843. Guess the Word

843. Guess the Word

This problem is an interactive problem new to the LeetCode platform.

We are given a word list of unique words, each word is 6 letters long, and one word in this list is chosen as secret.

You may call master.guess(word) to guess a word.  The guessed word should have type string and must be from the original list with 6 lowercase letters.

This function returns an integer type, representing the number of exact matches (value and position) of your guess to the secret word.  Also, if your guess is not in the given wordlist, it will return -1 instead.

For each test case, you have 10 guesses to guess the word. At the end of any number of calls, if you have made 10 or less calls to master.guess and at least one of these guesses was the secret, you pass the testcase.

Besides the example test case below, there will be 5 additional test cases, each with 100 words in the word list.  The letters of each word in those testcases were chosen independently at random from 'a' to 'z', such that every word in the given word lists is unique.

Example 1:
Input: secret = "acckzz", wordlist = ["acckzz","ccbazz","eiowzz","abcczz"]

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.

Note:  Any solutions that attempt to circumvent the judge will result in disqualification.

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扯淡题

思路是用MiniMax, 主要是求“稳”。有多种实现方式。
以下方法是，对每一个词，找出与他匹配数量为0的词的个数，然后用数量最小的那个词来猜
这里的解释不错： http://blog.vrqq.org/archives/363/

check zero match的原因：
“Nice Solution. Anyone who doesn't know why checking 0 match instead of 1,2,3...6 matches, please take a look at this comment. The probability of two words with 0 match is (25/26)^6 = 80%. That is to say, for a candidate word, we have 80% chance to see 0 match with the secret word. In this case, we had 80% chance to eliminate the candidate word and its "family" words which have at least 1 match. Additionally, in order to delete a max part of words, we select a candidate who has a big "family" (fewest 0 match).”
https://leetcode.com/problems/guess-the-word/discuss/133862/Random-Guess-and-Minimax-Guess-with-Comparison
通常情况下，猜一个词，返回为0的概率胃79% （25/26 ^ 6），这种情况下，wordlist范围缩小的力度也是最小的（因为任意2个词的match为0的几率高达79%）。我们要增大缩小的力度，减小match为0时新生成的list的大小。所以我们选出“找出与他匹配数量为0的词的个数，然后用数量最小的那个词来猜”

/**
 * // 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[] wordList, Master master) {
        List<String> list = Arrays.asList(wordList);
        
        while (!list.isEmpty()) {
            Map<String, Integer> zeroMatch = new HashMap<>();
            for (String cand : list) {
                for (String word : list) {
                    if (match(cand, word) == 0) zeroMatch.put(cand, zeroMatch.getOrDefault(cand, 0) + 1);
                }
            }    
            
            String guess = list.get(0);
            int min = 110;
            for (Map.Entry<String, Integer> entry : zeroMatch.entrySet()) {
                if (min > entry.getValue()) {
                    min = entry.getValue();
                    guess = entry.getKey();
                }
            }
            
            int m = master.guess(guess);
            if (m == 6) return;
            
            List<String> tmp = new ArrayList<>();
            for (String s : list) {
                if (m == match(s, guess)) {
                    tmp.add(s);    
                }
            }
            
            list = tmp;
        }
    }
    
    private int match(String a, String b) {
        int m = 0;
        for (int i = 0; i < a.length(); i++) {
            if (a.charAt(i) == b.charAt(i)) m++;
        }
        
        return m;
    }
}

另一种实现方式，LC官方解答：https://leetcode.com/problems/guess-the-word/solution/