diff --git a/Sprint-2/improve_with_precomputing/common_prefix/common_prefix.py b/Sprint-2/improve_with_precomputing/common_prefix/common_prefix.py index f4839e7..a2b0f74 100644 --- a/Sprint-2/improve_with_precomputing/common_prefix/common_prefix.py +++ b/Sprint-2/improve_with_precomputing/common_prefix/common_prefix.py @@ -3,22 +3,36 @@ def find_longest_common_prefix(strings: List[str]): """ - find_longest_common_prefix returns the longest string common at the start of any two strings in the passed list. + Return the longest common prefix shared by any two strings in the list. - 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. + If the list is empty, contains only one string, or no common prefixes + exist, an empty string is returned. """ + if len(strings) < 2: + return "" + + # Sorting places strings with similar prefixes next to each other. + # This reduces comparisons from O(n^2) to O(n log n). + strings.sort() + 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 + for i in range(len(strings) - 1): + # Compare only adjacent strings after sorting. + common = find_common_prefix(strings[i], strings[i + 1]) + if len(common) > len(longest): + longest = common + return longest def find_common_prefix(left: str, right: str) -> str: + """ + Return the common prefix between two strings. + """ 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] diff --git a/Sprint-2/improve_with_precomputing/count_letters/count_letters.py b/Sprint-2/improve_with_precomputing/count_letters/count_letters.py index 62c3ec0..6f39741 100644 --- a/Sprint-2/improve_with_precomputing/count_letters/count_letters.py +++ b/Sprint-2/improve_with_precomputing/count_letters/count_letters.py @@ -2,13 +2,28 @@ 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() + lower_case = set() + upper_case= set() + + +## Precompute lowercase and uppercase letters for letter in s: - if is_upper_case(letter): - if letter.lower() not in s: - only_upper.add(letter) - return len(only_upper) + if letter.islower(): + lower_case.add(letter) + elif letter.isupper(): + upper_case.add(letter) + + +# Count uppercase letters whose lowercase version never appears + count = 0 + for letter in upper_case: + if letter.lower() not in lower_case: + count += 1 + return count def is_upper_case(letter: str) -> bool: return letter == letter.upper() + +#The original solution was O(n²) because it repeatedly checked membership in the full string. +#By precomputing lowercase and uppercase sets, membership checks become O(1), so the whole function becomes O(n). \ No newline at end of file