How do you solve Design Task Manager on LeetCode?

Design Task Manager (LeetCode #3408) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(log n) time complexity and O(n) space complexity. Key insight: Using a priority queue allows efficient task execution.

What is the brute force solution for Design Task Manager?

The brute force approach for Design Task Manager is Brute Force, with O(n²) time complexity and O(1) space complexity. This approach involves using a simple list to manage tasks, where each operation directly manipulates the list. It is inefficient as it may require scanning the entire list for each operation. The optimal Optimal Solution approach improves this to O(log n).

What are the interview tips for Design Task Manager?

Explain your thought process clearly. Discuss trade-offs between different data structures. Consider edge cases in your implementation.

What are common mistakes when solving Design Task Manager?

Not updating the priority queue after editing a task. Failing to remove tasks from the data structures.

#3408

Design Task Manager

Medium

Hash TableDesignHeap (Priority Queue)Ordered SetHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(log n)Space O(n)

This approach uses a hash map to store tasks by user and a priority queue to efficiently retrieve and execute the highest priority task. This allows for faster operations compared to the brute force method.

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Algorithm

2 steps

1Step 1: Use a hash map to store tasks for each user and a max heap to manage task priorities.
2Step 2: For add, insert the task into the hash map and heap; for edit, update the priority in the heap; for rmv, remove the task from both structures; for execTop, pop the highest priority task from the heap.

solution.py20 lines

1import heapq
2class TaskManager:
3    def __init__(self, tasks):
4        self.user_tasks = {}
5        self.priority_queue = []
6        for uid, tid, prio in tasks:
7            self.add(uid, tid, prio)
8    def add(self, userId, taskId, priority):
9        self.user_tasks[taskId] = (userId, priority)
10        heapq.heappush(self.priority_queue, (-priority, taskId))
11    def edit(self, taskId, newPriority):
12        self.user_tasks[taskId] = (self.user_tasks[taskId][0], newPriority)
13        heapq.heappush(self.priority_queue, (-newPriority, taskId))
14    def rmv(self, taskId):
15        del self.user_tasks[taskId]
16    def execTop(self):
17        while self.priority_queue:
18            priority, taskId = heapq.heappop(self.priority_queue)
19            if taskId in self.user_tasks:
20                return taskId

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Complexity note: Using a priority queue allows efficient retrieval of the highest priority task, while the hash map provides O(1) access for task management.

1Using a priority queue allows efficient task execution.
2Hash maps provide quick access and modification of tasks.

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