L3 · FAANGGraphs~25 min · 5 tests

Word Ladder (Shortest Transformation)

Find the shortest chain of one-letter changes from one word to another using BFS with wildcard buckets in O(N·L²). Classic Amazon/Google hard, run live.

The problem

Transform begin into end by changing one letter at a time, where every intermediate word must be in words. Return the number of words in the shortest such sequence (including begin and end), or 0 if impossible.

Examples

Example 1

Input

ladder_length('hit', 'cog', ['hot', 'dot', 'dog', 'lot', 'log', 'cog'])

Expected output

Example 2

Input

ladder_length('hit', 'cog', ['hot', 'dot', 'dog', 'lot', 'log'])

Expected output

+ 3 hidden tests on Submit — no bridge word.

Edge cases to ask about

end not in the word list
begin == end
No path

Hints

0/3

How an interviewer scores this

0/9

1. Understand the problemRestate it, ask about edge cases and input size.
2. Write a brute-force solutionCorrect first, clever later.
3. Get the correct outputautoAll visible examples pass.
4. Analyse time complexityautoCommit to a Big-O in the Complexity Lab.
5. Analyse space complexityautoExtra memory, not the input.
6. Identify the bottleneckautoMeasure your code — where does it bend?
7. OptimiseUsually: a set or dict instead of a scan.
8. Explain why it worksSay it out loud to Solvi.
9. Test the edge casesautoSubmit passes every hidden test.

Python 3.13 · ladder_length

def ladder_length(begin, end, words):
    """Length of the shortest one-letter-change chain, or 0."""
    # Write your solution here
    pass

⌘/Ctrl + Enter runs the examples

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Complexity Lab

What does this cost as n grows?

Interviewers score the analysis as much as the code. Commit to an answer first — then check it, and read why.

Time complexity of the optimal solution

Space complexity (extra memory)

Pick both to reveal the answer.

From brute force to optimal

The progression an interviewer wants to hear, one step at a time.

Approach	Time	Space	Idea
Compare every pair of words to build edges	O(N² · L)	O(N²)
BFS + wildcard buckets	O(N · L²)	O(N · L²)	'h*t' groups hot, hit, hat — neighbours without pairwise comparison.
bestBidirectional BFS	O(N · L²)	O(N · L²)	Much smaller frontiers in practice.

Walkthrough of the optimal approach (try it yourself first)

Unweighted shortest path means BFS. The expensive part is finding neighbours: comparing every pair is O(N²·L). Instead, bucket each word under its L wildcard patterns (hot → *ot, h*t, ho*). Two words are neighbours exactly when they share a bucket.

Mark words seen on enqueue so each is queued once, and clear a bucket after using it so it's never scanned twice.

Complexity: O(N · L²) time, O(N · L²) space. Each of N words generates L wildcard patterns, each built in O(L).

Reveal the reference solution

from collections import defaultdict, deque

def ladder_length(begin, end, words):
    words = set(words)
    if end not in words:
        return 0
    buckets = defaultdict(list)
    for w in words | {begin}:
        for i in range(len(w)):
            buckets[w[:i] + "*" + w[i + 1:]].append(w)
    seen = {begin}
    queue = deque([(begin, 1)])
    while queue:
        word, steps = queue.popleft()
        if word == end:
            return steps
        for i in range(len(word)):
            key = word[:i] + "*" + word[i + 1:]
            for nxt in buckets[key]:
                if nxt not in seen:
                    seen.add(nxt)
                    queue.append((nxt, steps + 1))
            buckets[key] = []          # never scan this bucket again
    return 0

Follow-ups interviewers ask

Return all shortest ladders (Word Ladder II).
Bidirectional BFS.

Frequently asked interview questions

Core interview concepts, complexities, and follow-ups scored by hiring teams.

What is the time complexity of Word Ladder (Shortest Transformation) in Python?

The optimal solution runs in O(N · L²) time and O(N · L²) auxiliary space. Each of N words generates L wildcard patterns, each built in O(L).

What is the brute-force approach, and how do you optimise it?

Compare every pair of words to build edges: O(N² · L) time, O(N²) space. BFS + wildcard buckets: O(N · L²) time, O(N · L²) space. 'h*t' groups hot, hit, hat — neighbours without pairwise comparison. Bidirectional BFS: O(N · L²) time, O(N · L²) space. Much smaller frontiers in practice.

What follow-up questions do interviewers ask about Word Ladder (Shortest Transformation)?

Return all shortest ladders (Word Ladder II). Bidirectional BFS.