프로그래머스 / 광물 캐기

https://school.programmers.co.kr/learn/courses/30/lessons/172927

 

파이썬

제안사항을 확인하니 재귀를 이용해도 될 정도

def solution(picks, minerals):
    def solve(picks, minerals, fatigue):
        if sum(picks) == 0 or len(minerals) == 0:
            return fatigue
        result1 = result2 = result3 = float('inf')
        if picks[0] > 0:
            fatigues = {"diamond": 1, "iron": 1, "stone": 1}
            next_fatigue = sum(fatigues[mineral] for mineral in minerals[:5])
            result1 = solve([picks[0] - 1, picks[1], picks[2]], minerals[5:], fatigue + next_fatigue)
        if picks[1] > 0:
            fatigues = {"diamond": 5, "iron": 1, "stone": 1}
            next_fatigue = sum(fatigues[mineral] for mineral in minerals[:5])
            result2 = solve([picks[0], picks[1] - 1, picks[2]], minerals[5:], fatigue + next_fatigue)
        if picks[2] > 0:
            fatigues = {"diamond": 25, "iron": 5, "stone": 1}
            next_fatigue = sum(fatigues[mineral] for mineral in minerals[:5])
            result3 = solve([picks[0], picks[1], picks[2] - 1], minerals[5:], fatigue + next_fatigue)
        return min(result1, result2, result3)

    return solve(picks, minerals, 0)

이 정도면 깔끔

def solution(picks, minerals):
    def solve(picks, minerals, fatigue):
        if sum(picks) == 0 or len(minerals) == 0:
            return fatigue
        result = [float('inf')]
        for i, fatigues in enumerate(({"diamond": 1, "iron": 1, "stone": 1},
                                      {"diamond": 5, "iron": 1, "stone": 1},
                                      {"diamond": 25, "iron": 5, "stone": 1},)):
            if picks[i] > 0:
                next_picks = picks.copy()
                next_picks[i] -= 1
                next_fatigue = fatigue + sum(fatigues[mineral] for mineral in minerals[:5])
                result.append(solve(next_picks, minerals[5:], next_fatigue))
        return min(result)

    return solve(picks, minerals, 0)

굳이 이 정도는.... 

def solution(picks, minerals, fatigue=0):
    def get_next_picks(now_picks, i):
        next_picks = now_picks.copy()
        next_picks[i] -= 1
        return next_picks

    if sum(picks) == 0 or len(minerals) == 0:
        return fatigue
    return min([
        solution(get_next_picks(picks, i), minerals[5:], fatigue + sum(fatigues[mineral] for mineral in minerals[:5]))
        if picks[i] > 0 else float('inf')
        for i, fatigues in enumerate(({"diamond": 1, "iron": 1, "stone": 1},
                                      {"diamond": 5, "iron": 1, "stone": 1},
                                      {"diamond": 25, "iron": 5, "stone": 1}))])

 

코틀린

class Solution {
    private fun solve(picks: IntArray, minerals: List<String>, fatigue: Int): Int {
        if (picks.sum() == 0 || minerals.isEmpty()) return fatigue
        return listOf(
                mapOf("diamond" to 1, "iron" to 1, "stone" to 1),
                mapOf("diamond" to 5, "iron" to 1, "stone" to 1),
                mapOf("diamond" to 25, "iron" to 5, "stone" to 1),
        ).mapIndexed { index, fatigues ->
            if (picks[index] > 0) {
                val nextPicks = picks.clone()
                nextPicks[index] -= 1
                solve(nextPicks, minerals.drop(5), fatigue + minerals.take(5).sumOf { fatigues.getOrDefault(it, 0) })
            } else 2147483647
        }.minOrNull()!!
    }

    fun solution(picks: IntArray, minerals: Array<String>): Int {
        return solve(picks, minerals.asList(), 0)
    }
}