Question
Formatted question description: https://leetcode.ca/all/299.html
299 Bulls and Cows
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.
Algorithm
Two traversals, the first traversal finds all numbers with the same position and the same value, that is, bulls, and records the number of times that the numbers in the secret are not bulls. Then we traverse for the second time for the position that is not bulls in guess. If it exists in the hash table, cows increments by 1, and then the mapping value is reduced by 1.
To improve, it can be done with a single loop. When dealing with positions that are not bulls,
 If the mapping value of the current position of the secret is less than 0, it means that it has appeared in guess, cows is incremented by 1, and then the mapping value is incremented by 1.
 If the mapping value of the number at the current position of guess is greater than 0, it means that it has appeared in secret, cows is incremented by 1, and then the mapping value is reduced by 1.
Code
Java

public class Bulls_and_Cows { public class Solution { public String getHint(String secret, String guess) { if (secret == null  secret.length() == 0  guess == null  guess.length() == 0) { return "0A0B"; } int numBulls = 0; int numCows = 0; int[] hm = new int[256]; // Step 1: calculate the number of bulls and update the hm for (int i = 0; i < secret.length(); i++) { char s = secret.charAt(i); char g = guess.charAt(i); if (s == g) { numBulls++; } else { if (hm[s] < 0) { numCows++; } hm[s]++; if (hm[g] > 0) { numCows++; } hm[g]; } } return numBulls + "A" + numCows + "B"; } } }

// OJ: https://leetcode.com/problems/bullsandcows/ // Time: O(N) // Space: O(1) class Solution { public: string getHint(string secret, string guess) { int N = secret.size(), bull = 0, cow = 0, secretCnt[10] = {}, guessCnt[10] = {}; for (int i = 0; i < N; ++i) { if (secret[i] == guess[i]) ++bull; else { secretCnt[secret[i]  '0']++; guessCnt[guess[i]  '0']++; } } for (int i = 0; i < 10; ++i) cow += min(secretCnt[i], guessCnt[i]); return to_string(bull) + "A" + to_string(cow) + "B"; } };

from collections import Counter class Solution(object): def getHint(self, secret, guess): """ :type secret: str :type guess: str :rtype: str """ a = b = 0 ds = Counter() dg = Counter() for i in range(len(secret)): s = secret[i] g = guess[i] if secret[i] == guess[i]: a += 1 else: ds[s] += 1 dg[g] += 1 if ds[g] > 0: b += 1 dg[g] = 1 ds[g] = 1 if dg[s] > 0: b += 1 ds[s] = 1 dg[s] = 1 return "{}A{}B".format(a, b)