0621. Task Scheduler

https://leetcode.com/problems/task-scheduler

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.

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].

ac1: priority queue

class Solution {
    public int leastInterval(char[] tasks, int n) {
        // Edge cases
        if (tasks == null || tasks.length == 0) {
            return 0;
        }
        
        int[] count = new int[26];
        for (int i = 0; i < tasks.length; i++) {
            int idx = tasks[i] - 'A';
            count[idx]++;
        }
        
        PriorityQueue<Integer> q = new PriorityQueue<>((a, b) -> b - a);
        for(int cnt : count) {
            if (cnt > 0) q.offer(cnt);
        }
        
        int time = 0;
        while (!q.isEmpty()) {
            List<Integer> buffer = new ArrayList<>();
            // One shift of execution, (n + 1) times
            for (int i = 0; i <= n; i++) {
                if (!q.isEmpty()) {
                    // There is still valid task in queue
                    int taskCount = q.poll();
                    taskCount--; // Consume 1 task
                    if (taskCount > 0) buffer.add(taskCount);
                } else {
                    // No valid task in queue, no task in buffer, we are done.
                    if (buffer.size() == 0) break;
                }
                time++; // 1 unit of time passes
            }
            
            // Now we pass n time, tasks in the buffer can join the task queue again
            buffer.forEach(q::offer);
        }
        
        return time;
    }
}
// O(Nlog26) -> O(N) time, O(1) space.

ac2: greedy, count idle slots

class Solution {
    public int leastInterval(char[] tasks, int n) {
        // Edge cases
        if (tasks == null || tasks.length == 0) {
            return 0;
        }
        
        int[] count = new int[26];
        int topFrequency = 0;
        int topFrequencyCount = 0;
        for (int i = 0; i < tasks.length; i++) {
            int idx = tasks[i] - 'A';
            count[idx]++;
            if (count[idx] > topFrequency) {
                topFrequency = count[idx];
                topFrequencyCount = 1;
            } else if (topFrequency == count[idx]) {
                topFrequencyCount++;
            }
        }
        
        // [AB__AB__AB__]AB, n = 3
        int minRequiredTime = (topFrequency - 1) * (n + 1);
        int topFrequencyOccupiedTime = (topFrequency - 1) * topFrequencyCount;
        int emptySlot = minRequiredTime - topFrequencyOccupiedTime;
        int remainingTaskCount = tasks.length - topFrequency * topFrequencyCount;
        int idles = Math.max(0, emptySlot - remainingTaskCount);
        
        return tasks.length + idles;
    }
}
// O(N) time, O(1) space.
// Greedy: we need at least (topFrequency - 1) * (n + 1) time, always complete most frequent tasks. 

https://leetcode.com/problems/task-scheduler/discuss/104500/Java-O(n)-time-O(1)-space-1-pass-no-sorting-solution-with-detailed-explanation

It's hard to prove greedy solution though.