What is the optimal time complexity of solving Word Ladder?

The optimized solution achieves a time complexity of O(n) by using sorting, mathematical shortcuts, or hash table lookups.

What is the auxiliary space complexity for Word Ladder?

The space complexity is O(1) since we only use a minimal/constant amount of extra memory for tracking variables (or auxiliary space if dynamic structures are allocated).

What are the typical edge cases we must handle in Word Ladder?

Key edge cases include empty collections, negative or large bounds causing integer overflow, arrays with only one element, or inputs where target criteria are not met.

How can we optimize the brute force approach for Word Ladder?

Brute force methods often inspect all possible combinations. We optimize this by sorting the input list to enable binary search, or tracking processed items in a set to avoid redundant nested loop lookups.

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Word Ladder

Learn how to solve the 'Word Ladder' problem. This detailed resource details brute force and optimized approaches.

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Problem Statement

Easy

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 the number of words in the shortest transformation sequence from beginWord to endWord, or 0 if no such sequence exists.

Write a function ladderLength(beginWord: str, endWord: str, wordList: List[str]) -> int.

Constraints

•1 <= len(beginWord) <= 10
•endWord.length == beginWord.length
•1 <= len(wordList) <= 5000
•wordList[i].length == beginWord.length
•beginWord, endWord, and wordList[i] consist of lowercase English letters
•All the words in wordList are unique

Examples

Example 1

Input

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

Output

Explanation

One shortest transformation sequence is "hit" -> "hot" -> "dot" -> "dog" -> "cog", which is 5 words long.

Example 2

Input

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

Output

Explanation

The endWord "cog" is not in wordList, so there is no valid transformation sequence.

Need a Hint?

Represent graph node connections as an adjacency list/matrix, then use standard BFS or DFS graph traversal.

Edge Cases to Watch

Empty list or null input variables
Single item lists/arrays
Extremely large input bounds causing integer or stack overflow

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