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Improve with percomputing #5
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21 changes: 11 additions & 10 deletions
21
Sprint-2/improve_with_precomputing/common_prefix/common_prefix.py
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20 changes: 10 additions & 10 deletions
20
Sprint-2/improve_with_precomputing/count_letters/count_letters.py
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -1,14 +1,14 @@ | ||
| def count_letters(s: str) -> int: | ||
| def count_letters(text: 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: | ||
| if is_upper_case(letter): | ||
| if letter.lower() not in s: | ||
| only_upper.add(letter) | ||
| # Find all lowercase letters in the string | ||
| lower_set = set(char for char in text if char.islower()) | ||
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| # Find all uppercase letters in the string | ||
| upper_set = set(char for char in text if char.isupper()) | ||
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| # Count uppercase letters that don't have lowercase versions | ||
| only_upper = {char for char in upper_set if char.lower() not in lower_set} | ||
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| return len(only_upper) | ||
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| def is_upper_case(letter: str) -> bool: | ||
| return letter == letter.upper() | ||
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This is the key precomputing step. By building
lower_setonce here (O(n)), you turn every subsequent membership check into O(1) instead of O(n). The original code didletter.lower() not in swheresis a plain string — Python'sinoperator on strings is a substring search and costs O(n) per call, making the old function O(n²) overall. Your approach is a textbook example of precomputing a lookup structure.There was a problem hiding this comment.
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So, what is a better way to write it?