126. Word Ladder II #

Problem #

Given two words (beginWord and endWord), and a dictionary’s word list, find all shortest transformation sequence(s) from beginWord to endWord, such that:

1. Only one letter can be changed at a time
2. Each transformed word must exist in the word list. Note that beginWord is not a transformed word.

Note:

• Return an empty list if there is no such transformation sequence.
• All words have the same length.
• All words contain only lowercase alphabetic characters.
• You may assume no duplicates in the word list.
• You may assume beginWord and endWord are non-empty and are not the same.

Example 1:

Input:
beginWord = "hit",
endWord = "cog",
wordList = ["hot","dot","dog","lot","log","cog"]

Output:
[
  ["hit","hot","dot","dog","cog"],
  ["hit","hot","lot","log","cog"]
]

Example 2:

Input:
beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log"]

Output: []

Explanation: The endWord "cog" is not in wordList, therefore no possible transformation.

Problem Summary #

Given two words (beginWord and endWord) and a dictionary wordList, find all shortest transformation sequences from beginWord to endWord. The transformation must follow these rules:

1. Only one letter can be changed at a time.
2. Intermediate words in the transformation process must be words in the dictionary.

Notes:

• If no such transformation sequence exists, return an empty list.
• All words have the same length.
• All words consist only of lowercase letters.
• There are no duplicate words in the dictionary.
• You may assume beginWord and endWord are non-empty and are not the same.

Solution Approach #

• This problem is an enhanced version of Problem 127. In addition to finding the length of the path, it further requires outputting all paths. The solution approach is the same as Problem 127, also using BFS traversal.
• The current approach is not optimal. Could bidirectional BFS optimization, or Dijkstra’s algorithm, be considered?

Code #

package leetcode

func findLadders(beginWord string, endWord string, wordList []string) [][]string {
	result, wordMap := make([][]string, 0), make(map[string]bool)
	for _, w := range wordList {
		wordMap[w] = true
	}
	if !wordMap[endWord] {
		return result
	}
	// create a queue, track the path
	queue := make([][]string, 0)
	queue = append(queue, []string{beginWord})
	// queueLen is used to track how many slices in queue are in the same level
	// if found a result, I still need to finish checking current level cause I need to return all possible paths
	queueLen := 1
	// use to track strings that this level has visited
	// when queueLen == 0, remove levelMap keys in wordMap
	levelMap := make(map[string]bool)
	for len(queue) > 0 {
		path := queue[0]
		queue = queue[1:]
		lastWord := path[len(path)-1]
		for i := 0; i < len(lastWord); i++ {
			for c := 'a'; c <= 'z'; c++ {
				nextWord := lastWord[:i] + string(c) + lastWord[i+1:]
				if nextWord == endWord {
					path = append(path, endWord)
					result = append(result, path)
					continue
				}
				if wordMap[nextWord] {
					// different from word ladder, don't remove the word from wordMap immediately
					// same level could reuse the key.
					// delete from wordMap only when currently level is done.
					levelMap[nextWord] = true
					newPath := make([]string, len(path))
					copy(newPath, path)
					newPath = append(newPath, nextWord)
					queue = append(queue, newPath)
				}
			}
		}
		queueLen--
		// if queueLen is 0, means finish traversing current level. if result is not empty, return result
		if queueLen == 0 {
			if len(result) > 0 {
				return result
			}
			for k := range levelMap {
				delete(wordMap, k)
			}
			// clear levelMap
			levelMap = make(map[string]bool)
			queueLen = len(queue)
		}
	}
	return result
}