Guess the Word
HardUpdated: Aug 2, 2025
Practice on:
Problem
You are given an array of unique strings words where words[i] is six letters long. One word of words was chosen as a secret word.
You are also given the helper object Master. You may call
Master.guess(word) where word is a six-letter-long string, and it must be from words. Master.guess(word) returns:
-1ifwordis not fromwords, or- an integer representing the number of exact matches (value and position) of your guess to the secret word.
There is a parameter allowedGuesses for each test case where
allowedGuesses is the maximum number of times you can call
Master.guess(word).
For each test case, you should call Master.guess with the secret word without exceeding the maximum number of allowed guesses. You will get:
"Either you took too many guesses, or you did not find the secret word."if you calledMaster.guessmore thanallowedGuessestimes or if you did not callMaster.guesswith the secret word, or"You guessed the secret word correctly."if you calledMaster.guesswith the secret word with the number of calls toMaster.guessless than or equal toallowedGuesses.
The test cases are generated such that you can guess the secret word with a reasonable strategy (other than using the bruteforce method).
Examples
Example 1
Input: secret = "acckzz", words = ["acckzz","ccbazz","eiowzz","abcczz"], allowedGuesses = 10
Output: You guessed the secret word correctly.
Explanation:
master.guess("aaaaaa") returns -1, because "aaaaaa" is not in wordlist.
master.guess("acckzz") returns 6, because "acckzz" is secret and has all 6 matches.
master.guess("ccbazz") returns 3, because "ccbazz" has 3 matches.
master.guess("eiowzz") returns 2, because "eiowzz" has 2 matches.
master.guess("abcczz") returns 4, because "abcczz" has 4 matches.
We made 5 calls to master.guess, and one of them was the secret, so we pass the test case.
Example 2
Input: secret = "hamada", words = ["hamada","khaled"], allowedGuesses = 10
Output: You guessed the secret word correctly.
Explanation: Since there are two words, you can guess both.
Constraints
1 <= words.length <= 100words[i].length == 6words[i]consist of lowercase English letters.- All the strings of
wordlistare unique. secretexists inwords.10 <= allowedGuesses <= 30
Solution
Method 1 – Minimax with Match Counting
Intuition
To efficiently guess the secret word, we want to minimize the worst-case number of remaining candidates after each guess. For each guess, we can count how many words would remain for each possible match count, and pick the guess that minimizes the largest group. This is a minimax strategy. After each guess, we filter the word list to only those with the same number of matches as the guess result.
Approach
- For up to
allowedGuessestimes:- For each word in the current list, for each other word, count the number of matches at each position.
- For each word, record the largest group size of words with the same match count.
- Pick the word with the smallest largest group size as the next guess.
- Call
Master.guess(word)and get the number of matches. - If the guess is correct (6 matches), stop.
- Otherwise, filter the word list to only those with the same number of matches as the guess result.
- Repeat until the secret is found or guesses run out.
Code
C++
class Solution {
public:
void findSecretWord(vector<string>& words, Master& master) {
int n = words.size();
for (int t = 0; t < 10; ++t) {
vector<vector<int>> count(n, vector<int>(n));
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
int matches = 0;
for (int k = 0; k < 6; ++k) {
if (words[i][k] == words[j][k]) matches++;
}
count[i][j] = matches;
}
}
int min_max = n, guess_idx = 0;
for (int i = 0; i < n; ++i) {
vector<int> groups(7);
for (int j = 0; j < n; ++j) if (i != j) groups[count[i][j]]++;
int max_group = *max_element(groups.begin(), groups.end());
if (max_group < min_max) {
min_max = max_group;
guess_idx = i;
}
}
string guess = words[guess_idx];
int matches = master.guess(guess);
if (matches == 6) return;
vector<string> next;
for (auto& w : words) {
int m = 0;
