bpo-41972: Use the two-way algorithm for string searching (GH-22904)

Implement an enhanced variant of Crochemore and Perrin's Two-Way string searching algorithm, which reduces worst-case time from quadratic (the product of the string and pattern lengths) to linear. This applies to forward searches (like``find``, ``index``, ``replace``); the algorithm for reverse searches (like ``rfind``) is not changed.

Co-authored-by: Tim Peters <tim.peters@gmail.com>

Lib/test/string_tests.py

@@ -5,6 +5,7 @@ Common tests shared by test_unicode, test_userstring and test_bytes.
 import unittest, string, sys, struct
 from test import support
 from collections import UserList
+import random
 
 class Sequence:
     def __init__(self, seq='wxyz'): self.seq = seq
@@ -317,6 +318,44 @@ class BaseTest:
         else:
             self.checkraises(TypeError, 'hello', 'rindex', 42)
 
+    def test_find_periodic_pattern(self):
+        """Cover the special path for periodic patterns."""
+        def reference_find(p, s):
+            for i in range(len(s)):
+                if s.startswith(p, i):
+                    return i
+            return -1
+
+        rr = random.randrange
+        choices = random.choices
+        for _ in range(1000):
+            p0 = ''.join(choices('abcde', k=rr(10))) * rr(10, 20)
+            p = p0[:len(p0) - rr(10)] # pop off some characters
+            left = ''.join(choices('abcdef', k=rr(2000)))
+            right = ''.join(choices('abcdef', k=rr(2000)))
+            text = left + p + right
+            with self.subTest(p=p, text=text):
+                self.checkequal(reference_find(p, text),
+                                text, 'find', p)
+
+    def test_find_shift_table_overflow(self):
+        """When the table of 8-bit shifts overflows."""
+        N = 2**8 + 100
+
+        # first check the periodic case
+        # here, the shift for 'b' is N + 1.
+        pattern1 = 'a' * N + 'b' + 'a' * N
+        text1 = 'babbaa' * N + pattern1
+        self.checkequal(len(text1)-len(pattern1),
+                        text1, 'find', pattern1)
+
+        # now check the non-periodic case
+        # here, the shift for 'd' is 3*(N+1)+1
+        pattern2 = 'ddd' + 'abc' * N + "eee"
+        text2 = pattern2[:-1] + "ddeede" * 2 * N + pattern2 + "de" * N
+        self.checkequal(len(text2) - N*len("de") - len(pattern2),
+                        text2, 'find', pattern2)
+
     def test_lower(self):
         self.checkequal('hello', 'HeLLo', 'lower')
         self.checkequal('hello', 'hello', 'lower')