Word Ladder Problem: How Would You Solve It?
Learn how to solve the word ladder problem with BFS, why it guarantees the shortest path, and how to answer this interview question.
Expected Interview Answer
Word ladder is solved with BFS over an implicit graph where each word is a node and an edge connects two words differing by exactly one letter, because BFS explores by increasing transformation count and therefore guarantees the shortest transformation sequence from the start word to the target word.
You never build the graph explicitly. Instead, from the current word you generate every possible one-letter mutation across all 26 letters at each position, and check membership against a word set built from the dictionary in O(1) average time. Each mutation that exists in the dictionary and has not been visited becomes a neighbor, gets marked visited immediately to avoid re-queuing, and is pushed to the BFS queue with a distance one greater than the current word. The moment the target word is dequeued you can return its distance, since BFS guarantees the first time you reach a node is via the shortest path. A common optimization is bidirectional BFS, alternating expansion from both the start and target sets, which shrinks the search space from roughly 26^d to roughly two times 26^(d/2).
- Guarantees the shortest transformation sequence
- O(1) average dictionary membership checks via a hash set
- Bidirectional BFS cuts the effective search depth in half
- Generalizes to any single-character-mutation graph problem
AI Mentor Explanation
Picture a domestic league where each team's squad list is a word and you can change one player at a time to reach a rival team's exact lineup. A selector explores every one-player swap from the current squad, checking each resulting lineup against the list of squads that actually exist in the tournament, which mirrors the dictionary lookup. Squads are marked as already tried the moment they are queued, so the selector never re-explores the same lineup twice. Because the selector tries all one-swap lineups level by level before trying two-swap lineups, the first time the target squad appears it is guaranteed to be reachable with the fewest player changes possible.
Step-by-Step Explanation
Step 1
Build a dictionary set
Load all valid words into a hash set for O(1) average membership checks.
Step 2
BFS from the start word
Push the start word with distance 1, then explore all one-letter mutations at each level.
Step 3
Generate and validate neighbors
For each position, try all 26 letters, keep only mutations present in the dictionary and not yet visited.
Step 4
Return on target dequeue
The first time the target word is dequeued, its distance is the shortest transformation length; return 0 if the queue empties first.
What Interviewer Expects
- Recognize this as an implicit-graph shortest-path problem, not a string problem
- Justify BFS over DFS by explaining shortest-path guarantees
- Discuss neighbor generation cost: O(26 * L) per word, L being word length
- Mention bidirectional BFS as an optimization for large dictionaries
Common Mistakes
- Using DFS and expecting it to find the shortest path
- Forgetting to remove visited words from the dictionary set, causing infinite requeuing
- Rebuilding the dictionary as a list instead of a set, causing O(n) membership checks
- Not handling the case where the target word is not in the dictionary at all
Best Answer (HR Friendly)
“I treat this as a shortest-path problem on a hidden graph, where words are nodes and one-letter differences are edges, then I run BFS because it explores level by level and guarantees the fewest steps. I generate neighbors on the fly by trying every possible one-letter change and checking a hash set of valid words.”
Code Example
from collections import deque
from string import ascii_lowercase
def word_ladder_length(begin_word, end_word, word_list):
word_set = set(word_list)
if end_word not in word_set:
return 0
queue = deque([(begin_word, 1)])
visited = {begin_word}
while queue:
word, steps = queue.popleft()
if word == end_word:
return steps
for i in range(len(word)):
for ch in ascii_lowercase:
candidate = word[:i] + ch + word[i + 1:]
if candidate in word_set and candidate not in visited:
visited.add(candidate)
queue.append((candidate, steps + 1))
return 0Follow-up Questions
- How would you implement bidirectional BFS for this problem?
- How would you return the actual transformation path, not just its length?
- What is the time complexity in terms of word length L and dictionary size N?
- How would you handle a dictionary too large to fit in memory?
MCQ Practice
1. Why is BFS preferred over DFS for the word ladder problem?
BFS explores nodes in order of distance from the source, so the first time the target is reached it is via the shortest path.
2. What is the average-case cost of checking whether a candidate word is valid?
Storing the dictionary as a hash set gives O(1) average-case membership checks.
3. What technique reduces the effective search depth for large dictionaries?
Bidirectional BFS expands from both the start and target simultaneously, roughly halving the effective search depth.
Flash Cards
What kind of problem is word ladder really? — A shortest-path problem on an implicit graph where words are nodes and one-letter differences are edges.
Why use BFS instead of DFS here? — BFS guarantees the shortest transformation sequence because it explores level by level.
How are neighbors generated? — By trying all 26 letters at each character position and checking dictionary membership.
What optimization halves the effective search space? — Bidirectional BFS, expanding from both the start word and end word.