Create Word_Ladder.py

Author: IChowdhury01

Algorithms/Python/Graphs/Word_Ladder.py

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+class Solution:
+    def ladderLength(self, beginWord: str, endWord: str, wordList: List[str]) -> int:
+        '''
+        Make a directed graph connecting words with 1 character difference. Then do BFS until you reach endWord, or all nodes are traversed. Return 0 if you never reached endWord, otherwise return the level count.
+        O(N^2 * K) Time. 
+            K = length of words. N = number of words. 
+            Building adjacency list is O(N * K^2), because for each word we loop through each character and then perform splicing to create our pattern. Splicing and Inner loop are both O(K). 
+            Doing BFS on the graph is O(N^2 * K). BFS on fully connected graph = O(V+E) = O(V^2). We also generate patterns in O(K) time on each recursive call.
+            N >> K so the BFS TC takes precendence.
+        O(N * K^2) Space. Each word has length K. Each word is stored K times (for each possible pattern) in adjacency list. There are N words. 
+        Python: Create a Set/Queue with initial values by passing in a list, e.g. deque([beginWord])
+        '''
+        adj = defaultdict(list)
+        wordList.append(beginWord)
+        for word in wordList:
+            for i in range(len(word)):
+                pattern = word[:i] + "*" + word[i+1:]
+                adj[pattern].append(word)
+        
+        q = deque([beginWord])
+        visited = set([beginWord])
+        seqLen = 1
+        while q:
+            for i in range(len(q)):
+                word = q.popleft()
+                for j in range(len(word)):
+                    pattern = word[:j] + "*" + word[j+1:]
+                    for neighbor in adj[pattern]:
+                        if neighbor not in visited:
+                            q.append(neighbor)
+                            visited.add(neighbor)
+                            if neighbor == endWord:
+                                return seqLen + 1
+            seqLen += 1
+        return 0