321. 모의고사

프로그래머스

1. Python

def solution(answers):
    dict = {
        "one": [1, 2, 3, 4, 5],
        "two": [2, 1, 2, 3, 2, 4, 2, 5],
        "three":[3, 3, 1, 1, 2, 2, 4, 4, 5, 5]   
    }
    
    count = [0, 0, 0]
    
    for i in range(len(answers)):
        if dict["one"][i%5] == answers[i]:
            count[0] += 1
        if dict["two"][i%8] == answers[i]:
            count[1] += 1
        if dict["three"][i%10] == answers[i]:
            count[2] += 1
    
    answer = []
    m = max(count)
    for i in range(len(count)):
        if count[i] == m:
            answer.append(i+1)
    return answer
    

2. C++

#include <string>
#include <vector>
#include <algorithm>
using namespace std;

vector<int> solution(vector<int> answers) {
    vector<int> answer;
    int one[5] = {1, 2, 3, 4, 5};
    int two[8] = {2, 1, 2, 3, 2, 4, 2, 5};
    int three[10] = {3, 3, 1, 1, 2, 2, 4, 4, 5, 5};
    int count [3] = {0, 0, 0};
    int i;
        
    for(int i = 0; i < answers.size(); i++){
        if (one[i % 5] == answers[i]){
            count[0]++;
        }
        if(two[i % 8] == answers[i]){
            count[1]++;
        }
        if(three[i % 10] == answers[i]){
            count[2]++;
        }
    }
    
    int m = *max_element(count, count+3);
    
    for(i = 0; i < 3; i++){
        if(count[i] == m)
            answer.push_back(i+1);
        
    }
    
    
    return answer;
}

3. JavaScript

function solution(answers) {
    let one = [1, 2, 3, 4, 5];
    let two = [2, 1, 2, 3, 2, 4, 2, 5];
    let three = [3, 3, 1, 1, 2, 2, 4, 4, 5, 5];
    
    let count = [0, 0, 0];
    
    for (let i = 0; i < answers.length; i++){
        if (one[i % 5] == answers[i]){
            count[0]++;
        }
        if(two[i % 8] == answers[i]){
            count[1]++;
        }
        if(three[i % 10] == answers[i]){
            count[2]++;
        }
        
    }
    
    let m = Math.max.apply(null,count);
    
    return count.map((x, i) => m == x ? i+1 : null ).filter(
        (element, i) => element != null);
    
}