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17 changes: 15 additions & 2 deletions Sprint-2/improve_with_caches/fibonacci/fibonacci.py
Original file line number Diff line number Diff line change
@@ -1,4 +1,17 @@
#initialising a dictionary to store a copy of what we already know or have already computed
#caching reduces the fibonacci time complexity from exponential to linear.
cache = {}

def fibonacci(n):
if n <= 1:
if n in cache:
return cache[n]

if n <= 1: #when n is 0 or 1, we don’t need to calculate anything,just return th known value
return n
return fibonacci(n - 1) + fibonacci(n - 2)

result = fibonacci(n - 1) + fibonacci(n - 2)

#saving result in our dict before returning
cache[n] = result
return result

57 changes: 42 additions & 15 deletions Sprint-2/improve_with_caches/making_change/making_change.py
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@cjyuan cjyuan Feb 22, 2026

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Array creation is a relatively costly operation.

From line 42, 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 42 (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.

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  1. Thanks for the suggestion . I’ve updated the implementation so I no longer slice the coins list. Instead, I pass a start_index through the recursion, which avoids creating new subarrays entirely.
  2. I also switched the cache key to (total, start_index) rather than (total, tuple(coins)). This gives each of the 9 possible sub‑arrays a unique integer “identity” automatically, without needing a separate mapping
  3. This keeps the logic the same but removes the repeated list allocations and makes the memoisation faster and cheaper.

Original file line number Diff line number Diff line change
@@ -1,32 +1,59 @@
from typing import List

cache = {}

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


def ways_to_make_change_helper(total: int, coins: List[int]) -> int:
def ways_to_make_change_helper(total: int, coins: List[int],start_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:
#We found one valid way if its an exact match.
if total == 0:
return 1
# No coins left to try as we moved past the end of the list
if start_index >= len(coins):
return 0

# Cache key now uses (total, start_index) instead of hashing the whole coin list
key = (total, start_index)
if key in cache:
return cache[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

# Try using the current coin 1,2,3... times as long as we don't exceed the total.
coin = coins[start_index]
count_of_coin = 1

while coin * count_of_coin <= total:
total_from_coins = coin * count_of_coin

if total_from_coins == total:
# if exact match then one valid way

ways += 1
else:
# Remaining total, and move to the next coin (no slicing!)
intermediate = ways_to_make_change_helper(
total - total_from_coins,
coins,
start_index + 1
)
ways += intermediate

count_of_coin += 1

# Also consider the case where we skip this coin entirely
ways += ways_to_make_change_helper(total, coins, start_index + 1)

cache[key] = ways
return ways