How do you solve Word Ladder II on LeetCode?

Word Ladder II (LeetCode #126) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: The problem can be visualized as a graph where each word is a node and edges exist between words that differ by one letter.

What is the brute force solution for Word Ladder II?

The brute force approach for Word Ladder II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach tries to generate all possible transformations from the beginWord to the endWord by exploring every possible word in the wordList. It checks each transformation to see if it differs by one letter, which can be very inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Word Ladder II?

Word Ladder II uses the following concepts: Hash Table, String, Backtracking, Breadth-First Search. The recommended patterns to study are: Breadth-First Search, Graph Traversal, Backtracking.

What are the interview tips for Word Ladder II?

Clearly explain your thought process and how you are approaching the problem. Ask clarifying questions about the constraints and requirements before jumping into coding. Consider edge cases, such as when the endWord is not in the wordList, and handle them early.

What are common mistakes when solving Word Ladder II?

Failing to check if endWord is in the wordList at the beginning, leading to unnecessary computations. Not properly managing visited words, which can lead to infinite loops or incorrect paths.

#126

Word Ladder II

Hard

Hash TableStringBacktrackingBreadth-First SearchBreadth-First SearchGraph TraversalBacktracking

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a breadth-first search (BFS) to find the shortest paths from beginWord to endWord. It explores all possible transformations level by level, ensuring that we only consider the shortest paths and avoids revisiting nodes unnecessarily.

⚙️

Algorithm

3 steps

1Step 1: Initialize a queue with the beginWord and a set to track visited words.
2Step 2: Perform BFS to find all possible transformations, storing paths as we go.
3Step 3: Once we reach endWord, backtrack to construct all valid paths.

solution.py33 lines

1# Full working Python code
2from collections import defaultdict
3
4class Solution:
5    def findLadders(self, beginWord, endWord, wordList):
6        wordSet = set(wordList)
7        if endWord not in wordSet:
8            return []
9        res = []
10        self.bfs(beginWord, endWord, wordSet, res)
11        return res
12
13    def bfs(self, beginWord, endWord, wordSet, res):
14        queue = [[beginWord]]
15        visited = set([beginWord])
16        found = False
17
18        while queue and not found:
19            next_level = set()
20            for path in queue:
21                last_word = path[-1]
22                for i in range(len(last_word)):
23                    for c in 'abcdefghijklmnopqrstuvwxyz':
24                        new_word = last_word[:i] + c + last_word[i+1:]
25                        if new_word in wordSet:
26                            if new_word == endWord:
27                                found = True
28                                res.append(path + [new_word])
29                            if new_word not in visited:
30                                next_level.add(new_word)
31            visited.update(next_level)
32            queue = [path + [word] for path in queue for word in next_level]
33

ℹ

Complexity note: The time complexity is O(n) because we explore each word once and generate new words at each step. The space complexity is O(n) due to the storage of the queue and visited set.

1The problem can be visualized as a graph where each word is a node and edges exist between words that differ by one letter.
2Using BFS ensures that we find the shortest path first before exploring longer paths.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.