[Programmers-Java] 광물 캐기

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

접근

돌 곡괭이로 광물을 5개씩 캐서 피로도를 구하고 우선순위 큐에 삽입한다.
피로도가 큰 것부터 꺼내면서 다이아몬드 -> 철 -> 돌 곡괭이 순으로 광물을 캔다.

풀이

import java.util.*;

class Solution {
    private static final int[][] PICK_FATIGUE = {
            { 1, 1, 1 },
            { 5, 1, 1 },
            { 25, 5, 1 }
    }; // (1)

    public int solution(int[] picks, String[] minerals) {
        int pickCnt = Arrays.stream(picks).sum() * 5;
        if (pickCnt < minerals.length) { // (2)
            minerals = Arrays.copyOfRange(minerals, 0, pickCnt);
        }

        PriorityQueue<int[]> pq = new PriorityQueue<>((o1, o2) -> o2[1] - o1[1]); // (3)
        for (int i = 0; i < minerals.length; i += 5) { // (4)
            int fatigue = 0;
            for (int j = i; j < i + 5 && j < minerals.length; j++) {
                switch (minerals[j]) {
                    case "diamond" -> fatigue += PICK_FATIGUE[2][0];
                    case "iron" -> fatigue += PICK_FATIGUE[2][1];
                    case "stone" -> fatigue += PICK_FATIGUE[2][2];
                }
            }
            pq.offer(new int[] { i, fatigue }); // (5)
        }

        int p = 0;
        for (int i = 0; i < picks.length; i++) {
            if (picks[i] > 0) { // (6)
                p = i;
                break;
            }
        }
        int answer = 0;
        while (!pq.isEmpty()) {
            int m = pq.poll()[0];
            for (int j = m; j < m + 5 && j < minerals.length; j++) { // (7)
                switch (minerals[j]) {
                    case "diamond" -> answer += PICK_FATIGUE[p][0];
                    case "iron" -> answer += PICK_FATIGUE[p][1];
                    case "stone" -> answer += PICK_FATIGUE[p][2];
                }
            }
            if (--picks[p] == 0) { // (8)
                p++;
            }
        }

        return answer;
    }
}