revert test optimization

Author: mohammedahmed18

codeflash/discovery/functions_to_optimize.py

@@ -306,43 +306,25 @@ def levenshtein_distance(s1: str, s2: str) -> int:
     len1 = len(s1)
     len2 = len(s2)
     # Use a preallocated list instead of creating a new list every iteration
-
-    # Early exit for empty string cases
-    if len1 == 0:
-        return len2
-    if len2 == 0:
-        return len1
-
-    # Convert strings to lists for fast indexed access
-    s1_list = list(s1)
-    s2_list = list(s2)
-
-    # Preallocate and reuse arrays; avoid creating new ones every iteration
     previous = list(range(len1 + 1))
     current = [0] * (len1 + 1)
 
     for index2 in range(len2):
-        char2 = s2_list[index2]
+        char2 = s2[index2]
         current[0] = index2 + 1
-
-        # Remove redundant intermediate assignments for better cache locality
-        prev = previous
-        curr = current
-        s1_chars = s1_list
-        # Use local variables for frequently accessed values
         for index1 in range(len1):
-            # Unrolling char1 assignment and equality check
-            if s1_chars[index1] == char2:
-                curr[index1 + 1] = prev[index1]
+            char1 = s1[index1]
+            if char1 == char2:
+                current[index1 + 1] = previous[index1]
             else:
-                x = prev[index1]
-                y = prev[index1 + 1]
-                z = curr[index1]
-                min_xy = min(x, y)
-                min_xyz = min(z, min_xy)
-                curr[index1 + 1] = 1 + min_xyz
-
-        # Swap references rather than copying data
+                # Fast min calculation without tuple construct
+                a = previous[index1]
+                b = previous[index1 + 1]
+                c = current[index1]
+                min_val = min(b, a)
+                min_val = min(c, min_val)
+                current[index1 + 1] = 1 + min_val
+        # Swap references instead of copying
         previous, current = current, previous
     return previous[len1]