How do you solve Design Auction System on LeetCode?

Design Auction System (LeetCode #3815) 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 of hash maps for quick access to bids

What is the brute force solution for Design Auction System?

The brute force approach for Design Auction System is Brute Force, with O(n²) time complexity and O(1) space complexity. This approach involves maintaining a list of all bids for each item and searching through it to find the highest bid. It's straightforward but inefficient for larger datasets. The optimal Optimal Solution approach improves this to O(log n).

What are the interview tips for Design Auction System?

Clarify requirements and constraints Discuss trade-offs of different data structures Practice similar design problems

What are common mistakes when solving Design Auction System?

Not updating the heap after bid changes Confusing userId and bidAmount in comparisons

#3815

Design Auction System

Medium

Hash TableDesignHeap (Priority Queue)Ordered SetHash MapHeap

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)

Utilize a hash map to store bids and a max-heap to efficiently retrieve the highest bid. This allows for constant time updates and logarithmic time retrieval.

⚙️

Algorithm

3 steps

1Step 1: Use a hash map to store bids as (userId, bidAmount) pairs for each item.
2Step 2: Maintain a max-heap for each item to track the highest bid efficiently.
3Step 3: When adding/updating a bid, adjust the heap accordingly to ensure it reflects the current highest bid.

solution.py23 lines

1import heapq
2class AuctionSystem:
3    def __init__(self):
4        self.bids = {}
5        self.max_heap = {}
6
7    def addBid(self, userId, itemId, bidAmount):
8        if itemId not in self.bids:
9            self.bids[itemId] = {}
10            self.max_heap[itemId] = []
11        if userId in self.bids[itemId]:
12            self.bids[itemId][userId] = bidAmount
13        else:
14            self.bids[itemId][userId] = bidAmount
15        heapq.heappush(self.max_heap[itemId], (-bidAmount, userId))
16
17    def getHighestBidder(self, itemId):
18        while self.max_heap[itemId]:
19            bidAmount, userId = self.max_heap[itemId][0]
20            if self.bids[itemId][userId] == -bidAmount:
21                return userId
22            heapq.heappop(self.max_heap[itemId])
23        return -1

ℹ

Complexity note: The complexity is logarithmic for adding bids due to the heap operations, while space complexity is linear due to storing bids and heaps.

1Use of hash maps for quick access to bids
2Max-heap allows efficient retrieval of highest bids

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