LeetCode 2365. Task Scheduler II

原题链接在这里：https://leetcode.com/problems/task-scheduler-ii/

题目：

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

题解：

Have a map to record the next available date we could do the task. The key is the task.

Get the maximum of next available date and res + 1. Res + 1 is the next day, maybe current day is 8, the next day is 9 is already larger than next available date we could do this task again.

Upate the map with res + space + 1 is the next available date.

Time Complexity: O(n). n = tasks.length.

Space: O(n).

AC Java:

 1 class Solution {
 2     public long taskSchedulerII(int[] tasks, int space) {
 3         long res = 0L;
 4         HashMap<Integer, Long> nextTime = new HashMap<>();
 5         for(int t : tasks){
 6             res = Math.max(res + 1, nextTime.getOrDefault(t, 0L));
 7             nextTime.put(t, res + space + 1);
 8         }
 9         
10         return res;
11     }
12 }

类似Task Scheduler.