How do you solve Task Scheduler II on LeetCode?

Task Scheduler II (LeetCode #2365) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a hashmap to track the last completion day of tasks allows for efficient checking of whether breaks are needed.

What is the brute force solution for Task Scheduler II?

The brute force approach for Task Scheduler II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves simulating each day and deciding to either complete a task or take a break. This is straightforward but inefficient, as it checks every possible day for each task type. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Task Scheduler II?

Task Scheduler II uses the following concepts: Array, Hash Table, Simulation. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Task Scheduler II?

Always think about how to optimize your solution after writing the brute force version. Make sure to explain your thought process clearly, especially how you handle the timing of breaks. Practice similar problems to recognize patterns, such as using hashmaps for tracking state.

What are common mistakes when solving Task Scheduler II?

Failing to account for the space requirement correctly, leading to incorrect day counts. Not using a hashmap effectively to store last completion days, resulting in unnecessary iterations.

#2365

Task Scheduler II

Medium

ArrayHash TableSimulationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a hashmap to track the last day a task was completed, allowing us to efficiently calculate the required breaks without simulating each day. This reduces unnecessary checks and speeds up the process.

⚙️

Algorithm

4 steps

1Step 1: Initialize a day counter and a hashmap to track the last completion day of each task type.
2Step 2: Iterate through the tasks array.
3Step 3: For each task, check when it was last completed and calculate the necessary breaks to ensure the 'space' requirement is met.
4Step 4: Update the day counter and the last completion day for the task.

solution.py10 lines

1def minDays(tasks, space):
2    days = 0
3    last_completed = {}
4    for task in tasks:
5        days += 1
6        if task in last_completed:
7            days += max(0, space - (days - last_completed[task]))
8        last_completed[task] = days
9    return days
10

ℹ

Complexity note: The complexity is linear because we only make a single pass through the tasks array, and the hashmap operations (insert and lookup) are average O(1).

1Using a hashmap to track the last completion day of tasks allows for efficient checking of whether breaks are needed.
2Taking breaks as late as possible helps minimize the total number of days.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.