What is the time complexity of Stock Price Fluctuation ?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Stock Price Fluctuation ?

This problem uses Hash Table, Design, Heap (Priority Queue), Data Stream, Ordered Set. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Stock Price Fluctuation ?

Insight 1: Using a hash map allows for O(1) average time complexity for updating the price at a given timestamp, handling corrections efficiently. Insight 2: Priority queues (heaps) are suitable for maintaining maximum and minimum prices. However, due to the possibility of price corrections, lazy deletion is needed, where stale entries in the heaps are removed only when they are accessed.

Which companies ask LeetCode 2034. Stock Price Fluctuation in interviews?

This problem is frequently asked at Atlassian, Mixpanel, MongoDB, Ripple.

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Stock Price Fluctuation - LeetCode 2034 Solution

Stock Price Fluctuation - Complete Solution Guide

Stock Price Fluctuation is LeetCode problem 2034, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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. Unfortunately due to the volatile nature of the stock market, the records do not come in order. Even worse, some records may be incorrect. Another record with the same timestamp may appear later in the stream correcting the price of the previous wrong record. Design an algorithm that: Updates the price of the stock at a particular timestamp, cor

Detailed Explanation

The problem requires you to design a data structure that simulates a stock price tracking system. The system should handle updates to stock prices at specific timestamps, even if the updates are out of order or correct previous entries. You need to efficiently retrieve the latest price, the maximum price, and the minimum price observed so far. The key challenge is handling potentially unordered and incorrect data while providing fast access to these aggregate values.

Solution Approach

The solution utilizes a hash map (`time_to_price`) to store the price at each timestamp. Two priority queues (max-heap and min-heap) are used to track the maximum and minimum prices, respectively. The 'update' function updates the hash map and adds new price/timestamp pairs to both heaps. The 'current' function simply retrieves the price from the hash map using the latest timestamp. The 'maximum' and 'minimum' functions lazily remove stale entries from the heaps until a valid maximum or minimum price is found, that is price corresponding to that timestamp should exist in time_to_price map with the same price value.

Step-by-Step Algorithm

Step 1: Initialize the `StockPrice` class with a hash map `time_to_price` to store timestamp-price pairs, a max-heap `max_heap`, a min-heap `min_heap`, and a variable `latest_timestamp` to keep track of the most recent timestamp.
Step 2: Implement the `update` function: Insert or update the `time_to_price` map with the given timestamp and price. Update `latest_timestamp` if the current timestamp is greater. Push the price-timestamp pair into both `max_heap` and `min_heap`.
Step 3: Implement the `current` function: Return the price associated with the `latest_timestamp` from the `time_to_price` map.
Step 4: Implement the `maximum` function: While the `max_heap` is not empty, peek at the top element (maximum price). Check if the price from the heap matches the price stored in `time_to_price` for the given timestamp. If they match, return the price. If they don't match (meaning the price has been updated), pop the element from the heap and continue the loop.
Step 5: Implement the `minimum` function: While the `min_heap` is not empty, peek at the top element (minimum price). Check if the price from the heap matches the price stored in `time_to_price` for the given timestamp. If they match, return the price. If they don't match (meaning the price has been updated), pop the element from the heap and continue the loop.

Key Insights

Insight 1: Using a hash map allows for O(1) average time complexity for updating the price at a given timestamp, handling corrections efficiently.
Insight 2: Priority queues (heaps) are suitable for maintaining maximum and minimum prices. However, due to the possibility of price corrections, lazy deletion is needed, where stale entries in the heaps are removed only when they are accessed.
Insight 3: Maintaining a separate 'latest_timestamp' variable simplifies the 'current' price retrieval operation.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Hash Table, Design, Heap (Priority Queue), Data Stream, Ordered Set.

Companies

Asked at: Atlassian, Mixpanel, MongoDB, Ripple.

Stock Price Fluctuation | LeetCode 2034

Medium

Hash Table Design Heap (Priority Queue)Data Stream Ordered Set

Time: O(n)Space: O(n)

Solution Code

Available in:

import java.util.HashMap;
import java.util.PriorityQueue;
import java.util.Map;

class StockPrice {

    private Map<Integer, Integer> timeToPrice;
    private PriorityQueue<int[]> maxHeap;
    private PriorityQueue<int[]> minHeap;
    private int latestTimestamp;

    public StockPrice() {
        timeToPrice = new HashMap<>();
        maxHeap = new PriorityQueue<>((a, b) -> b[0] - a[0]);
        minHeap = new PriorityQueue<>((a, b) -> a[0] - b[0]);
        latestTimestamp = 0;
    }

    public void update(int timestamp, int price) {
        timeToPrice.put(timestamp, price);
        latestTimestamp = Math.max(latestTimestamp, timestamp);

        maxHeap.offer(new int[] {price, timestamp});
        minHeap.offer(new int[] {price, timestamp});
    }

    public int current() {
        return timeToPrice.get(latestTimestamp);
    }

    public int maximum() {
        while (!maxHeap.isEmpty()) {
            int[] top = maxHeap.peek();
            int price = top[0];
            int timestamp = top[1];
            if (price == timeToPrice.get(timestamp)) {
                return price;
            }
            maxHeap.poll();
        }
        return -1; // Should not happen if updates are valid
    }

    public int minimum() {
        while (!minHeap.isEmpty()) {
            int[] top = minHeap.peek();
            int price = top[0];
            int timestamp = top[1];
            if (price == timeToPrice.get(timestamp)) {
                return price;
            }
            minHeap.poll();
        }
        return -1; // Should not happen if updates are valid
    }
}

/**
 * Your StockPrice object will be instantiated and called as such:
 * StockPrice obj = new StockPrice();
 * obj.update(timestamp,price);
 * int param_2 = obj.current();
 * int param_3 = obj.maximum();
 * int param_4 = obj.minimum();
 */

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Problem Info

Description

Problem 2034: Stock Price Fluctuation - Detailed explanation and solution approaches.

