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2365. Task Scheduler II

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

Solutions

Solution 1: Hash Table + Simulation

We can use a hash table $day$ to record the next time each task can be executed. Initially, all values in $day$ are $0$. We use a variable $ans$ to record the current time.

We iterate through the array $tasks$. For each task $task$, we increment the current time $ans$ by one, indicating that one day has passed since the last task execution. If $day[task] > ans$ at this time, it means that task $task$ can only be executed on the $day[task]$ day. Therefore, we update the current time $ans = \max(ans, day[task])$. Then we update the value of $day[task]$ to $ans + space + 1$, indicating that the next time task $task$ can be executed is at $ans + space + 1$.

After the iteration, we return $ans$.

The time complexity is $O(n)$ and the space complexity is $O(n)$, where $n$ is the length of the array $tasks$.

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class Solution:
    def taskSchedulerII(self, tasks: List[int], space: int) -> int:
        day = defaultdict(int)
        ans = 0
        for task in tasks:
            ans += 1
            ans = max(ans, day[task])
            day[task] = ans + space + 1
        return ans

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class Solution {
    public long taskSchedulerII(int[] tasks, int space) {
        Map<Integer, Long> day = new HashMap<>();
        long ans = 0;
        for (int task : tasks) {
            ++ans;
            ans = Math.max(ans, day.getOrDefault(task, 0L));
            day.put(task, ans + space + 1);
        }
        return ans;
    }
}

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class Solution {
public:
    long long taskSchedulerII(vector<int>& tasks, int space) {
        unordered_map<int, long long> day;
        long long ans = 0;
        for (int& task : tasks) {
            ++ans;
            ans = max(ans, day[task]);
            day[task] = ans + space + 1;
        }
        return ans;
    }
};

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func taskSchedulerII(tasks []int, space int) (ans int64) {
    day := map[int]int64{}
    for _, task := range tasks {
        ans++
        if ans < day[task] {
            ans = day[task]
        }
        day[task] = ans + int64(space) + 1
    }
    return
}

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function taskSchedulerII(tasks: number[], space: number): number {
    const day = new Map<number, number>();
    let ans = 0;
    for (const task of tasks) {
        ++ans;
        ans = Math.max(ans, day.get(task) ?? 0);
        day.set(task, ans + space + 1);
    }
    return ans;
}

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