[프로그래머스] 성격 유형 검사하기

문제출처 :
https://school.programmers.co.kr/learn/courses/30/lessons/118666?language=javascript

1차작성 코드

function updateChoice(choice) {
    if(choice <= 3) {
        return -1*(4-choice)
    } else if (choice > 4) {
        return choice-4
    } else {
        return 0
    }
}
function solution(survey, choices) {
    let choice = choices.map((choose) => updateChoice(choose))
    let score_dict = {"RT":0,"CF":0,"JM":0,"AN":0,}
    survey.forEach((ques,idx) => {
        if(ques == "AN" || ques == "NA") {
            score_dict["AN"] += ques[0] == "A" ? choice[idx] : -1*choice[idx]
        } else if(ques == "RT" || ques == "TR") {
            score_dict["RT"] += ques[0] == "R" ? choice[idx] : -1*choice[idx]
        } else if(ques == "FC" || ques == "CF") {
            score_dict["CF"] += ques[0] == "C" ? choice[idx] : -1*choice[idx]
        } else {
            score_dict["JM"] += ques[0] == "J" ? choice[idx] : -1*choice[idx]
        }
    })
    let result = ""
    for(mbti in score_dict) {
        if(score_dict[mbti] <= 0) {
            result += mbti[0]
        } else {
            result += mbti[1]
        }
    }
    return result;
}

2차작성 코드

function solution(survey, choices) {
    let score_dict = {"A":0,"N":0,"C":0,"F":0,"M":0,"J":0,"R":0,"T":0} 
    survey.forEach((ques,idx) => {
        const acc = Math.abs(4-choices[idx])
        if(choices[idx] > 4) {
            score_dict[ques[1]] += acc
        } else {
            score_dict[ques[0]] += acc
        }
    })
    let result = ""
    result += score_dict["R"]-score_dict["T"]>=0 ? "R" : "T"
    result += score_dict["C"]-score_dict["F"]>=0 ? "C" : "F"
    result += score_dict["J"]-score_dict["M"]>=0 ? "J" : "M"
    result += score_dict["A"]-score_dict["N"]>=0 ? "A" : "N"
    return result;
}

총평 : 두 코드의 시간복잡도는 확연히 차이가 난다. 경우의 수를 고려해보고 좀 더 빠르다고 생각하는 것에 집중해보자.