Topics

Hash Table Design Heap (Priority Queue)Data Stream Ordered Set

Companies

AtlassianMixpanelMongoDBRipple

Stats

Difficulty:Medium

Languages:4

Companies:4

Problem #:2034

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

5 steps

1
Step 1: Initialize the `StockPrice` class with a hash map `time_to_price` to store timestamp-price pairs, a max-heap `max_heap`, a min-heap `min_heap`, and a variable `latest_timestamp` to keep track of the most recent timestamp.
2
Step 2: Implement the `update` function: Insert or update the `time_to_price` map with the given timestamp and price. Update `latest_timestamp` if the current timestamp is greater. Push the price-timestamp pair into both `max_heap` and `min_heap`.
3
Step 3: Implement the `current` function: Return the price associated with the `latest_timestamp` from the `time_to_price` map.
4
Step 4: Implement the `maximum` function: While the `max_heap` is not empty, peek at the top element (maximum price). Check if the price from the heap matches the price stored in `time_to_price` for the given timestamp. If they match, return the price. If they don't match (meaning the price has been updated), pop the element from the heap and continue the loop.
5
Step 5: Implement the `minimum` function: While the `min_heap` is not empty, peek at the top element (minimum price). Check if the price from the heap matches the price stored in `time_to_price` for the given timestamp. If they match, return the price. If they don't match (meaning the price has been updated), pop the element from the heap and continue the loop.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The hash map provides O(1) average time complexity for updates. The priority queues maintain the maximum and minimum prices in O(log n) time per update. The space complexity is determined by the number of unique timestamps and prices stored.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Ordered Set (Sorted List)

Instead of using heaps, one could use an ordered set (e.g., a self-balancing binary search tree or sorted list) to store all prices. This would allow for efficient retrieval of the minimum and maximum prices. Using a Red-Black tree data structure or similar can guarantee logarithmic time complexity for insertion, deletion, and retrieval.

Time: O(log n) for update, maximum, and minimum, O(1) for current.Space: O(n)

Trade-offs: This approach avoids the need for lazy deletion, but it requires maintaining a sorted data structure which might have slightly higher overhead than heaps in some cases. It could be beneficial if maximum and minimum are called very frequently.

✨

Key Insights

1Insight 1: Using a hash map allows for O(1) average time complexity for updating the price at a given timestamp, handling corrections efficiently.
2Insight 2: Priority queues (heaps) are suitable for maintaining maximum and minimum prices. However, due to the possibility of price corrections, lazy deletion is needed, where stale entries in the heaps are removed only when they are accessed.
3Insight 3: Maintaining a separate 'latest_timestamp' variable simplifies the 'current' price retrieval operation.

📚

Key Terms Explained

For Students

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Edge Cases

Edge case 1: Correcting a previously recorded price. The hash map ensures that the latest price is always stored for a given timestamp, effectively handling corrections.
Edge case 2: When the maximum or minimum price in the heap is no longer valid (due to updates), the lazy deletion mechanism in the `maximum` and `minimum` functions ensures that stale entries are discarded until a valid price is found.
Edge case 3: Empty StockPrice object: The code assumes that the `current`, `maximum`, and `minimum` methods will be called only after `update` has been called at least once. Adding checks for empty `time_to_price` in these functions would make the code more robust.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling price corrections correctly. Failing to update the price in the hash map leads to incorrect results for `current`, `maximum`, and `minimum`.

✗Common mistake 2: Not implementing lazy deletion in the priority queues. Without lazy deletion, the `maximum` and `minimum` functions might return incorrect results if the maximum or minimum price has been updated.

✗Common mistake 3: Failing to consider the order of operations in the heapq implementation when using a max heap. You either need to negate the price or use a custom comparator to implement max-heap using heapq.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Stock price tracking systems, as the problem describes, need to maintain up-to-date information about stock prices, even when data arrives out of order or with corrections.

🏦Real-world scenario 2: Monitoring systems for sensor data, where sensor readings might be delayed, incorrect, or arrive out of order. The system needs to track the latest readings, as well as the maximum and minimum values observed over a period of time.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

Hash tables trade space for time. Before using one, consider if the problem constraints allow the extra memory. In interviews, always mention this trade-off.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Don't forget to handle key collisions and null/undefined keys. Consider what happens with duplicate values.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 25, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 14

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 7

❓

Frequently Asked Questions

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Stock Price Fluctuation "?

For this medium-level Hash Table problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether design techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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