621. Task Scheduler
Tags: java, algorithms, programming, 100daysofcode
Published: 2021-12-29 16:45:04

# Description

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*.

<aside>
💡 **Note:**

</aside>

## 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]`.

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# Solutions

## Solution 1

### Intuition


### Code

```java
public int leastInterval(char[] tasks, int n) {
    int[] frequencies = new int[26];
    int max = Integer.MIN_VALUE;
    for (int task : tasks) {
        frequencies[task - 'A']++;
        max = Math.max(max, frequencies[task - 'A']);
    }
    int maxCount = 0;
    for (int frequency : frequencies) {
        if (frequency == max) {
            maxCount++;
        }
    }
    return Math.max(tasks.length, (max - 1) * (n + 1) + maxCount);
}
```

### Complexity

- Time: O(n)
- Space: O(1)

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## Solution 2

### Intuition

1. create the task frequencies;
2. you need build a heap from big to small
3. process each task in slots
4. remember the last slot, you need to check if heap is empty, and add working time rather than the whole slot 

### Code

```java
public int leastInterval(char[] tasks, int n) {
    HashMap<Character, Integer> frequencies = new HashMap<>();
    for (char task : tasks) {
        frequencies.put(task, frequencies.getOrDefault(task, 0) + 1);
    }
    PriorityQueue<Integer> heap = new PriorityQueue<>((a, b) -> b - a);
    heap.addAll(frequencies.values());

    int slot = n + 1, time = 0;
    while (!heap.isEmpty()) {
        int workingTime = 0;
        ArrayList<Integer> unfinishedTasks = new ArrayList<>();
        for (int i = 0; i < slot; i++) {
            if (!heap.isEmpty()) {
								// process task: task - 1
                unfinishedTasks.add(heap.poll() - 1);
                workingTime++;
            }
        }
        for (Integer unfinishedTask : unfinishedTasks) {
            if (unfinishedTask > 0) {
                heap.add(unfinishedTask);
            }
        }
        time += heap.isEmpty() ? workingTime : slot;
    }
    return time;
}
```

### Complexity

- Time: O(n)
- Space: O(1)

---