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This approach leverages Breadth First Search (BFS), where each word is a node in the graph, and an edge exists between two nodes if they differ by exactly one letter. The goal is to find the shortest path from beginWord to endWord.
Start BFS from the beginWord, exploring transformations by changing each letter and checking if the new word exists in the set of available words. Polished implementation should consider preprocessing the word list for efficient lookup.
Time Complexity: O(N * m^2), where N is the number of words in the list and m is the length of each word. Each word pair is compared in the worst case.
Space Complexity: O(N), where N is the number of nodes (words) we store in the queue.
1from collections import deque
2
3def ladderLength(beginWord, endWord, wordList):
4 wordSet = set(wordList)
5 if endWord not in wordSet:
6 return 0
7 q = deque([beginWord])
8 ladder = 1
9
10 while q:
11 levelSize = len(q)
12 for _ in range(levelSize):
13 current = q.popleft()
14 for i in range(len(current)):
15 for c in 'abcdefghijklmnopqrstuvwxyz':
16 nextWord = current[:i] + c + current[i+1:]
17 if nextWord == endWord:
18 return ladder + 1
19 if nextWord in wordSet:
20 wordSet.remove(nextWord)
21 q.append(nextWord)
22 ladder += 1
23 return 0
24
25beginWord = "hit"
26endWord = "cog"
27wordList = ["hot", "dot", "dog", "lot", "log", "cog"]
28print(ladderLength(beginWord, endWord, wordList))This Python solution employs BFS utilizing a deque for efficient popping from the front. Each character in a word is replaced to generate transformations. Successful transformations (existing in the word set) are queued. Direct transformation to endWord increases the level count by one, indicating a complete path.