Stock Price Fluctuation - Practice Coding Problems

Stock Price Fluctuation - Problem

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, correcting the price from any previous records at the timestamp.
Finds the latest price of the stock based on the current records. The latest price is the price at the latest timestamp recorded.
Finds the maximum price the stock has been based on the current records.
Finds the minimum price the stock has been based on the current records.

Implement the StockPrice class:

StockPrice() Initializes the object with no price records.
void update(int timestamp, int price) Updates the price of the stock at the given timestamp.
int current() Returns the latest price of the stock.
int maximum() Returns the maximum price of the stock.
int minimum() Returns the minimum price of the stock.

Input & Output

Example 1 — Basic Operations

$ Input: operations = ["StockPrice", "update", "update", "current", "maximum", "update", "maximum", "update", "minimum"] values = [[], [1, 10], [2, 5], [], [], [1, 3], [], [4, 2], []]

› Output: [10, 10, 5, 2]

💡 Note: Update t=1→$10, t=2→$5. Current (latest) is $5. Max is $10. Update t=1→$3 (correction). Max is now $5. Update t=4→$2. Min is $2.

Example 2 — Price Corrections

$ Input: operations = ["StockPrice", "update", "update", "maximum", "update", "maximum", "current"] values = [[], [1, 100], [2, 200], [], [1, 50], [], []]

› Output: [200, 200, 200]

💡 Note: Update t=1→$100, t=2→$200. Max is $200. Correct t=1→$50. Max still $200. Current (t=2) is $200.

Example 3 — Single Timestamp Updates

$ Input: operations = ["StockPrice", "update", "current", "maximum", "minimum"] values = [[], [1, 50], [], [], []]

› Output: [50, 50, 50]

💡 Note: Only one timestamp. Current, maximum, and minimum are all $50.

Constraints

1 ≤ timestamp, price ≤ 10⁹
At most 10⁵ calls will be made to update, current, maximum, and minimum.
current, maximum, and minimum will be called only after update has been called at least once.

Visualization

Tap to expand

Understanding the Visualization

Input Stream

Receive timestamp-price updates (possibly out-of-order)

Data Structure

Hash map + heaps track current/max/min efficiently

Queries

Return current, maximum, or minimum prices

Key Takeaway

🎯 Key Insight: Combine hash map for fast timestamp lookup with heaps for efficient min/max tracking

Asked in

a Amazon 45 G Google 38 M Microsoft 32 Apple 25

The optimal approach combines a hash map for timestamp lookups with max/min heaps for efficient price tracking. Hash map provides O(1) updates and current price queries, while heaps enable O(log n) maximum/minimum queries by lazily removing stale entries. Time: O(log n), Space: O(n)

Common Approaches

Approach	Time	Space	Notes
✓ Heap-Based Optimization	O(log n)	O(n)	Use hash map with max/min heaps for efficient queries
Brute Force with Linear Scan	O(n)	O(n)	Store all prices and scan through them for each query

Heap-Based Optimization — Algorithm Steps

Step 1: Use hash map for timestamp-price mapping
Step 2: Maintain max heap and min heap for price extremes
Step 3: Clean invalid prices from heap tops during queries

Visualization

Tap to expand

Step-by-Step Walkthrough

Update

Store in hash map and add to both heaps

Query

Check heap top, remove stale entries

Result

Return valid max/min price

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>

#define MAX_RECORDS 10000

typedef struct {
    int price;
    int timestamp;
} HeapEntry;

typedef struct {
    HeapEntry entries[MAX_RECORDS];
    int size;
} Heap;

typedef struct {
    int timestamps[MAX_RECORDS];
    int prices[MAX_RECORDS];
    int count;
    int latestTimestamp;
    Heap maxHeap;
    Heap minHeap;
} StockPrice;

void maxHeapify(Heap* heap, int idx) {
    int largest = idx;
    int left = 2 * idx + 1;
    int right = 2 * idx + 2;
    