for (int k = 0; k < 6; ++k) if (w[k] == guess[k]) m++;
if (m == matches) next.push_back(w);
}
words = next;
}
}
};
Python
class Solution:
def findSecretWord(self, words: list[str], master: 'Master') -> None:
from collections import Counter
for _ in range(10):
count = [[sum(a == b for a, b in zip(w1, w2)) for w2 in words] for w1 in words]
min_max, guess_idx = float('inf'), 0
for i, row in enumerate(count):
groups = Counter(row)
max_group = max(groups[j] for j in range(7) if j != 6)
if max_group < min_max:
min_max = max_group
guess_idx = i
guess = words[guess_idx]
matches = master.guess(guess)
if matches == 6:
return
words = [w for w in words if sum(a == b for a, b in zip(w, guess)) == matches]
Java
class Solution {
public void findSecretWord(String[] words, Master master) {
int n = words.length;
for (int t = 0; t < 10; t++) {
int[][] count = new int[n][n];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
int matches = 0;
for (int k = 0; k < 6; k++) {
if (words[i].charAt(k) == words[j].charAt(k)) matches++;
}
count[i][j] = matches;
}
}
int minMax = n, guessIdx = 0;
for (int i = 0; i < n; i++) {
int[] groups = new int[7];
for (int j = 0; j < n; j++) if (i != j) groups[count[i][j]]++;
int maxGroup = 0;
for (int g = 0; g < 6; g++) maxGroup = Math.max(maxGroup, groups[g]);
if (maxGroup < minMax) {
minMax = maxGroup;
guessIdx = i;
}
}
String guess = words[guessIdx];
int matches = master.guess(guess);
if (matches == 6) return;
List<String> next = new ArrayList<>();
for (String w : words) {
int m = 0;
for (int k = 0; k < 6; k++) if (w.charAt(k) == guess.charAt(k)) m++;
if (m == matches) next.add(w);
}
words = next.toArray(new String[0]);
}
}
}
Kotlin
class Solution {
fun findSecretWord(words: Array<String>, master: Master) {
var ws = words.toList()
repeat(10) {
val count = ws.map { w1 -> ws.map { w2 -> w1.zip(w2).count { it.first == it.second } } }
var minMax = Int.MAX_VALUE
var guessIdx = 0
for ((i, row) in count.withIndex()) {
val groups = IntArray(7)
for (j in row.indices) if (i != j) groups[row[j]]++
val maxGroup = groups.take(6).maxOrNull() ?: 0
if (maxGroup < minMax) {
minMax = maxGroup
guessIdx = i
}
}
val guess = ws[guessIdx]
val matches = master.guess(guess)
if (matches == 6) return
ws = ws.filter { w -> w.zip(guess).count { it.first == it.second } == matches }
}
}
}
Rust
impl Solution {
pub fn find_secret_word(words: Vec<String>, master: &mut Master) {
let mut ws = words;
for _ in 0..10 {
let n = ws.len();
let mut count = vec![vec![0; n]; n];
for i in 0..n {
for j in 0..n {
count[i][j] = ws[i].chars().zip(ws[j].chars()).filter(|(a, b)| a == b).count();
}
}
let mut min_max = n;
let mut guess_idx = 0;
for i in 0..n {
let mut groups = vec![0; 7];
for j in 0..n {
if i != j { groups[count[i][j]] += 1; }
}
let max_group = *groups.iter().take(6).max().unwrap();
if max_group < min_max {
min_max = max_group;
guess_idx = i;
}
}
let guess = &ws[guess_idx];
let matches = master.guess(guess.clone());
if matches == 6 { return; }
ws = ws.into_iter().filter(|w| w.chars().zip(guess.chars()).filter(|(a, b)| a == b).count() == matches).collect();
}
}
}
TypeScript
class Solution {
findSecretWord(words: string[], master: Master): void {
for (let t = 0; t < 10; t++) {
const n = words.length;
const count: number[][] = Array.from({length: n}, () => Array(n).fill(0));
for (let i = 0; i < n; i++) {
for (let j = 0; j < n; j++) {
let matches = 0;
for (let k = 0; k < 6; k++) {
if (words[i][k] === words[j][k]) matches++;
}
count[i][j] = matches;
}
}
let minMax = n, guessIdx = 0;
for (let i = 0; i < n; i++) {
const groups = Array(7).fill(0);
for (let j = 0; j < n; j++) if (i !== j) groups[count[i][j]]++;
const maxGroup = Math.max(...groups.slice(0, 6));
if (maxGroup < minMax) {
minMax = maxGroup;
guessIdx = i;
}
}
const guess = words[guessIdx];
const matches = master.guess(guess);
if (matches === 6) return;
words = words.filter(w => w.split('').filter((c, k) => c === guess[k]).length === matches);
}
}
}
Complexity
- ⏰ Time complexity:
O(n^2 * k)— For each guess, we compare all pairs of words of lengthk=6. - 🧺 Space complexity:
O(n^2)— For the match count matrix.