Question
Given a char array representing tasks CPU need to do. It contains capital letters A to Z where different letters represent different tasks.Tasks could be done without original order. Each task could be done in one interval. For each interval, CPU could finish one task or just be idle.
However, there is a non-negative cooling interval n that means between two same tasks, there must be at least n intervals that CPU are doing different tasks or just be idle.
You need to return the least number of intervals 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.
Note:
- The number of tasks is in the range [1, 10000].
- The integer n is in the range [0, 100].
Solution
It’s obvious that we should always process the task which has largest amount left.
Put tasks in a PriorityQueue in descending order. Start to process tasks from front of the queue. If amount left > 0, put it into a frozenTimes HashMap. If there’s task which frozen time expired, put it into the queue and wait to be processed.
Repeat step the process till there is no task left.
Code
class Solution {
public int leastInterval(char[] tasks, int n) {
Map<Character, Integer> counts = new HashMap<>();
Map<Integer, Character> frozenTimes = new HashMap<>();
// q should only contains task that could be selected at the moment
PriorityQueue<Character> q = new PriorityQueue<Character>((a, b) -> counts.get(b)-counts.get(a));
for(int i=0; i<tasks.length; i++) {
if(!counts.containsKey(tasks[i])) {
counts.put(tasks[i], 0);
}
counts.put(tasks[i], counts.get(tasks[i])+1);
}
for(char task: counts.keySet()) {
q.offer(task);
}
int worktime = 0;
while(!q.isEmpty() || !counts.isEmpty()) {
worktime ++;
if(!q.isEmpty()) {
char task = q.poll();
int count = counts.get(task);
if(count == 1) {
counts.remove(task);
}
else {
counts.put(task, count-1);
// mark the time when the task is frozen
frozenTimes.put(worktime, task);
}
}
// release the task when it's frozen at "n times" ago
if(frozenTimes.containsKey(worktime - n)) {
q.offer(frozenTimes.get(worktime - n));
frozenTimes.remove(worktime - n);
}
}
return worktime;
}
}