London | 26-SDC-Mar | beko | Sprint 2 | improve code with precomputing

Sprint-2/improve_with_precomputing/common_prefix/common_prefix.py

@@ -8,11 +8,14 @@ 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.
     """
     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
+    sorted_string = sorted(strings)
+
+    for i in range(len(sorted_string) - 1):
+        common = find_common_prefix(
+            sorted_string[i], sorted_string[i + 1]
+        )
+        if len(common) > len(longest):
+            longest = common
     return longest
 
 
@@ -22,3 +25,7 @@ def find_common_prefix(left: str, right: str) -> str:
         if left[i] != right[i]:
             return left[:i]
     return left[:min_length]
+
+
+# Before: the logic force all-pairs comparison, and that results in O(n^2 * m) where n is the number of strings and m is the average string length.
+# After: I sort once, then compare only with adjacent pairs, improving it to O(n log n * m).

Sprint-2/improve_with_precomputing/count_letters/count_letters.py

@@ -2,13 +2,14 @@ def count_letters(s: str) -> int:
     """
     count_letters returns the number of letters which only occur in upper case in the passed string.
     """
+    lowercase_letters = {letter.lower() for letter in s if letter.islower()}
+
     only_upper = set()
     for letter in s:
-        if is_upper_case(letter):
-            if letter.lower() not in s:
-                only_upper.add(letter)
+        if letter.isupper() and letter.lower() not in lowercase_letters:
+            only_upper.add(letter)
     return len(only_upper)
 
 
-def is_upper_case(letter: str) -> bool:
-    return letter == letter.upper()
+# Before: O(n^2) for each uppercase letter, we checked if the lowercase form was in the string s itself, an O(n) scan repeated per letter.
+# After: precompute a set of lowercase letters seen in 's' once (O(n)), so each presence check becomes O(1), making the overall complexity O(n).