Leetcode #2034: Stock Price Fluctuation
In this guide, we solve Leetcode #2034 **Stock Price Fluctuation ** in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given a stream of records about a particular stock. Each record contains a timestamp and the corresponding price of the stock at that timestamp.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Design, Hash Table, Data Stream, Ordered Set, Heap (Priority Queue)
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
Input
["StockPrice", "update", "update", "current", "maximum", "update", "maximum", "update", "minimum"]
[[], [1, 10], [2, 5], [], [], [1, 3], [], [4, 2], []]
Output
[null, null, null, 5, 10, null, 5, null, 2]
Explanation
StockPrice stockPrice = new StockPrice();
stockPrice.update(1, 10); // Timestamps are [1] with corresponding prices [10].
stockPrice.update(2, 5); // Timestamps are [1,2] with corresponding prices [10,5].
stockPrice.current(); // return 5, the latest timestamp is 2 with the price being 5.
stockPrice.maximum(); // return 10, the maximum price is 10 at timestamp 1.
stockPrice.update(1, 3); // The previous timestamp 1 had the wrong price, so it is updated to 3.
// Timestamps are [1,2] with corresponding prices [3,5].
stockPrice.maximum(); // return 5, the maximum price is 5 after the correction.
stockPrice.update(4, 2); // Timestamps are [1,2,4] with corresponding prices [3,5,2].
stockPrice.minimum(); // return 2, the minimum price is 2 at timestamp 4.
Python Solution
class StockPrice:
def __init__(self):
self.d = {}
self.ls = SortedList()
self.last = 0
def update(self, timestamp: int, price: int) -> None:
if timestamp in self.d:
self.ls.remove(self.d[timestamp])
self.d[timestamp] = price
self.ls.add(price)
self.last = max(self.last, timestamp)
def current(self) -> int:
return self.d[self.last]
def maximum(self) -> int:
return self.ls[-1]
def minimum(self) -> int:
return self.ls[0]
# Your StockPrice object will be instantiated and called as such:
# obj = StockPrice()
# obj.update(timestamp,price)
# param_2 = obj.current()
# param_3 = obj.maximum()
# param_4 = obj.minimum()
Complexity
The time complexity is O(log n). The space complexity is , where is the number of update operations.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.