From 66b6ecd19370cbbf26e11abb014c6d9af6f61c49 Mon Sep 17 00:00:00 2001
From: ivan <ivan1016017@gmail.com>
Date: Fri, 24 Jul 2026 05:14:12 -0600
Subject: [PATCH] adding updates

---
 .../round_1/25_word_ladder.py                 | 43 +++++++++++++++++++
 1 file changed, 43 insertions(+)
 create mode 100644 src/my_project/interviews/google_top_exercises/round_1/25_word_ladder.py

diff --git a/src/my_project/interviews/google_top_exercises/round_1/25_word_ladder.py b/src/my_project/interviews/google_top_exercises/round_1/25_word_ladder.py
new file mode 100644
index 00000000..ae56a692
--- /dev/null
+++ b/src/my_project/interviews/google_top_exercises/round_1/25_word_ladder.py
@@ -0,0 +1,43 @@
+from typing import List
+from collections import defaultdict, deque
+
+
+class Solution:
+    def ladderLength(self, beginWord: str, endWord: str, wordList: List[str]) -> int:
+        """BFS over a generic-pattern graph.
+
+        Build buckets keyed by patterns like 'h*t' so every word that differs
+        by a single letter shares a bucket. BFS from beginWord finds the
+        shortest transformation, counting words in the sequence.
+        """
+
+        if endWord not in wordList:
+            return 0
+
+        # Map each wildcard pattern -> list of words matching it.
+        patterns = defaultdict(list)
+        for word in wordList:
+            for i in range(len(word)):
+                pattern = word[:i] + "*" + word[i + 1:]
+                patterns[pattern].append(word)
+
+        # BFS. Level = number of words in the sequence so far (beginWord counts as 1).
+        queue = deque([(beginWord, 1)])
+        visited = {beginWord}
+
+        while queue:
+            word, level = queue.popleft()
+
+            if word == endWord:
+                return level
+
+            for i in range(len(word)):
+                pattern = word[:i] + "*" + word[i + 1:]
+                for neighbor in patterns[pattern]:
+                    if neighbor not in visited:
+                        visited.add(neighbor)
+                        queue.append((neighbor, level + 1))
+                # Clear the bucket so it is not scanned again by another word.
+                patterns[pattern] = []
+
+        return 0