Skip to content

2365. Task Scheduler II

Difficulty Topics

Description

You are given a 0-indexed array of positive integers tasks, representing tasks that need to be completed in order, where tasks[i] represents the type of the ith task.

You are also given a positive integer space, which represents the minimum number of days that must pass after the completion of a task before another task of the same type can be performed.

Each day, until all tasks have been completed, you must either:

  • Complete the next task from tasks, or
  • Take a break.

Return the minimum number of days needed to complete all tasks.

 

Example 1:

Input: tasks = [1,2,1,2,3,1], space = 3
Output: 9
Explanation:
One way to complete all tasks in 9 days is as follows:
Day 1: Complete the 0th task.
Day 2: Complete the 1st task.
Day 3: Take a break.
Day 4: Take a break.
Day 5: Complete the 2nd task.
Day 6: Complete the 3rd task.
Day 7: Take a break.
Day 8: Complete the 4th task.
Day 9: Complete the 5th task.
It can be shown that the tasks cannot be completed in less than 9 days.

Example 2:

Input: tasks = [5,8,8,5], space = 2
Output: 6
Explanation:
One way to complete all tasks in 6 days is as follows:
Day 1: Complete the 0th task.
Day 2: Complete the 1st task.
Day 3: Take a break.
Day 4: Take a break.
Day 5: Complete the 2nd task.
Day 6: Complete the 3rd task.
It can be shown that the tasks cannot be completed in less than 6 days.

 

Constraints:

  • 1 <= tasks.length <= 105
  • 1 <= tasks[i] <= 109
  • 1 <= space <= tasks.length

Solution

task-scheduler-ii.py
class Solution:
    def taskSchedulerII(self, tasks: List[int], space: int) -> int:
        n = len(tasks)
        last = defaultdict(lambda: -1)
        day = 1

        for task in tasks:
            if last[task] == -1 or day > last[task] + space + 1:
                last[task] = day
            else:
                day = last[task] + space + 1
                last[task] = day

            day += 1

        return day - 1