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..aa5cdfc 100644
--- a/Sprint-2/improve_with_precomputing/common_prefix/common_prefix.py
+++ b/Sprint-2/improve_with_precomputing/common_prefix/common_prefix.py
@@ -7,18 +7,22 @@ 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 puts string with similar prefixes next to each other and reduces the complexity from O(n^2)to O(nlogn)
+    strings.sort() 
     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]) #we only compare the ones next to each other and use find_common_prefix to get their shared prefix
+        if len(common)> len(longest):
+            longest = common
     return longest
 
 
+
 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]
+        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]
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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..6f39741 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 +2,28 @@ 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()
+    lower_case = set()
+    upper_case= set()
+
+
+## Precompute lowercase and uppercase letters
     for letter in s:
-        if is_upper_case(letter):
-            if letter.lower() not in s:
-                only_upper.add(letter)
-    return len(only_upper)
+        if letter.islower():
+            lower_case.add(letter)
+        elif letter.isupper():
+            upper_case.add(letter)
+
+
+# Count uppercase letters whose lowercase version never appears
+    count = 0
+    for letter in upper_case:
+        if letter.lower() not in lower_case:
+            count += 1
+    return count
 
 
 def is_upper_case(letter: str) -> bool:
     return letter == letter.upper()
+
+#The original solution was O(n²) because it repeatedly checked membership in the full string.
+#By precomputing lowercase and uppercase sets, membership checks become O(1), so the whole function becomes O(n).
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