Guess the Word
Time O(n) · Space O(n) · Official statement on LeetCode
Solutions
// Time: O(n)
// Space: O(n)
/**
* // This is the Master's API interface.
* // You should not implement it, or speculate about its implementation
* class Master {
* public:
* int guess(string word);
* };
*/
class Solution {
public:
void findSecretWord(vector<string>& wordlist, Master& master) {
vector<int> possible(size(wordlist));
iota(begin(possible), end(possible), 0);
int n = 0;
while (n < 6) {
auto guess = find_guess_with_most_frequent_chars(wordlist, possible);
n = master.guess(wordlist[guess]);
vector<int> new_possible;
for (const auto& j : possible) {
if (match(wordlist[guess], wordlist[j]) == n) {
new_possible.emplace_back(j);
}
}
possible = move(new_possible);
}
}
private:
int find_guess_with_most_frequent_chars(
const vector<string>& wordlist,
const vector<int>& possible) {
vector<vector<int>> count(6, vector<int>(26));
for (int i = 0; i < 6; ++i) {
for (const auto& p : possible) {
++count[i][wordlist[p][i] - 'a'];
}
}
int guess = 0, max_score = 0;
for (const auto& p : possible) {
int score = 0;
for (int i = 0; i < 6; ++i) {
score += count[i][wordlist[p][i] - 'a'];
}
if (score > max_score) {
max_score = score;
guess = p;
}
}
return guess;
}
int match(const string& a, const string& b) {
int matches = 0;
for (int i = 0; i < size(a); ++i) {
if (a[i] == b[i]) {
++matches;
}
}
return matches;
}
};
// Time: O(n^2)
// Space: O(n)
class Solution2 {
public:
void findSecretWord(vector<string>& wordlist, Master& master) {
vector<vector<int>> H(size(wordlist), vector<int>(size(wordlist)));
for (int i = 0; i < size(wordlist); ++i) {
for (int j = 0; j < size(wordlist); ++j) {
H[i][j] = match(wordlist[i], wordlist[j]);
}
}
vector<int> possible(size(wordlist));
iota(begin(possible), end(possible), 0);
int n = 0;
while (n < 6) {
auto guess = solve(H, possible);
n = master.guess(wordlist[guess]);
vector<int> new_possible;
for (const auto& j : possible) {
if (H[guess][j] == n) {
new_possible.emplace_back(j);
}
}
possible = move(new_possible);
}
}
private:
int solve(const vector<vector<int>>& H,
const vector<int>& possible) {
int min_max_size = size(possible);
int best_guess = -1;
for (const auto& guess : possible) {
vector<vector<int>> groups(7);
for (const auto& j : possible) {
if (j != guess) {
groups[H[guess][j]].emplace_back(j);
}
}
int max_group_i = 0;
for (int i = 0; i < size(groups); ++i) {
if (size(groups[i]) > size(groups[max_group_i])) {
max_group_i = i;
}
}
if (size(groups[max_group_i]) < min_max_size) {
min_max_size = size(groups[max_group_i]);
best_guess = guess;
}
}
return best_guess;
}
int match(const string& a, const string& b) {
int matches = 0;
for (int i = 0; i < size(a); ++i) {
if (a[i] == b[i]) {
++matches;
}
}
return matches;
}
};
// Time: O(n^2)
// Space: O(n)
class Solution3 {
public:
void findSecretWord(vector<string>& wordlist, Master& master) {
vector<vector<int>> H(size(wordlist), vector<int>(size(wordlist)));
for (int i = 0; i < size(wordlist); ++i) {
for (int j = 0; j < size(wordlist); ++j) {
H[i][j] = match(wordlist[i], wordlist[j]);
}
}
vector<int> possible(size(wordlist));
iota(begin(possible), end(possible), 0);
int n = 0;
while (n < 6) {
auto guess = solve(H, possible);
n = master.guess(wordlist[guess]);
vector<int> new_possible;
for (const auto& j : possible) {
if (H[guess][j] == n) {
new_possible.emplace_back(j);
}
}
possible = move(new_possible);
}
}
private:
int solve(const vector<vector<int>>& H,
const vector<int>& possible) {
int min_max_size = size(possible);
int best_guess = -1;
for (const auto& guess : possible) {
vector<vector<int>> groups(7);
for (const auto& j : possible) {
if (j != guess) {
groups[H[guess][j]].emplace_back(j);
}
}
int max_group_i = 0; // assumed the size of the other groups equals to the size of 0-group
if (size(groups[max_group_i]) < min_max_size) {
min_max_size = size(groups[max_group_i]);
best_guess = guess;
}
}
return best_guess;
}
int match(const string& a, const string& b) {
int matches = 0;
for (int i = 0; i < size(a); ++i) {
if (a[i] == b[i]) {
++matches;
}
}
return matches;
}
};
Beginner Explanation
What is Guess the Word?
Guess the Word (LeetCode #843) is a Hard problem that primarily trains greedy.
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 greedy.
- 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: MinMax, Math.
AlgoForge explanations are original teaching notes. Always open the official problem statement on LeetCode for constraints and examples.
Interview Walkthrough
Interview approach for Guess the Word
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 greedy.
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(n)) and space (O(n)) 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(n) time and O(n) space.
Pattern focus: greedy
Use the pattern as a checklist:
- greedy — 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(n) |
| Space | O(n) |
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 Guess the Word
- Skipping edge cases — empty collections, single-element inputs, max constraints.
- Wrong invariant for greedy — 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 greedy:
- 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: greedy.
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 Guess the Word in a second language (cpp, python).
- Drill 3–5 more problems tagged greedy.
- 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 greedy 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
Guess the Word (#843) — Hard. Pattern: greedy. Complexity: O(n) time / O(n) space. Re-derive the invariant before coding.
FAQs
What is the time complexity of Guess the Word?+
The reference solutions aim for O(n) time and O(n) space. Always re-derive complexity from the code you write in the interview.
What pattern does Guess the Word use?+
It primarily maps to greedy, within the broader topic of greedy.
Is Guess the Word 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/guess-the-word/