Introduction

In this blog post, we will explore how to solve the LeetCode problem "901" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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 is 2, then the span of today is 4 because starting from today, the price of the stock was less than or equal 2 for 4 consecutive 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 is 8, then the span of today is 3 because starting from today, the price of the stock was less than or equal 8 for 3 consecutive 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 is price. 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 StockSpanner stockSpanner = new StockSpanner(); stockSpanner.next(100); // return 1 stockSpanner.next(80); // return 1 stockSpanner.next(60); // return 1 stockSpanner.next(70); // return 2 stockSpanner.next(60); // return 1 stockSpanner.next(75); // return 4, because the last 4 prices (including today's price of 75) were less than or equal to today's price. stockSpanner.next(85); // return 6 Constraints: 1 <= price <= 105 At most 104 calls will be made to next.

Explanation

Here's the breakdown of the solution:

• Monotonic Stack: The core idea is to use a stack to maintain a decreasing sequence of prices. The stack stores pairs of (price, span).
• Span Calculation: When a new price arrives, we pop elements from the stack whose prices are less than or equal to the current price. The span is the sum of the spans of the popped elements plus 1 (for the current day).

• Stack Update: Finally, we push the current price and its calculated span onto the stack.

• Runtime & Storage Complexity: O(1) average time complexity for next() because each price is pushed and popped at most once from the stack. O(N) storage complexity, where N is the number of calls to next(), for storing the stack.

Code

    class StockSpanner:

    def __init__(self):
        self.stack = []  # (price, span)

    def next(self, price: int) -> int:
        span = 1
        while self.stack and self.stack[-1][0] <= price:
            span += self.stack[-1][1]
            self.stack.pop()
        self.stack.append((price, span))
        return span