Word Ladder II
Time O(b^(d/2)) · Space O(w * l) · Official statement on LeetCode
Solutions
// Time: O(b^(d/2)), b is the branch factor of bfs, d is the result depth
// Space: O(w * l), w is the number of words, l is the max length of words
class Solution {
public:
vector<vector<string>> findLadders(string beginWord, string endWord, vector<string>& wordList) {
unordered_set<string> words(cbegin(wordList), cend(wordList));
if (!words.count(endWord)) {
return {};
}
unordered_map<string, unordered_set<string>> tree;
unordered_set<string> left = {beginWord}, right = {endWord};
bool is_found = false, is_reversed = false;
while (!empty(left)) {
for (const auto& word : left) {
words.erase(word);
}
unordered_set<string> new_left;
for (const auto& word : left) {
auto new_word = word;
for (int i = 0; i < size(new_word); ++i) {
char prev = new_word[i];
for (int j = 0; j < 26; ++j) {
new_word[i] = 'a' + j;
if (!words.count(new_word)) {
continue;
}
if (right.count(new_word)) {
is_found = true;
} else {
new_left.emplace(new_word);
}
if (!is_reversed) {
tree[new_word].emplace(word);
} else {
tree[word].emplace(new_word);
}
}
new_word[i] = prev;
}
}
if (is_found) {
break;
}
left = move(new_left);
if (size(left) > size(right)) {
swap(left, right);
is_reversed = !is_reversed;
}
}
return backtracking(tree, beginWord, endWord);
}
private:
vector<vector<string>> backtracking(
const unordered_map<string, unordered_set<string>>& tree,
const string& beginWord,
const string& word) {
vector<vector<string>> result;
if (word == beginWord) {
result.emplace_back(vector<string>({beginWord}));
} else {
if (tree.count(word)) {
for (const auto& new_word : tree.at(word)) {
if (word == new_word) {
continue;
}
auto paths = backtracking(tree, beginWord, new_word);
for (auto& path : paths) {
path.emplace_back(word);
result.emplace_back(move(path));
}
}
}
}
return result;
}
};
Beginner Explanation
What is Word Ladder II?
Word Ladder II (LeetCode #126) is a Hard problem that primarily trains backtracking.
How to think about it
- Restate the goal in your own words before coding.
- Work a tiny example by hand so the invariant becomes obvious.
- Identify the pattern — this problem aligns with dfs backtracking.
- Only then translate the idea into code.
Why this problem matters
Hard problems force you to combine patterns and prove complexity carefully — interview gold. Official solution notes mention: Bi-BFS.
AlgoForge explanations are original teaching notes. Always open the official problem statement on LeetCode for constraints and examples.
Interview Walkthrough
Interview approach for Word Ladder II
Opening (30–60 seconds)
- Clarify inputs/outputs and edge cases (empty input, single element, duplicates, overflow).
- State a brute force so the interviewer knows you can solve it naively.
- Propose the optimal direction tied to dfs backtracking.
Core solution narrative
- Define the state you track (pointers, DP cell, set membership, stack top, etc.).
- Explain the transition when you process the next element.
- Call out time (O(b^(d/2))) and space (O(w * l)) before coding.
- Code cleanly; narrate variable names.
What interviewers listen for
- Correctness on edge cases
- Complexity honesty
- Ability to discuss trade-offs (e.g., hash map space vs. sort + two pointers)
Follow-up questions they may ask
- Can you solve it with less memory?
- What if the input stream is infinite / doesn't fit in RAM?
- How would tests look for adversarial inputs?
Optimized Approach
Optimized solution notes
The reference solutions on AlgoForge target O(b^(d/2)) time and O(w * l) space.
Pattern focus: dfs backtracking
Use the pattern as a checklist:
- dfs backtracking — confirm the invariant holds after each step
Multiple methods appear in the source solutions — compare them and explain when each is preferable.
Implementation tips
- Prefer readable names over micro-optimizations in interviews.
- Extract helpers only when they clarify (e.g., expand-around-center, DFS visit).
- After AC-level logic, re-scan for off-by-one and null checks.
Complexity Analysis
Complexity
| Measure | Bound |
|---|---|
| Time | O(b^(d/2)) |
| Space | O(w * l) |
How to justify this in an interview
- Time: count loops, map/set operations, and recursive branching; state average vs worst case if relevant.
- Space: include hash maps, recursion stack, and output allocation when the problem asks for it.
If your implementation differs from the reference, re-derive big-O from your code — never memorize a complexity you cannot defend.
Common Mistakes
Common mistakes on Word Ladder II
- Skipping edge cases — empty collections, single-element inputs, max constraints.
- Wrong invariant for dfs backtracking — updating state too early or too late.
- Mutating input unexpectedly when the problem forbids it.
- Off-by-one in windows, ranges, or binary search bounds.
- Ignoring overflow / precision for integer arithmetic problems.
- Overengineering — jumping to an advanced structure when a simpler approach works.
Alternative Approaches
Alternatives
The source file includes more than one method. Compare:
- Primary optimized path — best complexity for typical interviews.
- Secondary approach — often brute force, sorting-based, or space-optimized variant.
Practice articulating when you would pick each (constraints, readability, follow-ups).
Edge Cases
Edge cases checklist
- Minimum input size
- Maximum input size / time limits
- Duplicates and already-sorted input
- Negative numbers / zeros (if applicable)
- Disconnected structures (graphs/trees)
- Single path vs branching recursion depth
Pattern Recognition
Spotting this pattern
Signal phrases that point to dfs backtracking:
- Sorted input or ability to sort without changing the answer class
- Need for contiguous subarray / substring → consider sliding window
- Need for O(1) membership → hash set/map
- Optimal substructure + overlapping subproblems → DP
- Connectivity / components → graph DFS/BFS or Union-Find
Primary topics: backtracking.
Follow-up Interview Questions
Follow-ups
- How does the solution change if the input is a stream?
- Can you solve it in-place?
- What if duplicates must be handled differently?
- How would you parallelize the approach?
- Design tests that would break a buggy implementation.
Practice Recommendations
What to practice next
- Re-solve Word Ladder II in a second language (cpp, python).
- Drill 3–5 more problems tagged backtracking.
- Teach the solution out loud in under 5 minutes.
- Add this problem to your revision calendar in 3 days and 14 days.
Visualization
Study checklist
- Read the official problem statement on LeetCode
- Solve on paper / whiteboard first
- Implement the dfs backtracking approach
- Verify edge cases from the checklist
- State time and space complexity aloud
- Compare with the AlgoForge reference solution
- Schedule a revision session
Revision notes
Word Ladder II (#126) — Hard. Pattern: dfs backtracking. Complexity: O(b^(d/2)) time / O(w * l) space. Re-derive the invariant before coding.
FAQs
What is the time complexity of Word Ladder II?+
The reference solutions aim for O(b^(d/2)) time and O(w * l) space. Always re-derive complexity from the code you write in the interview.
What pattern does Word Ladder II use?+
It primarily maps to dfs backtracking, within the broader topic of backtracking.
Is Word Ladder II good for interviews?+
Yes — as a Hard problem it is a solid practice target. Pair it with related problems in the same pattern family for spaced repetition.
Where can I read the official statement?+
Open the official LeetCode page for constraints and examples: https://leetcode.com/problems/word-ladder-ii/