Word Search II
Time O(m * n * 3^h) · Space O(t) · Official statement on LeetCode
Solutions
// Time: O(m * n * 4 * 3^(h - 1)) ~= O(m * n * 3^h), h is the height of trie
// Space: O(t), t is the number of nodes in trie
class Solution {
private:
struct TrieNode {
bool isString = false;
unordered_map<char, TrieNode *> leaves;
bool Insert(const string& s) {
auto* p = this;
for (const auto& c : s) {
if (p->leaves.find(c) == p->leaves.cend()) {
p->leaves[c] = new TrieNode;
}
p = p->leaves[c];
}
// s already existed in this trie.
if (p->isString) {
return false;
} else {
p->isString = true;
return true;
}
}
~TrieNode() {
for (auto& kv : leaves) {
if (kv.second) {
delete kv.second;
}
}
}
};
public:
/**
* @param board: A list of lists of character
* @param words: A list of string
* @return: A list of string
*/
vector<string> findWords(vector<vector<char>>& board, vector<string>& words) {
unordered_set<string> ret;
vector<vector<bool>> visited(board.size(), vector<bool>(board[0].size(), false));
string cur;
TrieNode trie;
for (const auto& word : words) {
trie.Insert(word);
}
for (int i = 0; i < board.size(); ++i) {
for (int j = 0; j < board[0].size(); ++j) {
findWordsDFS(board, visited, &trie, i, j, cur, ret);
}
}
return vector<string>(ret.begin(), ret.end());
}
void findWordsDFS(vector<vector<char>> &grid,
vector<vector<bool>> &visited,
TrieNode *trie,
int i,
int j,
string cur,
unordered_set<string> &ret) {
// Invalid state.
if (!trie || i < 0 || i >= grid.size() || j < 0 || j >= grid[0].size()) {
return;
}
// Not in trie or visited.
if (!trie->leaves[grid[i][j] ] || visited[i][j]) {
return;
}
// Get next trie nodes.
TrieNode *nextNode = trie->leaves[grid[i][j]];
// Update current string.
cur.push_back(grid[i][j]);
// Find the string, add to the answers.
if (nextNode->isString) {
ret.insert(cur);
}
// Marked as visited.
visited[i][j] = true;
// Try each direction.
const vector<pair<int, int>> directions{{0, -1}, {0, 1},
{-1, 0}, {1, 0}};
for (const auto& d : directions) {
findWordsDFS(grid, visited, nextNode,
i + d.first, j + d.second, cur, ret);
}
visited[i][j] = false;
}
};
Beginner Explanation
What is Word Search II?
Word Search II (LeetCode #212) 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 trie and 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: Trie, DFS.
AlgoForge explanations are original teaching notes. Always open the official problem statement on LeetCode for constraints and examples.
Interview Walkthrough
Interview approach for Word Search 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 trie and 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(m * n * 3^h)) and space (O(t)) 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(m * n * 3^h) time and O(t) space.
Pattern focus: trie and dfs backtracking
Use the pattern as a checklist:
- trie — confirm the invariant holds after each step
- 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(m * n * 3^h) |
| Space | O(t) |
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 Search II
- Skipping edge cases — empty collections, single-element inputs, max constraints.
- Wrong invariant for trie and 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 trie and 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 Search 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 trie and 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 Search II (#212) — Hard. Pattern: trie and dfs backtracking. Complexity: O(m * n * 3^h) time / O(t) space. Re-derive the invariant before coding.
FAQs
What is the time complexity of Word Search II?+
The reference solutions aim for O(m * n * 3^h) time and O(t) space. Always re-derive complexity from the code you write in the interview.
What pattern does Word Search II use?+
It primarily maps to trie and dfs backtracking, within the broader topic of backtracking.
Is Word Search 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-search-ii/