How do you solve Design Event Manager on LeetCode?

Design Event Manager (LeetCode #3885) 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: Use a priority queue for efficient retrieval of the highest priority event.

What is the time complexity of Design Event Manager?

The optimal solution for Design Event Manager runs in O(log n) time and uses O(n) space. Priority queue operations are logarithmic, and we maintain a hash map for constant time access.

What is the brute force solution for Design Event Manager?

The brute force approach for Design Event Manager is Brute Force, with O(n²) time complexity and O(1) space complexity. Iterate through the list of events to find the highest priority each time. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(log n).

What are the interview tips for Design Event Manager?

Explain your thought process clearly. Discuss trade-offs between different data structures. Practice coding on a whiteboard or in a shared document.

What are common mistakes when solving Design Event Manager?

Forgetting to check if the priority in the queue matches the current priority. Not handling the case where multiple events have the same priority correctly.

#3885

Design Event Manager

Medium

ArrayHash 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)

💡

Intuition

Time O(log n)Space O(n)

Use a priority queue to efficiently manage events by priority, and a hash map to track current priorities.

⚙️

Algorithm

3 steps

1Step 1: Store events in a priority queue sorted by priority and eventId.
2Step 2: Use a hash map to keep track of the current priority of each eventId.
3Step 3: In pollHighest, pop from the queue until the top matches the priority in the hash map.

solution.py16 lines

1import heapq
2class EventManager:
3    def __init__(self, events):
4        self.event_map = {event[0]: event[1] for event in events}
5        self.pq = [(event[1], event[0]) for event in events]
6        heapq.heapify(self.pq)
7    def updatePriority(self, eventId, newPriority):
8        self.event_map[eventId] = newPriority
9        heapq.heappush(self.pq, (newPriority, eventId))
10    def pollHighest(self):
11        while self.pq:
12            priority, eventId = heapq.heappop(self.pq)
13            if self.event_map[eventId] == priority:
14                del self.event_map[eventId]
15                return eventId
16        return -1

ℹ

Complexity note: Priority queue operations are logarithmic, and we maintain a hash map for constant time access.

1Use a priority queue for efficient retrieval of the highest priority event.
2Maintain a hash map to track current priorities for quick updates.

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