Leetcode: Bulls and Cows

Difficulty: Easy

You are playing the following Bulls and Cows game with your friend: You write a 4-digit secret number and ask your friend to guess it. Each time your friend guesses a number, you give a hint. The hint tells your friend how many digits are in the correct positions (called "bulls") and how many digits are in the wrong positions (called "cows"). Your friend will use those hints to find out the secret number.

For example:Secret number: "1807" Friend's guess: "7810"
Hint: 1 bull and 3 cows. (The bull is 8, the cows are 0, 1 and 7.)


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. In the above example, your function should return "1A3B".

Please note that both secret number and friend's guess may contain duplicate digits, for example:Secret number: "1123" Friend's guess: "0111"
In this case, the 1st 1 in friend's guess is a bull, the 2nd or 3rd 1 is a cow, and your function should return "1A1B".



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




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Make sure you understand the problem:

how many digits are in the correct positions (called "bulls") and how many digits are in the wrong positions (called "cows")!!

Only when they have the same number but not in the same place, in such case there digits can be counted as cows. 

1. Count the number of bulls in the first run. In the same time, build the frequency map for secret array (bull digits not counted in!) and build a tmp string to store all digits without the bulls. 

2. based on the tmp string, deduct the frequency map to count the # of cows. 

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//codes

class Solution {

 public:

        string getHint(string secret, string guess) {

               //find bulls

               unordered_map<char, int> count;

               unsigned bull = 0, cow = 0;

               string tmp;//store all potential cows in guess

               for (int i = 0; i<secret.size(); i++){

                      if (secret[i] == guess[i]) bull++;

                      else{

                            count[secret[i]]++;//store all potential cows in secret

                            tmp += guess[i];

                      }

               }

               //count cow

               for (int i = 0; i<tmp.size(); i++){

                      if (count.find(tmp[i]) != count.end() && count[tmp[i]]>0){

                             cow++;

                            count[tmp[i]]--;

                      }

               }

               //output

               string bu = to_string(bull), co = to_string(cow);

               string final = bu + 'A' + co + 'B';

               return final;

        }

 };