package com.flang.queue;

import java.util.*;

/*
Given a characters array tasks, representing the tasks a CPU needs to do, where each letter represents a different task.
Tasks could be done in any order. Each task is done in one unit of time. For each unit of time, the CPU could complete
either one task or just be idle.

However, there is a non-negative integer n that represents the cooldown period between two same tasks (the same letter
in the array), that is that there must be at least n units of time between any two same tasks.

Return the least number of units of times that the CPU will take to finish all the given tasks.


Example 1:

Input: tasks = ["A","A","A","B","B","B"], n = 2
Output: 8
Explanation:
A -> B -> idle -> A -> B -> idle -> A -> B
There is at least 2 units of time between any two same tasks.

 */
public class TaskSchedulerGreedy {
    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);
        }
        PriorityQueue<Integer> maxHeap = new PriorityQueue<>((a, b) -> b - a);
        maxHeap.addAll(map.values());

        int cycle = 0;
        while (!maxHeap.isEmpty()) {
            List<Integer> temp = new ArrayList();
            for (int i = 0; i < n + 1; i++) {
                if (!maxHeap.isEmpty()) {
                    temp.add(maxHeap.remove());
                }
            }
            for (int i : temp) {
                if (--i > 0) {
                    maxHeap.add(i);
                }
            }
            // if maxHeap is empty, so we wait whatever the size of task we run.
            // Otherwise, we know we have to wait full cool down period.
            cycle += maxHeap.isEmpty() ? temp.size() : n + 1;
        }
        return cycle;
    }
}

package com.flang.queue;

import java.util.*;

/*
Given a characters array tasks, representing the tasks a CPU needs to do, where each letter represents a different task.
Tasks could be done in any order. Each task is done in one unit of time. For each unit of time, the CPU could complete
either one task or just be idle.

However, there is a non-negative integer n that represents the cooldown period between two same tasks (the same letter
in the array), that is that there must be at least n units of time between any two same tasks.

Return the least number of units of times that the CPU will take to finish all the given tasks.


Example 1:

Input: tasks = ["A","A","A","B","B","B"], n = 2
Output: 8
Explanation:
A -> B -> idle -> A -> B -> idle -> A -> B
There is at least 2 units of time between any two same tasks.

 */
public class TaskSchedulerGreedy {
    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);
        }
        // create a maxHeap to keep track most frequent tasks
        PriorityQueue<Integer> maxHeap = new PriorityQueue<>((a, b) -> b - a);
        maxHeap.addAll(map.values());

        int cycle = 0;
        while (!maxHeap.isEmpty()) {
            List<Integer> temp = new ArrayList();
            for (int i = 0; i < n + 1; i++) {
                if (!maxHeap.isEmpty()) {
                    temp.add(maxHeap.remove());
                }
            }
            for (int i : temp) {
                // if i-1 is not 0, we know the current task is still need to be run
                // so we add it back the maxHeap
                if (--i > 0) {
                    maxHeap.add(i);
                }
            }
            // if maxHeap is empty, so we wait whatever the size of task we run.
            // Otherwise, we know we have to wait full cool down period.
            cycle += maxHeap.isEmpty() ? temp.size() : n + 1;
        }
        return cycle;
    }
}

Update TaskSchedulerGreedy.java