330. 위장

프로그래머스

1. Python

from collections import Counter
def solution(clothes):
    answer = 1
    dict = Counter()
    for a, b in clothes:
        dict[b] += 1
    
    for k, v in dict.items():
        answer *= (v + 1) # 안 입는 경우 더해주기
    
    return answer - 1
  

2. C++

#include <string>
#include <vector>
#include <unordered_map>

using namespace std;

int solution(vector<vector<string>> clothes) {
    int answer = 1;

    unordered_map <string, int> dict;
    for(int i = 0; i < clothes.size(); i++)
        dict[clothes[i][1]]++;
    for(auto it = dict.begin(); it != dict.end(); it++)
        answer *= (it->second+1);

    
    /*
     for(pair<string,int> elem : um){
        cout<<"key : "<<elem.first<<" value : "<<elem.second<<endl;
    }
    
    
     for(auto elem : um){
        cout<<"key : "<<elem.first<<" value : "<<elem.second<<endl;
    }
    
    */
    return answer - 1;
}

3. JavaScript

function solution(clothes) {
    var answer = 1;
    let dict = {};
    for(let [a, b] of clothes){
        if (!dict.hasOwnProperty(b)){
            dict[b] = 1;
        }else{
            dict[b]++;
        }
    }
    
    for(let a in dict){
        answer *= (dict[a] + 1)
    }
    return answer - 1;
}