Description

You are playing the following Bulls and Cows game with your friend: You write down a number and ask your friend to guess what the number is. Each time your friend makes a guess, you provide a hint that indicates how many digits in said guess match your secret number exactly in both digit and position (called "bulls") and how many digits match the secret number but locate in the wrong position (called "cows"). Your friend will use successive guesses and hints to eventually derive the secret number.

Write a function to return a hint according to the secret number and friend's guess, use A to indicate the bulls and B to indicate the cows.

Please note that both secret number and friend's guess may contain duplicate digits.

Example 1:

Input: secret = "1807", guess = "7810"

Output: "1A3B"

Explanation: 1 bull and 3 cows. The bull is 8, the cows are 0, 1 and 7.

Example 2:

Input: secret = "1123", guess = "0111"

Output: "1A1B"

Explanation: The 1st 1 in friend's guess is a bull, the 2nd or 3rd 1 is a cow.

**Note: **You may assume that the secret number and your friend's guess only contain digits, and their lengths are always equal.

Solution

HashMap

原本以为扫两边就可以，第一遍扫secret统计need，第二遍直接得出bulls and cows。后来发现不行，考虑这个test case:

secret = "001", guess = "121"

应该返回"1A0B"，而非"0A1B"，因为最后面的1 match上了。所以match未必是按照顺序来的。
所以变成了扫三遍，先统计bulls，再统计cows...

class Solution {
    public String getHint(String secret, String guess) {
        int bulls = 0;
        int cows = 0;
        Map<Character, Integer> need = new HashMap<>();
        
        for (int i = 0; i < secret.length(); ++i) {
            need.put(secret.charAt(i), need.getOrDefault(secret.charAt(i), 0) + 1);
        }
        
        for (int i = 0; i < secret.length(); ++i) {
            if (secret.charAt(i) == guess.charAt(i)) {
                ++bulls;
                need.put(guess.charAt(i), need.get(guess.charAt(i)) - 1);
            }
        }
        
        for (int i = 0; i < secret.length(); ++i) {
            if (secret.charAt(i) != guess.charAt(i) 
                && need.getOrDefault(guess.charAt(i), 0) > 0) {
                ++cows;
                need.put(guess.charAt(i), need.get(guess.charAt(i)) - 1);
            }
        }
        
        return bulls + "A" + cows + "B";
    }
}

HashTable, O(n), S(n)

其实上面的做法中，第一遍扫描只是为了填充need map。由于题目给定digit范围，所以可以用int[] need代替map。

class Solution {
    public String getHint(String secret, String guess) {
        int bulls = 0;
        int cows = 0;
        int n = secret.length();
        int[] sarr = new int[10];   // 10 possible digits
        int[] garr = new int[10];
        
        for (int i = 0; i < n; ++i) {
            if (secret.charAt(i) == guess.charAt(i)) {
                ++bulls;
            } else {
                ++sarr[secret.charAt(i) - '0'];
                ++garr[guess.charAt(i) - '0'];
            }
        }
        
        for (int i = 0; i < sarr.length; ++i) { // iterate sarr and garr!
            cows += Math.min(sarr[i], garr[i]);
        }
        
        return bulls + "A" + cows + "B";
    }
}