Intuition
The number of $\bf{\it{Bulls}}$ is exactly the number of cases at position $j$, the value in "Secret" is the same as the value in "Guess".
In the rest of the characters, count the occurrences of each value in "Secret" and "Guess". The lower number of occurrences is
the number of $\bf{\it{Cows}}$ for each specific value.
Algorithm

Loop over each character in strings "Secret" and "Guess"

for character index $j$, first check if they are equal. If so, it is a Bull.

Otherwise, check if the characters from both "Secret" and "Guess" have been recorded or not. If so, add one more occurrence correspondingly.
Use a pair to keep the number of occurrences in "Secret" (pair>first) and "Guess" (pair>second). 
After looping over the strings, count the number of cows.
C++
class Solution {
public:
string getHint(string secret, string guess) {
// map key: character (value) either in Secret or Guess
// map value: first > count in Secret, second > count in Guess
typedef unordered_map<int, pair<int,int>> IntPair;
IntPair countMap;
int iA = 0, n = secret.size();
for(int j=0; j<n; j++){
if(secret[j]==guess[j]) iA++;
else{
int id = secret[j]  '0';
if(countMap.find(id) != countMap.end()) countMap[id].first += 1;
else countMap.insert( make_pair(id, make_pair(1,0) ) );
id = guess[j]  '0';
if(countMap.find(id) != countMap.end()) countMap[id].second += 1;
else countMap.insert( make_pair(id, make_pair(0,1) ) );
}
}
// count the number of cows.
int iB = 0;
for(IntPair::iterator it = countMap.begin(); it!= countMap.end(); it++){
iB += min( (it>second).first, (it>second).second );
}
string result = to_string(iA) + "A" + to_string(iB) + "B";
return result;
}
};
Complexity Analysis

Time complexity : $O(n)$. Each string is looped once.

Space complexity : $O(n)$. We need $O(n)$ space for the counting map to record the occurrences of each value from input strings.