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London | 25-SDC-July | Mikiyas Gebremichael | Sprint 2 | Improve code with precomputing #38
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38 changes: 15 additions & 23 deletions
38
Sprint-2/improve_with_precomputing/common_prefix/common_prefix.py
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -1,24 +1,16 @@ | ||
| from typing import List | ||
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| 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. | ||
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| 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. | ||
| """ | ||
| 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 | ||
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| 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] | ||
| def find_longest_common_prefix(strings: List[str]) -> str: | ||
| if len(strings) < 2: | ||
| return "" | ||
| #Find shortest string limit for prefix | ||
| min_len = min(len(s) for s in strings) | ||
| if min_len == 0: | ||
| return "" | ||
| #Compare character by character across all strings | ||
| for i in range(min_len): | ||
| char_set = set(s[i] for s in strings) | ||
| if len(char_set) > 1: #mismatch found | ||
| return strings[0][:i] | ||
| return strings[0][:min_len] | ||
| #Originally Compares every pair O(n² * m) | ||
| #now compare column wise across all strings O(n * m) | ||
27 changes: 14 additions & 13 deletions
27
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 +1,15 @@ | ||
| from collections import Counter | ||
| 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: | ||
| if is_upper_case(letter): | ||
| if letter.lower() not in s: | ||
| only_upper.add(letter) | ||
| return len(only_upper) | ||
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| def is_upper_case(letter: str) -> bool: | ||
| return letter == letter.upper() | ||
| #Returns the number of letters which only occur in uppercase in the passed string | ||
| letters = Counter(s) #count occurrences of each character | ||
|
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. This works, but is doing more work than is needed. We don't care if T occurs 3 times or 1, if t doesn't. What could you use instead of a Counter that would skip the counting process but still remove duplicates? |
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| only_upper = 0 | ||
| for letter in letters: | ||
| if letter.isupper(): | ||
| if letter.lower() not in letters: | ||
| only_upper += 1 | ||
| return only_upper | ||
| #The first implementation was O(n² * m) in the worst case comparing every pair of strings character by character. | ||
| # Now, | ||
| #Counter scans the string once O(n) | ||
| #Checking letter.lower() in letters O(1) per letter | ||
| #Total time O(n + u) where u is the number of unique letters | ||
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No, this is wrong. If the input is ["","string_one","string_two"] we want to return "string_" whereas you'll return ""