Bulls and Cows

• Time : $O(n)$
• Space : $O(1)$

string Solution::solve(string A, string B) {
    unordered_map<char, int> mp1, mp2;
    int b = 0, c = 0;
    for(int i = 0; i < A.size(); i++) {
        if(A[i] == B[i]) b += 1;
        else {
            mp1[A[i]] += 1;
            mp2[B[i]] += 1;
        }
    }
    for(auto [k,v] : mp2) {
        c += min(v,mp1[k]);
    }
    return to_string(b) + "A" + to_string(c) + "B";
}