What is the brute force approach and complexity for Online Stock Span?

Store all prices; each call counts backwards while prices are ≤ today with time complexity O(N) and space complexity O(N).

What is the optimal approach and complexity for Online Stock Span?

Each stack entry stands for a whole run of days; popped spans fold into the new one with optimal time complexity O(1) and space complexity O(N).

How do you identify the pattern for Online Stock Span?

Look for key signals in the prompt: "one day at a time" indicates An online stream: you can't look ahead, and you don't want to store and rescan the whole history.; "days in a row … less than or equal to today" indicates Same boundary as the array version: stop at the previous strictly greater price.; "return the span on each call" indicates Each call must be fast on average → monotonic stack whose entries carry their span..

Online Stock Span

Medium·LeetCode #901 Pro OnlyBasic Reinforcement

LeetCode

Problem Statement

Build a class StockSpanner that receives a stock's price one day at a time.

Each call next(price) gives you today's price and must return today's span: the number of days in a row, ending today and going backwards, on which the price was less than or equal to today's price. Today itself counts, so the span is at least 1.

You do not get the prices in advance — you must answer each call before you see the next price.

Input

A sequence of calls: StockSpanner() once, then next(price) many times.

Output

For each next(price) call, an integer: today's span.

Example 1

Input: StockSpanner(), next(100), next(80), next(60), next(70), next(60), next(75), next(85)

Output: null, 1, 1, 1, 2, 1, 4, 6

For 75: going back, 60, 70, 60 are all ≤ 75, and 80 is higher, so the span is 4. For 85: everything back to 100 is ≤ 85, so the span is 6.

Constraints

• 1 ≤ price ≤ 10^5
• At most 10^4 calls to `next`

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“one day at a time”

→ An online stream: you can't look ahead, and you don't want to store and rescan the whole history.

“days in a row … less than or equal to today”

→ Same boundary as the array version: stop at the previous strictly greater price.

“return the span on each call”

→ Each call must be fast on average → monotonic stack whose entries carry their span.

Approach Complexity Ladder

Brute Force — keep every price, walk back each callO(N)O(N)

Store all prices; each call counts backwards while prices are ≤ today

Optimal — stack of (price, span) pairsO(1)O(N)

Each stack entry stands for a whole run of days; popped spans fold into the new one

Brute Force Intuition

Keep every price in a vector. On each next(price), append it, then walk back from the second-last price while prices are ≤ price, counting as you go.

Recognition rule for this sub-pattern:

"How many consecutive previous items are ≤ me?" Popped items fold their span into the new one.

Remember this · Monotonic Stack

Trigger

Next/previous greater or smaller element, spans, histogram areas, "for each element, how far does it dominate?"

Think

When a new element arrives, which older elements can never be the answer again?

Key question

Is my stack increasing or decreasing, and what do I record when I pop?

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