[알고리즘] Count the Number of Consistent Strings

포인트

1. 연관되어 있는 단어, 포함되어 있는 단어를 찾는 문제, 적절히 set과 map 을 사용하여 해결하자

🧶문서는 항상 수정 될 수 있습니다. 비판은 환영합니다. 

c++/cpp

#include <bits/stdc++.h>

using namespace std;

class Solution
{
public:
    int countConsistentStrings(string allowed, vector<string> &words)
    {
        int count = 0;
        for(auto ele1: words){
            unordered_set<char> set(ele1.begin(), ele1.end());

            for(auto ele2 : set){
                if(allowed.find(ele2)==string::npos){
                    count+=1;
                    break;
                }
            }
        }
        return words.size() - count;
    }

     int countConsistentStrings2(string allowed, vector<string>& words) {
        // support variable
        int res = words.size();
        unordered_set<char> set;
        // populating alpha
        for (char c: allowed) set.insert(c);
        // parsing all the words to see if each character is in alpha
        for (string word: words) {
            // in case we find an unallowed character, we decrease res and break
            for (char c: word) if (set.find(c)==set.end()) {
                res--;
                break;
            }
        }
        return res;
    }
};

python

class Solution:
    def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
        count = 0
        for ele1 in words:
            temp_set = set(ele1)
            
            for ele2 in temp_set:
                if allowed.find(ele2) == -1:
                    count+=1
                    break
        return len(words) - count

    def countConsistentStrings2(self, allowed: str, words: List[str]) -> int:
        return sum(not set(w) - set(allowed) for w in words)

    def countConsistentStrings3(self, allowed: str, words: List[str]) -> int:
        res = len(words)
        m_set = set(allowed)
        
        for ele1 in words:
            for ele2 in ele1:
                if ele2 not in m_set:
                    res-=1
                    break
        return res