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..0a6e883 100644
--- a/Sprint-2/improve_with_precomputing/common_prefix/common_prefix.py
+++ b/Sprint-2/improve_with_precomputing/common_prefix/common_prefix.py
@@ -7,12 +7,24 @@ 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 len(strings) < 2:
+        return ""
+    
+    # Sorting first (O(n log n * k)) means the longest common prefix between
+    # ANY two strings in the list is guaranteed to be between two ADJACENT
+    # strings after sorting. This lets us check only n-1 adjacent pairs
+    # instead of comparing every pair (O(n^2 * k) in the original version),
+    # reducing the overall complexity to O(n log n * k).
+    
+    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(strings) - 1):
+        common = find_common_prefix(strings[i], strings[i + 1])
+        if len(common) > len(longest):
+            longest = common
     return longest
 
 
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..6b2a621 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 +2,15 @@ def count_letters(s: str) -> int:
     """
     count_letters returns the number of letters which only occur in upper case in the passed string.
     """
+    # Precompute a set of all characters in s: O(n) once.
+    # This turns the `letter.lower() not in s` check (O(n) linear scan,
+    # making the whole function O(n^2)) into an O(1) set lookup,
+    # bringing the overall complexity down to O(n).
+    letters = set(s)
     only_upper = set()
+    
     for letter in s:
-        if is_upper_case(letter):
-            if letter.lower() not in s:
+        if letter.isupper():
+            if letter.lower() not in letters:
                 only_upper.add(letter)
     return len(only_upper)
-
-
-def is_upper_case(letter: str) -> bool:
-    return letter == letter.upper()