73. Task Scheduler

Topic :

priority queue

Difficulty :

medium

Problem Link :

problem statement

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.

Example 2:

Input: tasks = ["A","A","A","B","B","B"], n = 0
Output: 6
Explanation: On this case any permutation of size 6 would work since n = 0.
["A","A","A","B","B","B"]
["A","B","A","B","A","B"]
["B","B","B","A","A","A"]
...
And so on.

Example 3:

Input: tasks = ["A","A","A","A","A","A","B","C","D","E","F","G"], n = 2
Output: 16
Explanation: 
One possible solution is
A -> B -> C -> A -> D -> E -> A -> F -> G -> A ->
idle -> idle -> A -> idle -> idle -> A

Constraints:

1 <= task.length <= 104
tasks[i] is upper-case English letter.
The integer n is in the range [0, 100].

solution

import java.io.*;
import java.util.*;
class TaskScheduler {
    public static void main(String args[])
    {
        char tasks[]={'A','A','B','B','A','B'};
        int n=2;
        System.out.println(leastInterval(tasks,n));
    }
    static int leastInterval(char[] tasks, int n) 
    {
        // store the frequency of each task
        HashMap<Character,Integer> map=new HashMap<>();
        for(char ch : tasks)
            map.put(ch,map.getOrDefault(ch,0)+1);
        
        // sort in desecnding order
        //so task with max freq is at top
        // max heap is formed
        PriorityQueue<Map.Entry<Character,Integer>> pq=new PriorityQueue<>((a,b)->(b.getValue()-a.getValue()));
        pq.addAll(map.entrySet());
        
        int time=0;
        
        while(!pq.isEmpty()) // while there are still tasks to perform
        {
            int interval=n+1; //time slot 
            
            //stores the updated freq of the performed tasks
            ArrayList<Map.Entry<Character,Integer>> temp=new ArrayList<>();
            
            while(interval>0 && !pq.isEmpty())
            {
                Map.Entry<Character,Integer> currTask=pq.poll();
                currTask.setValue(currTask.getValue()-1);
                temp.add(currTask);
                // 1 task is performed
                interval--;
                time++;
            }
            
            // add the tasks that are yet to completed
            for( Map.Entry<Character,Integer> task : temp)
            {
                if(task.getValue()>0)
                    pq.offer(task);
            }
            
            //if suppose all tasks were done and pq didnot need to be updated then our job is done
            if(pq.isEmpty()) break;
            
            //if we have performed task that could be done yet interval>0 
            // there for the machine has to be idle for time = interval
            
            if(interval>0) time+=interval;
        }
        
        return time;
        
    }
}