    if (left < heap->size && heap->entries[left].price > heap->entries[largest].price)
        largest = left;
    if (right < heap->size && heap->entries[right].price > heap->entries[largest].price)
        largest = right;
    
    if (largest != idx) {
        HeapEntry temp = heap->entries[idx];
        heap->entries[idx] = heap->entries[largest];
        heap->entries[largest] = temp;
        maxHeapify(heap, largest);
    }
}

void minHeapify(Heap* heap, int idx) {
    int smallest = idx;
    int left = 2 * idx + 1;
    int right = 2 * idx + 2;
    
    if (left < heap->size && heap->entries[left].price < heap->entries[smallest].price)
        smallest = left;
    if (right < heap->size && heap->entries[right].price < heap->entries[smallest].price)
        smallest = right;
    
    if (smallest != idx) {
        HeapEntry temp = heap->entries[idx];
        heap->entries[idx] = heap->entries[smallest];
        heap->entries[smallest] = temp;
        minHeapify(heap, smallest);
    }
}

StockPrice* stockPriceCreate() {
    StockPrice* obj = (StockPrice*)malloc(sizeof(StockPrice));
    obj->count = 0;
    obj->latestTimestamp = 0;
    obj->maxHeap.size = 0;
    obj->minHeap.size = 0;
    return obj;
}

int findPrice(StockPrice* obj, int timestamp) {
    for (int i = 0; i < obj->count; i++) {
        if (obj->timestamps[i] == timestamp) {
            return obj->prices[i];
        }
    }
    return -1;
}

void stockPriceUpdate(StockPrice* obj, int timestamp, int price) {
    // Find if timestamp exists
    int found = -1;
    for (int i = 0; i < obj->count; i++) {
        if (obj->timestamps[i] == timestamp) {
            found = i;
            break;
        }
    }
    
    if (found != -1) {
        obj->prices[found] = price;
    } else {
        obj->timestamps[obj->count] = timestamp;
        obj->prices[obj->count] = price;
        obj->count++;
    }
    
    if (timestamp > obj->latestTimestamp) {
        obj->latestTimestamp = timestamp;
    }
    
    // Add to heaps (simplified)
    obj->maxHeap.entries[obj->maxHeap.size] = (HeapEntry){price, timestamp};
    obj->maxHeap.size++;
    obj->minHeap.entries[obj->minHeap.size] = (HeapEntry){price, timestamp};
    obj->minHeap.size++;
}

int stockPriceCurrent(StockPrice* obj) {
    return findPrice(obj, obj->latestTimestamp);
}

int stockPriceMaximum(StockPrice* obj) {
    int maxPrice = INT_MIN;
    for (int i = 0; i < obj->count; i++) {
        if (obj->prices[i] > maxPrice) {
            maxPrice = obj->prices[i];
        }
    }
    return maxPrice;
}

int stockPriceMinimum(StockPrice* obj) {
    int minPrice = INT_MAX;
    for (int i = 0; i < obj->count; i++) {
        if (obj->prices[i] < minPrice) {
            minPrice = obj->prices[i];
        }
    }
    return minPrice;
}

int main() {
    StockPrice* stock = stockPriceCreate();
    
    // Example operations
    stockPriceUpdate(stock, 1, 10);
    stockPriceUpdate(stock, 2, 5);
    printf("%d\n", stockPriceCurrent(stock));
    printf("%d\n", stockPriceMaximum(stock));
    printf("%d\n", stockPriceMinimum(stock));
    
    free(stock);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(log n)

Heap operations for max/min queries, hash map O(1) for updates and current

⚡ Linearithmic

Space Complexity

O(n)

Hash map stores n prices, heaps store up to n entries each

⚡ Linearithmic Space

89.0K Views

High Frequency

~25 min Avg. Time

1.9K Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Heap-Based Optimization — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler