901. Online Stock Span (Python)
Related Topic
Description
Write a class StockSpanner which 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 today is defined as the maximum number of consecutive days (starting from today and going backwards) for which the price of the stock was less than or equal to today’s price.
For example, if the price of a stock over the next 7 days were [100, 80, 60, 70, 60, 75, 85], then the stock spans would be [1, 1, 1, 2, 1, 4, 6].
Sample I/O
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:
First, S = StockSpanner() is initialized. Then:
S.next(100) is called and returns 1,
S.next(80) is called and returns 1,
S.next(60) is called and returns 1,
S.next(70) is called and returns 2,
S.next(60) is called and returns 1,
S.next(75) is called and returns 4,
S.next(85) is called and returns 6.
Note that (for example) S.next(75) returned 4, because the last 4 prices
(including today's price of 75) were less than or equal to today's price.
Note
- Calls to StockSpanner.next(int price) will have 1 <= price <= 10^5.
- There will be at most 10000 calls to StockSpanner.next per test case.
- There will be at most 150000 calls to StockSpanner.next across all test cases.
- The total time limit for this problem has been reduced by 75% for C++, and 50% for all other languages.
Methodology
Append the price and day (init to 1) to the stack. For each price, if the price of top in stack is smaller than current price, then pop the top and append the new price with day + top’s day. return day as result.
class StockSpanner:
def __init__(self):
self.stack = []
def next(self, price: int) -> int:
day = 1
while self.stack and self.stack[-1][0] <= price:
day += self.stack.pop()[1]
self.stack.append((price, day))
return day
# Your StockSpanner object will be instantiated and called as such:
# obj = StockSpanner()
# param_1 = obj.next(price)
BigO
We iterate the string once the time complexity is O(N).