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NW| 26-Jul-SDC | Ahmad Hmedan | Sprint 2 | Improve code with precomputing #236

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NW| 26-Jul-SDC | Ahmad Hmedan | Sprint 2 | Improve code with precomputing #236

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2754820
refactor: precomputing prefixes for faster lookup.
AhmadHmedann Oct 4, 2026
347b775
refactor: precomputing unique letters for faster lookup
AhmadHmedann Oct 4, 2026
7e39c41
chore: remove commented-out code
AhmadHmedann Oct 4, 2026
5269203
style: format code
AhmadHmedann Oct 5, 2026
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21 changes: 8 additions & 13 deletions Sprint-2/improve_with_precomputing/common_prefix/common_prefix.py
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Original file line number Diff line number Diff line change
Expand Up @@ -7,18 +7,13 @@ def find_longest_common_prefix(strings: List[str]):

In the event that an empty list, a list containing one string, or a list of strings with no common prefixes is passed, the empty string will be returned.
"""
seen = set()
longest = ""
for string_index, string in enumerate(strings):
for other_string in strings[string_index+1:]:
common = find_common_prefix(string, other_string)
if len(common) > len(longest):
longest = common
return longest

for string in strings:
for i in range(1, len(string) + 1):
prefix = string[:i]
if prefix in seen and len(prefix) > len(longest):
longest = prefix
Comment on lines +10 to +16

@cjyuan cjyuan Oct 10, 2026

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Can you use complexity to explain how the new implementation is better than the original implementation?

@AhmadHmedann AhmadHmedann Oct 11, 2026

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Original:

  • Time Complexity: O(N² × K)
  • Space Complexity: O(N + K)

New:

  • Time Complexity: O(N × K²)
  • Space Complexity: O(N × K)

Where:

  • N is the number of strings in the list.
  • K is the maximum length of a string.

@cjyuan cjyuan Oct 11, 2026

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The time complexity analyses are correct.

Both the original and the refactored code require more space. If string length, $K$, is to be taken into consideration, an array of strings would need O(Nk) space.

@AhmadHmedann AhmadHmedann Oct 11, 2026

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Do you mean the input space complexity?

I was calculating the auxiliary space complexity, excluding the input memory. Should I make that clear in my analysis?

I also realised that because of prefix = string[:i] and storing all the prefixes in a set, the auxiliary space complexity of the new implementation should be O(N × K²), not O(N × K).

Is my understanding correct?

@cjyuan cjyuan Oct 11, 2026

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Do you mean the input space complexity?

Yes

Should I make that clear in my analysis?

No need. Since your space complexity analyses were wrong. I was just trying to clarify.

I also realised that because of prefix = string[:i] and storing all the prefixes in a set, the auxiliary space complexity of the new implementation should be O(N × K²), not O(N × K).

Is my understanding correct?

Correct.


def find_common_prefix(left: str, right: str) -> str:
min_length = min(len(left), len(right))
for i in range(min_length):
if left[i] != right[i]:
return left[:i]
return left[:min_length]
seen.add(prefix)
return longest
5 changes: 3 additions & 2 deletions Sprint-2/improve_with_precomputing/count_letters/count_letters.py
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Expand Up @@ -3,9 +3,10 @@ def count_letters(s: str) -> int:
count_letters returns the number of letters which only occur in upper case in the passed string.
"""
only_upper = set()
for letter in s:
unique_letters = set(s)
for letter in unique_letters:
if is_upper_case(letter):
if letter.lower() not in s:
if letter.lower() not in unique_letters:
only_upper.add(letter)
return len(only_upper)

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