Stock Price Fluctuation

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

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

✓ Brute Force with Linear Scan

⏱️ Time: O(n) Space: O(n)

Use a hash map to store timestamp-price pairs. For each query operation, scan through all stored prices to find the maximum, minimum, or current price.

Heap-Based Optimization

⏱️ Time: O(log n) Space: O(n)

Combine a hash map for timestamp lookups with max and min heaps to track price extremes. Handle price updates by marking old prices as invalid in heaps.

Brute Force with Linear Scan — Algorithm Steps

Step 1: Store timestamp-price pairs in hash map
Step 2: For current(), find the price at maximum timestamp
Step 3: For maximum()/minimum(), scan all prices

Visualization

Tap to expand

Step-by-Step Walkthrough

Store Updates

Hash map stores timestamp-price pairs

Query Operations

Scan all values for max/min/current

Result

Return found value

Code -

solution.c — C

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

#define MAX_RECORDS 100000
#define MAX_LINE 1000000

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

typedef struct {
    Record records[MAX_RECORDS];
    int count;
    int latestTimestamp;
} StockPrice;

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

void stockPriceUpdate(StockPrice* obj, int timestamp, int price) {
    // Find if timestamp already exists
    for (int i = 0; i < obj->count; i++) {
        if (obj->records[i].timestamp == timestamp) {
            obj->records[i].price = price;
            return;
        }
    }
    
    // Add new record
    obj->records[obj->count].timestamp = timestamp;
    obj->records[obj->count].price = price;
    obj->count++;
    
    if (timestamp > obj->latestTimestamp) {
        obj->latestTimestamp = timestamp;
    }
}

int stockPriceCurrent(StockPrice* obj) {
    for (int i = 0; i < obj->count; i++) {
        if (obj->records[i].timestamp == obj->latestTimestamp) {
            return obj->records[i].price;
        }
    }
    return 0;
}

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

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

char* parseOperations(char* line, char operations[][20], int* opCount) {
    *opCount = 0;
    int len = strlen(line);
    int start = 1; // Skip first [
    int inQuote = 0;
    int opStart = -1;
    
    for (int i = 1; i < len - 1; i++) {
        if (line[i] == '"') {
            if (!inQuote) {
                inQuote = 1;
                opStart = i + 1;
            } else {
                inQuote = 0;
                int opLen = i - opStart;
                strncpy(operations[*opCount], line + opStart, opLen);
                operations[*opCount][opLen] = '\0';
                (*opCount)++;
            }
        }
    }
    
    return NULL;
}

void parseValues(char* line, int values[][2], int* valCount) {
    *valCount = 0;
    int len = strlen(line);
    
    for (int i = 0; i < len; i++) {
        if (line[i] == '[' && i + 1 < len && line[i + 1] != ']') {
            // Found start of non-empty array
            int j = i + 1;
            int nums[2];
            int numCount = 0;
            char numStr[20];
            int numPos = 0;
            
            while (j < len && line[j] != ']') {
                if (line[j] >= '0' && line[j] <= '9') {
                    numStr[numPos++] = line[j];
                } else if (line[j] == ',' || line[j] == ']') {
                    if (numPos > 0) {
                        numStr[numPos] = '\0';
                        nums[numCount++] = atoi(numStr);
                        numPos = 0;
                    }
                }
                j++;
            }
            
            if (numPos > 0) {
                numStr[numPos] = '\0';
                nums[numCount++] = atoi(numStr);
            }
            
            if (numCount >= 2) {
                values[*valCount][0] = nums[0];
                values[*valCount][1] = nums[1];
            }
            (*valCount)++;
            i = j;
        } else if (line[i] == '[' && i + 1 < len && line[i + 1] == ']') {
            // Empty array
            (*valCount)++;
            i++;
        }
    }
}

int main() {
    static char line1[MAX_LINE], line2[MAX_LINE];
    static char operations[1000][20];
    static int values[1000][2];
    int opCount, valCount;
    
    fgets(line1, MAX_LINE, stdin);
    fgets(line2, MAX_LINE, stdin);
    
    // Remove newlines
    line1[strcspn(line1, "\n")] = 0;
    line2[strcspn(line2, "\n")] = 0;
    
    parseOperations(line1, operations, &opCount);
    parseValues(line2, values, &valCount);
    
    StockPrice* stock = stockPriceCreate();
    int results[1000];
    int resultCount = 0;
    
    for (int i = 0; i < opCount; i++) {
        if (strcmp(operations[i], "StockPrice") == 0) {
            // Constructor
        } else if (strcmp(operations[i], "update") == 0) {
            stockPriceUpdate(stock, values[i][0], values[i][1]);
        } else if (strcmp(operations[i], "current") == 0) {
            results[resultCount++] = stockPriceCurrent(stock);
        } else if (strcmp(operations[i], "maximum") == 0) {
            results[resultCount++] = stockPriceMaximum(stock);
        } else if (strcmp(operations[i], "minimum") == 0) {
            results[resultCount++] = stockPriceMinimum(stock);
        }
    }
    
    printf("[");
    for (int i = 0; i < resultCount; i++) {
        printf("%d", results[i]);
        if (i < resultCount - 1) printf(",");
    }
    printf("]\n");
    
    free(stock);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Each query scans through all n stored prices

✓ Linear Growth

Space Complexity

O(n)

Hash map stores up to n timestamp-price pairs

⚡ 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

Brute Force with Linear Scan — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler