Word Ladder - Video Solutions

Problem Overview: Given a beginWord, endWord, and a dictionary of valid words, compute the length of the shortest transformation sequence where only one character can change at a time and every intermediate word must exist in the dictionary.

Approach 1: Graph Construction + Breadth‑First Search (O(N^2 * L) time, O(N^2) space)

Treat every word as a node in a graph. Two nodes connect if the words differ by exactly one character. Building this graph requires comparing every pair of words and checking if they differ by one position, which costs O(N^2 * L) where N is the number of words and L is the word length. After constructing the adjacency list, run Breadth‑First Search from beginWord to find the shortest path to endWord. BFS guarantees the minimum number of transformations because it explores level by level. This approach is conceptually simple but wastes time building a dense graph that may never be fully explored.

Approach 2: BFS with Word Transformations (O(N * L * 26) time, O(N) space)

The optimized solution avoids explicitly building the entire graph. Store the dictionary in a hash table for O(1) lookups. Starting from beginWord, perform Breadth‑First Search. For each word popped from the queue, iterate through every character position and try replacing it with letters 'a' to 'z'. Each generated candidate word represents a possible transformation. If the word exists in the dictionary and hasn't been visited, push it into the queue and remove it from the set to prevent revisits. Each word can generate up to L * 26 neighbors, and each valid word is processed once, giving overall time complexity O(N * L * 26) with O(N) memory for the queue and visited structure. This method implicitly explores the graph without building it.

Recommended for interviews: The BFS transformation approach is the expected solution. It shows you recognize that the problem is a shortest‑path search in an unweighted graph and that you can generate neighbors dynamically using string manipulation. A brute-force graph build demonstrates understanding of the problem structure, but interviewers typically look for the optimized BFS that combines hashing and on‑the‑fly neighbor generation.