Online Stock Span
Problem Description
Design an algorithm that collects daily price quotes for some stock and returns the span of that stock's price for the current day.
The span of the stock's price in one day is the maximum number of consecutive days (starting from that day and going backward) for which the stock price was less than or equal to the price of that day.
- For example, if the prices of the stock in the last four days is
[7,2,1,2]and the price of the stock today is2, then the span of today is4because starting from today, the price of the stock was less than or equal2for4consecutive days. - Also, if the prices of the stock in the last four days is
[7,34,1,2]and the price of the stock today is8, then the span of today is3because starting from today, the price of the stock was less than or equal8for3consecutive days.
Implement the StockSpanner class:
StockSpanner()Initializes the object of the class.int next(int price)Returns the span of the stock's price given that today's price isprice.
Example 1:
Input
["StockSpanner", "next", "next", "next", "next", "next", "next", "next"]
[[], [100], [80], [60], [70], [60], [75], [85]]
Output
[null, 1, 1, 1, 2, 1, 4, 6]
Explanation
Constraints:
- 1 <= price <= 105
- At most
104calls will be made tonext.
Solve it on LeetCode
Approaches
1. Brute Force
Intuition:
The brute force approach involves keeping track of all the stock prices encountered so far and then, for each new price, iterating backwards through the stored prices to calculate the span directly. This means, for each price, we need to traverse backwards until a price higher than the current price is found.
Steps:
- Initialize a list to store the stock prices.
- For each new price, traverse backwards in the list:
- Increase the span as long as prices are less than or equal to the current price.
- Return the span once a higher price is found or the list is exhausted.
Code:
- Time Complexity: O(n^2) in the worst case for
ncalls to thenextmethod because we might need to traverse all previous prices for each call. - Space Complexity: O(n) for storing all the prices.
2. Using Stack
Intuition:
To optimize the previous approach, we can utilize a stack to maintain a history of price indices that help in quickly determining the span. We store prices along with their respective spans in a stack. Each time a price is added, all prices less than or equal to the current price are popped from the stack, effectively calculating the span in constant amortized time.
Steps:
- Use a stack to store pairs of prices and their spans.
- For each incoming price:
- Pop elements from the stack while they are less than or equal to the current price and accumulate their spans.
- The span for the current price is the accumulated span plus one (for the current price itself).
- Push the current price and its calculated span onto the stack.
Code:
- Time Complexity: O(1) on average for each call to
next, because each element is pushed and popped from the stack at most once. - Space Complexity: O(n) for storing elements in the stack where
nis the number of prices recorded.