Word Ladder II Double BFS
A transformation sequence from word beginWord to word endWord using a dictionary wordList is a sequence of words beginWord -> s1 -> s2 -> ... -> sk such that:
Every adjacent pair of words differs by a single letter.
Every si for 1 <= i <= k is in wordList. Note that beginWord does not need to be in wordList.
sk == endWord
Given two words, beginWord and endWord, and a dictionary wordList, return all the shortest transformation sequences from beginWord to endWord, or an empty list if no such sequence exists. Each sequence should be returned as a list of the words [beginWord, s1, s2, ..., sk].
from collections import defaultdict
class Solution:
def findLadders(self, beginWord: str, endWord: str, wordList: List[str]) -> List[List[str]]:
# edge case
if endWord not in wordList:
return []
# 1) build neighbor list for first bfs
if beginWord not in wordList:
wordList.append(beginWord)
unseen = set(wordList)
word_size = len(beginWord)
neighbors = defaultdict(list)
for word in wordList:
for i in range(word_size):
neighbors[f'{word[:i]}*{word[i+1:]}'].append(word)
# 2) do first bfs and build reversed neighbors list for second bfs
reverse_neighbors = defaultdict(list)
n_t_h = [beginWord]
unseen.remove(beginWord)
while n_t_h:
new_seen = set()
for word in n_t_h:
for i in range(word_size):
for neighbor in neighbors[f'{word[:i]}*{word[i+1:]}']:
if neighbor in unseen:
reverse_neighbors[neighbor].append(word)
new_seen.add(neighbor)
n_t_h = list(new_seen)
unseen -= new_seen
if reverse_neighbors[endWord]:
break
# if endWord does not have reversed neigbors it is not reachable so return empty list
if not reverse_neighbors[endWord]:
return []
# 3) do second bfs
paths = [[endWord]]
while True:
new_paths = []
for path in paths:
last_node = path[-1]
for reverse_neighbor in reverse_neighbors[last_node]:
new_paths.append(path + [reverse_neighbor])
paths = new_paths
if paths[0][-1] == beginWord:
break
# 4) reverse the paths
result = []
for path in paths:
path.reverse()
result.append(path)
return result
Related Problems
There is an integer array nums sorted in ascending order (with distinct values).
Prior to being passed to your function, nums is possibly rotated at an unknown pivot index k (1 <= k < nums.length) such that the resulting array is [nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]] (0-indexed). For example, [0,1,2,4,5,6,7] might be rotated at pivot index 3 and become [4,5,6,7,0,1,2].
Given the array nums after the possible rotation and an integer target, return the index of target if it is in nums, or -1 if it is not in nums.
You must write an algorithm with O(log n) runtime complexity.
Given an array of distinct integers candidates and a target integer target, return a list of all unique combinations of candidates where the chosen numbers sum to target. You may return the combinations in any order.
The same number may be chosen from candidates an unlimited number of times. Two combinations are unique if the frequency of at least one of the chosen numbers is different.
There are a total of numCourses courses you have to take, labeled from 0 to numCourses - 1. You are given an array prerequisites where prerequisites[i] = [ai, bi] indicates that you must take course bi first if you want to take course ai.
For example, the pair [0, 1], indicates that to take course 0 you have to first take course 1.
Return true if you can finish all courses. Otherwise, return false.
Given an array nums of n integers where nums[i] is in the range [1, n], return an array of all the integers in the range [1, n] that do not appear in nums.