[LeetCode] Online Stock Span

Problem: 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:

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

Approach: This is a monotonic stack problem. We keep popping the elements from stack till the current price is greater than or equal to the top of stack and return the difference of the sequences.

The problem here if we just record the the price and not the sequence in which the values are coming we will never be able to get the number of elements less than or equal to current price until we store every price. But instead of storing every price, we  can store the pair of price and the sequence number in the stack.

We just need to be careful while storing the price, we need to store the sequence number of the last popped price from stack. What it is indicating is that the current price is the maximum price till this particular sequence number (reversed).

That's all!

Implementation  in C#:

public class StackElement

{

    public int Value {get; set;}

    public int Sequence {get; set;}

    public StackElement(int val, int seq)

    {

        this.Value = val;

        this.Sequence = seq;

    }

}

public class StockSpanner

{

    public StockSpanner()

    {

        this.stack = new Stack<StackElement>();    

    }

   

    public int Next(int price)

    {

        int retVal = 1;

        ++this.seqNo;

        int seq = this.seqNo;

        StackElement se = null;

        while(stack.Count > 0 &&

              price >= stack.Peek().Value)

        {

            se = stack.Pop();

        }

        if (se != null)

        {

            retVal = seqNo - se.Sequence + 1;

            seq = se.Sequence;

        }

        stack.Push(new StackElement(price, seq));

        return retVal;

    }

    private Stack<StackElement> stack;

    private int seqNo;

}

Complexity: O(n)