ||
| return 1 | ||
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| #If total becomes negative or we run out of coins, this way failed | ||
| if total < 0 or coin_index >= len(coins): | ||
| return 0 | ||
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| 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] | ||
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| current_coin = coins[coin_index] | ||
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| # 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) | ||
| ) | ||
| # store the result in the cache before the return | ||
| change_cache[cache_key] = ways | ||
| return ways | ||
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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: