901. Online Stock Span

1. 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 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.

2. Example

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

3. Constraints

• 1 <= price <= $10^5$
• At most $10^4$ calls will be made to next.

4. Solutions

Monotonic Stack

n is the number of next calls
Time complexity: O(n)
Space complexity: O(n)

class StockSpanner {
public:
    StockSpanner() {
        // Sentinel (day = 0, price = +inf): prevents the stack from becoming empty.
        // It also guarantees the top always has a day index smaller than any real day.
        dec_stock_index.emplace(index, numeric_limits<int>::max());
        ++index;
    }

    int next(int price) {
        // Maintain a monotonic decreasing stack by price.
        // Pop all previous days whose price <= current price, because they cannot
        // be the nearest previous greater price for future queries.
        while (dec_stock_index.top().second <= price) {
            dec_stock_index.pop();
        }

        int span = index - dec_stock_index.top().first;

        dec_stock_index.emplace(index, price);
        ++index;

        return span;
    }

private:
    int index = 0;
    // Each element is (day_index, price)
    stack<pair<int, int>, vector<pair<int, int>>> dec_stock_index;
};