London | 26-Jul-SDC | Roman Sanaye | Sprint 2 | Improve code with pre-computing

Sprint-2/improve_with_precomputing/common_prefix/common_prefix.py

@@ -7,12 +7,21 @@ 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.
     """
+    if not strings or len(strings) < 2:
+        return ""
+
+    # Pre-computation: Sort the strings lexicographically.
+    # The longest common prefix of any two strings in the collection
+    # will always be shared by at least one pair of adjacent strings in the sorted list.
+    sorted_strings = sorted(strings)
+
     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(sorted_strings) - 1):
+        common = find_common_prefix(sorted_strings[i], sorted_strings[i + 1])
+        if len(common) > len(longest):
+            longest = common
+
     return longest
 
 
@@ -22,3 +31,16 @@ def find_common_prefix(left: str, right: str) -> str:
         if left[i] != right[i]:
             return left[:i]
     return left[:min_length]
+
+
+# Time Complexity Comparison:
+# Original version: It used a nested loop to check every possible pair of strings.
+# This gave it a slow quadratic time complexity (O(N^2)), which gets very heavy
+# and slow as the list grows larger.
+#
+# New version: We sort the strings alphabetically first. When a list is sorted,
+# any two strings with the longest common prefix will always be right next to each other.
+# Because of this, we only need a single loop to check neighbor pairs.
+#
+# Result: This changes the time complexity from O(N^2) to O(N log N),
+# making the code much faster and more efficient.

Sprint-2/improve_with_precomputing/count_letters/count_letters.py

@@ -1,14 +1,32 @@
+from typing import Set
+
+
 def count_letters(s: str) -> int:
     """
-    count_letters returns the number of letters which only occur in upper case in the passed string.
+    Returns the number of alphabetic letters that only occur in upper case in the passed string.
+    Optimized to O(N) time complexity using sets.
     """
-    only_upper = set()
-    for letter in s:
-        if is_upper_case(letter):
-            if letter.lower() not in s:
-                only_upper.add(letter)
+    if not s:
+        return 0
+
+    # Collect all unique upper-case and lower-case characters
+    upper_chars: Set[str] = {char for char in s if char.isupper()}
+    lower_chars: Set[str] = {char.lower() for char in s if char.islower()}
+
+    # Find upper-case letters whose lower-case versions do not exist in the string
+    only_upper = {char for char in upper_chars if char.lower() not in lower_chars}
+
     return len(only_upper)
 
 
-def is_upper_case(letter: str) -> bool:
-    return letter == letter.upper()
+# Time and Space Complexity Comparison:
+#
+# Original version:
+# - Time Complexity: O(N^2). It used a loop to look at every character, and inside the loop, it searched the entire string s (letter.lower() not in s). This nested search makes it quadratic and slow for long texts.
+# - Space Complexity: O(N). It stored unique letters in a set.
+#
+# Refactored version (with Sets):
+# - Time Complexity: O(N). It passes through the string to collect uppercase and lowercase characters into sets first. Because checking a set takes O(1) time, looking up the lowercase letter is instant.
+# - Space Complexity: O(N). It uses extra memory to store unique characters in sets, but this is a minor trade-off for a massive increase in speed.
+#
+# Result: The time complexity drops from O(N^2) to O(N), making the code much faster and more efficient for large inputs.