Word Ladder
Find the shortest transformation sequence from beginWord to endWord.
Pattern fit
Every valid transformation is an edge in an implicit unweighted graph, so shortest steps means BFS almost immediately.
Key observation
The hard part is not BFS itself but generating neighbors efficiently without trying every word naively.
Target complexity
O(N * L * alphabet) / O(N)
How to break down the solution cleanly
Every valid transformation is an edge in an implicit unweighted graph, so shortest steps means BFS almost immediately.
The hard part is not BFS itself but generating neighbors efficiently without trying every word naively.
Identify the graph model and how neighbors are generated.
Choose BFS for shortest unweighted distance or DFS for structure exploration.
Reference implementation
Python# Generic pattern template
from collections import deque
def bfs(start):
queue = deque([start])
visited = {start}
while queue:
node = queue.popleft()
for nxt in graph[node]:
if nxt not in visited:
visited.add(nxt)
queue.append(nxt)
def dfs(node):
visited.add(node)
for nxt in graph[node]:
if nxt not in visited:
dfs(nxt)
Common pitfalls
Generating neighbors too expensively and timing out.
Marking visited too late and revisiting the same word many times.
Common follow-ups
How does bidirectional BFS improve this?
What exactly is the graph node in this problem?
Continue with related problems
Build repeatable depth inside the BFS / DFS cluster before moving on.