[2021 KAKAO BLIND RECURITMENT] 메뉴 리뉴얼

코스개수를 순회하면서 주문할 수 있는 개수에 따른 combination을 만들어주고 가장 많이 나온 아이를 answer에 넣어주는 방법.

from itertools import combinations
from collections import Counter

def solution(orders, course):
    answer = []
    for c in course: 
        temp = []

        for order in orders:
            combi = combinations(sorted(order), c)
            temp += combi
        counter = Counter(temp)
        if len(counter) != 0 and max(counter.values()) != 1:
            answer += [''.join(f) for f in counter if counter[f] == max(counter.values())]

    return sorted(answer)

JavaScript

function solution (orders, course) {
    const ordered = {};
    const candidates = {};
    const maxNum = Array(10 + 1).fill(0)
    const createSet = (arr, start, len, foods) => {
        if (len == 0) {
            ordered[foods] = (ordered[foods] || 0) +  1;
            if (ordered[foods] > 1) candidates[foods] = ordered[foods]
            maxNum[foods.length] = Math.max(maxNum[foods.length], ordered[foods])
            return 
        }

        for (let i = start; i < arr.length; i++) {
            createSet(arr, i+1, len-1, foods + arr[i])
        }
    }

    orders.forEach((order) => {
        const sorted = order.split('').sort()
        course.forEach((len) => {
            createSet(sorted, 0, len, '')
        })
    })

    const launched = Object.keys(candidates).filter((food) => maxNum[food.length] === candidates[food])

    return launched.sort()
}