LeetCode #843 — HARD

Guess the Word

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

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:

-1 if word is not from words, 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 called Master.guess more than allowedGuesses times or if you did not call Master.guess with the secret word, or
"You guessed the secret word correctly." if you called Master.guess with the secret word with the number of calls to Master.guess less than or equal to allowedGuesses.

The test cases are generated such that you can guess the secret word with a reasonable strategy (other than using the bruteforce method).

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 words.
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 <= 100
words[i].length == 6
words[i] consist of lowercase English letters.
All the strings of words are unique.
secret exists in words.
10 <= allowedGuesses <= 30

Step 01

Brute Force Baseline

Problem summary: 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: -1 if word is not from words, 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 called Master.guess more than allowedGuesses times or if you did not call Master.guess with the

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Math

Example 1

"acckzz"
["acckzz","ccbazz","eiowzz","abcczz"]
10

Example 2

"hamada"
["hamada","khaled"]
10

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #843: Guess the Word
/**
 * // This is the Master's API interface.
 * // You should not implement it, or speculate about its implementation
 * interface Master {
 *   public int guess(String word) {}
 * }
 */
class Solution {
  public void findSecretWord(String[] words, Master master) {
    Random rand = new Random();

    for (int i = 0; i < 10; ++i) {
      final String guessedWord = words[rand.nextInt(words.length)];
      final int matches = master.guess(guessedWord);
      if (matches == 6)
        break;
      List<String> updated = new ArrayList<>();
      for (final String word : words)
        if (getMatches(guessedWord, word) == matches)
          updated.add(word);
      words = updated.toArray(new String[0]);
    }
  }

  private int getMatches(final String s1, final String s2) {
    int matches = 0;
    for (int i = 0; i < s1.length(); ++i)
      if (s1.charAt(i) == s2.charAt(i))
        ++matches;
    return matches;
  }
}

// Accepted solution for LeetCode #843: Guess the Word
// Auto-generated Go example from java.
func exampleSolution() {
}
// Reference (java):
// // Accepted solution for LeetCode #843: Guess the Word
// /**
//  * // This is the Master's API interface.
//  * // You should not implement it, or speculate about its implementation
//  * interface Master {
//  *   public int guess(String word) {}
//  * }
//  */
// class Solution {
//   public void findSecretWord(String[] words, Master master) {
//     Random rand = new Random();
// 
//     for (int i = 0; i < 10; ++i) {
//       final String guessedWord = words[rand.nextInt(words.length)];
//       final int matches = master.guess(guessedWord);
//       if (matches == 6)
//         break;
//       List<String> updated = new ArrayList<>();
//       for (final String word : words)
//         if (getMatches(guessedWord, word) == matches)
//           updated.add(word);
//       words = updated.toArray(new String[0]);
//     }
//   }
// 
//   private int getMatches(final String s1, final String s2) {
//     int matches = 0;
//     for (int i = 0; i < s1.length(); ++i)
//       if (s1.charAt(i) == s2.charAt(i))
//         ++matches;
//     return matches;
//   }
// }

# Accepted solution for LeetCode #843: Guess the Word
# """
# This is Master's API interface.
# You should not implement it, or speculate about its implementation
# """
# Class Master:
#   def guess(self, word: str) -> int:

class Solution:
  def findSecretWord(self, words: list[str], master: 'Master') -> None:
    for _ in range(10):
      guessedWord = words[random.randint(0, len(words) - 1)]
      matches = master.guess(guessedWord)
      if matches == 6:
        break
      words = [
          word for word in words
          if sum(c1 == c2 for c1, c2 in zip(guessedWord, word)) == matches]

// Accepted solution for LeetCode #843: Guess the Word
struct Solution;

use std::cell::Cell;
use std::collections::HashSet;
use std::iter::FromIterator;

use rand::{thread_rng, Rng};

fn exact_matches(a: &str, b: &str) -> usize {
    a.chars().zip(b.chars()).filter(|(a, b)| a == b).count()
}

impl Solution {
    fn find_secret_word(words: Vec<String>, master: &Master) {
        let n = words.len();
        let mut matrix: Vec<Vec<usize>> = vec![vec![0; n]; n];
        let mut excluded: Vec<bool> = vec![false; n];
        for i in 0..n {
            for j in i + 1..n {
                let count = exact_matches(&words[i], &words[j]);
                matrix[i][j] = count;
                matrix[j][i] = count;
            }
        }
        let mut rng = thread_rng();

        for _ in 0..10 {
            let mut id = rng.gen_range(0, n);
            while excluded[id] {
                id = rng.gen_range(0, n);
            }
            let count = master.guess(words[id].to_string()) as usize;
            if count == 6 {
                break;
            }
            for i in 0..n {
                if matrix[id][i] < count {
                    excluded[i] = true;
                }
            }
        }
    }
}

struct Master {
    list: HashSet<String>,
    secret: String,
    call_count: Cell<usize>,
    guessed: Cell<bool>,
}

impl Master {
    fn new(secret: String, words: Vec<String>) -> Self {
        let call_count = Cell::new(0);
        let guessed = Cell::new(false);
        let list: HashSet<String> = HashSet::from_iter(words);
        Master {
            list,
            secret,
            call_count,
            guessed,
        }
    }
    fn guess(&self, word: String) -> i32 {
        self.call_count.set(self.call_count.get() + 1);
        if word == self.secret {
            self.guessed.set(true);
        }
        if self.list.contains(&word) {
            exact_matches(&self.secret, &word) as i32
        } else {
            -1
        }
    }
    fn pass(&self) -> bool {
        self.call_count.get() <= 10 && self.guessed.get()
    }
}

#[test]
fn test() {
    let secret = "acckzz".to_string();
    let wordlist = vec_string!["acckzz", "ccbazz", "eiowzz", "abcczz"];
    let master = Master::new(secret, wordlist.to_vec());
    Solution::find_secret_word(wordlist, &master);
    assert_eq!(master.pass(), true);
}

// Accepted solution for LeetCode #843: Guess the Word
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #843: Guess the Word
// /**
//  * // This is the Master's API interface.
//  * // You should not implement it, or speculate about its implementation
//  * interface Master {
//  *   public int guess(String word) {}
//  * }
//  */
// class Solution {
//   public void findSecretWord(String[] words, Master master) {
//     Random rand = new Random();
// 
//     for (int i = 0; i < 10; ++i) {
//       final String guessedWord = words[rand.nextInt(words.length)];
//       final int matches = master.guess(guessedWord);
//       if (matches == 6)
//         break;
//       List<String> updated = new ArrayList<>();
//       for (final String word : words)
//         if (getMatches(guessedWord, word) == matches)
//           updated.add(word);
//       words = updated.toArray(new String[0]);
//     }
//   }
// 
//   private int getMatches(final String s1, final String s2) {
//     int matches = 0;
//     for (int i = 0; i < s1.length(); ++i)
//       if (s1.charAt(i) == s2.charAt(i))
//         ++matches;
//     return matches;
//   }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.