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2365. Task Scheduler II
Description
You are given a 0indexed array of positive integers tasks
, representing tasks that need to be completed in order, where tasks[i]
represents the type of the i^{th}
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 <= 10^{5}
1 <= tasks[i] <= 10^{9}
1 <= space <= tasks.length
Solutions

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; } }

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; } };

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

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 }

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; }