LeetCode 2365. Task Scheduler II Solution in Java, C++, Python & More | Explanation + Code

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.

PythonJavaC++GoTypeScript

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(code-box)

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; } }(code-box)

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; } };(code-box)

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 }(code-box)

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; }(code-box)