LeetCode 621. Task Scheduler

Description

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

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].

Explanation

The most frequent task M which appears M times decides the max total time.

  1. total M – 1 numbers of idle timeeach length is N, the total idle time: (M – 1) * N.
  2. The tasks run M times. Therefore, plus the idle time, the total time becomes: (M – 1) * N + M.
  3. For all other tasks OTHER_M which have same frequency as task M, those tasks can be put into the max task’s intervals. Only the last time, without idle time, need to be separately counted. The time is OTHER_M – 1.

In total the time is: (M – 1) * n + M + OTHER_M – 1.

Python Solution

class Solution:
    def leastInterval(self, tasks: List[str], n: int) -> int:
        counter = collections.Counter(tasks)
            
        task_counts = list(counter.values())
        
        max_count = max(task_counts)
        
        
        max_count_times = task_counts.count(max_count)
        return max(len(tasks), (max_count- 1) * n + max_count + (max_count_times - 1))        
  • Time Complexity: O(N).
  • Space Complexity: O(N).

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