Determine if Two Strings Are Close

두 문자열 데이터를 받고 해당 데이터들이 Close인지 아닌지 판단하는 문제

Operation 1: Swap any two existing characters.
For example, abcde -> aecdb

Operation 2: Transform every occurrence of one existing character into another existing character, and do the same with the other character.
For example, aacabb -> bbcbaa (all a's turn into b's, and all b's turn into a's)

Close 조건은 한 문자열을 위의 두 operation을 사용하여 다른 문자열과 같도록 만들 수 있다면 true이다.

class Solution {
public:
    bool closeStrings(string word1, string word2) {

        int length = word1.size();
        if (length != word2.size())
        {
            return false;
        }

        unordered_map<char, int> table1{};
        unordered_map<char, int> table2{};

        for (int i = 0; i < length; i++)
        {
            table1[word1[i]]++;
            table2[word2[i]]++;
        }

        length = table1.size();
        if (length != table2.size())
        {
            return false;
        }

        map<int, int> frequencyTable1{};
        map<int, int> frequencyTable2{};

        {
            auto it1 = table1.begin();
            auto it2 = table2.begin();
            for (int i = 0; i < length; i++)
            {
                if (table2.count(it1->first) == 0
                 || table1.count(it2->first) == 0)
                {
                    return false;
                }

                frequencyTable1[it1->second]++;
                frequencyTable2[it2->second]++;

                it1++;
                it2++;
            }
        }

        length = frequencyTable1.size();
        if (length != frequencyTable2.size())
        {
            return false;
        }

        {
            auto it1 = frequencyTable1.begin();
            auto it2 = frequencyTable2.begin();
            for (int i = 0; i < length; i++)
            {
                if ((it1->first != it2->first)
                || (it1->second != it2->second))
                {
                    return false;
                }

                it1++;
                it2++;
            }
        }
        
        return true;
    }
};