Skip to content

Glasgow | 25-SDC-Nov | Nataliia Volkova | Sprint 2 | improve_with_caches #106

New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

Sign up for GitHub

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

Open
Nataliia74 wants to merge 3 commits into
base: main
Choose a base branch
from
Open

Glasgow | 25-SDC-Nov | Nataliia Volkova | Sprint 2 | improve_with_caches #106

Show file tree
Hide file tree
Changes from 2 commits
Commits
Show all changes
3 commits
Select commit Hold shift + click to select a range
2da2744
fibonacci
Nataliia74 Feb 15, 2026
62e1e71
making_change
Nataliia74 Feb 15, 2026
403c901
simplify complexity, math instead recursion on 0 total, and 0 number…
Nataliia74 Apr 21, 2026
File filter

Filter by extension

Filter by extension
Conversations
Failed to load comments.
Loading
Jump to
Jump to file
Failed to load files.
Loading
Diff view
Diff view
12 changes: 10 additions & 2 deletions Sprint-2/improve_with_caches/fibonacci/fibonacci.py
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -1,4 +1,12 @@
cache = {}

def fibonacci(n):
if n in cache:
return cache[n]
if n <= 1:
return n
return fibonacci(n - 1) + fibonacci(n - 2)
result = n
else:
result = fibonacci(n - 1) + fibonacci(n - 2)

cache[n] = result
return result
18 changes: 16 additions & 2 deletions Sprint-2/improve_with_caches/making_change/making_change.py
View file
Open in desktop
Copy link
Copy Markdown

@cjyuan cjyuan Feb 22, 2026

Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

Array creation is a relatively costly operation.

From line 41, we know coins can only be one of the following 9 arrays:

[200, 100, 50, 20, 10, 5, 2, 1]
[100, 50, 20, 10, 5, 2, 1]
[50, 20, 10, 5, 2, 1]
...
[1]
[]

We could further improve the performance if we can

  • avoid repeatedly creating the same sub-arrays at line 41 (e.g. use another cache), and
  • create key as (total, a_unique_integer_identifying_the_subarray) instead of as (total, tuple of coins)
    • There are only a small number of different subarrays. We can easily assign each subarray a unique integer.

Copy link
Copy Markdown
Author

@Nataliia74 Nataliia74 Apr 21, 2026

Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

Eliminated slicing, which is costly and creates a new array every recursion, and also many duplicates; the tuple conversion was costly as well. Changed by exchanging the tuple at the index an array.

Original file line number Diff line number Diff line change
@@ -1,5 +1,6 @@
from typing import List

cache = {}

def ways_to_make_change(total: int) -> int:
"""
Expand All @@ -8,14 +9,25 @@ 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])



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

if total == 1 or len(coins) == 1:
result = 1
cache[key] = result
return result
Copy link
Copy Markdown

@cjyuan cjyuan Feb 22, 2026

Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

When total == 1 or len(coins) == 1 is true, the result may not be 1 if the last coin is not 1 (e.g. the initial list of coins given is different)

On the other hand, when len(coins) == 1, there is simple way to determine if the current total (any value) can be made by the coin value.

Copy link
Copy Markdown
Author

@Nataliia74 Nataliia74 Apr 21, 2026

Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

Hello CJ Yuan, I understood that I can have 2 possible results, not just 1. So, the best founded solution is not to use recursion, but simple math to check if the total is divisible by that left type coin.

Copy link
Copy Markdown

@cjyuan cjyuan Apr 21, 2026 •
edited
Loading

Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

What do you mean by 2 possible results? 0 or 1?

Yes, this approach:

simple math to check if the total is divisible by that left type coin

Copy link
Copy Markdown
Author

@Nataliia74 Nataliia74 Apr 22, 2026 •
edited
Loading

Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

Hello CJ Yuan, I mean 0 or 1. In this case, a better approach is to use math.


ways = 0
for coin_index in range(len(coins)):
Expand All @@ -29,4 +41,6 @@ def ways_to_make_change_helper(total: int, coins: List[int]) -> int:
intermediate = ways_to_make_change_helper(total - total_from_coins, coins=coins[coin_index+1:])
ways += intermediate
count_of_coin += 1
return ways

cache[key] = ways
return ways