621. Task Scheduler

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# 題目敘述

You are given an array of CPU tasks , each represented by letters A to Z, and a cooling time, n . Each cycle or interval allows the completion of one task. Tasks can be completed in any order, but there's a constraint: identical tasks must be separated by at least n intervals due to cooling time.

​Return the minimum number of intervals required to complete all tasks.

# Example 1

Input: tasks = ["A","A","A","B","B","B"], n = 2
Output: 8
Explanation: A possible sequence is: A -> B -> idle -> A -> B -> idle -> A -> B.
After completing task A, you must wait two cycles before doing A again. The same applies to task B. In the 3rd interval, neither A nor B can be done, so you idle. By the 4th cycle, you can do A again as 2 intervals have passed.

# Example 2

Input: tasks = ["A","C","A","B","D","B"], n = 1
Output: 6
Explanation: A possible sequence is: A -> B -> C -> D -> A -> B.
With a cooling interval of 1, you can repeat a task after just one other task.

# Example 3

Input: tasks = ["A","A","A", "B","B","B"], n = 3
Output: 10
Explanation: A possible sequence is: A -> B -> idle -> idle -> A -> B -> idle -> idle -> A -> B.
There are only two types of tasks, A and B, which need to be separated by 3 intervals. This leads to idling twice between repetitions of these tasks.

# 解題思路

# Solution

import java.util.HashMap;

import java.util.Map;

class Solution {

    public int leastInterval(char[] tasks, int n) {

        Map<Character, Integer> map = new HashMap<>();

        for (char c : tasks) {

            map.put(c, map.getOrDefault(c, 0) + 1);

        }

        int max = 0;

        int maxCount = 0;

        for (int count : map.values()) {

            if (count > max) {

                max = count;

                maxCount = 1;

            } else if (count == max) {

                maxCount++;

            }

        }

        int block = max - 1;

        int blockLength = n - (maxCount - 1);

        int emptySlots = block * blockLength;

        int availableTasks = tasks.length - max * maxCount;

        int idles = Math.max(0, emptySlots - availableTasks);

        return tasks.length + idles;

    }

}