Stock Span Problem Using Stack in DSA Rust - Deep Dive

Overview - Stock Span Problem Using Stack

What is it?

The Stock Span Problem is about finding how many consecutive days before today the stock price was less than or equal to today's price. We use a stack to efficiently track these days. This helps us quickly calculate the span for each day without checking all previous days one by one.

Why it matters

Without this method, calculating the stock span for each day would take a lot of time, especially for many days. Using a stack makes it fast and efficient, saving time and computing power. This is important in real-world stock analysis where quick decisions matter.

Where it fits

Before learning this, you should understand basic arrays and stacks. After this, you can explore other stack problems like Next Greater Element or advanced data structures for range queries.

Mental Model

Core Idea

Use a stack to remember days with higher stock prices to quickly find how far back the current price is greater or equal.

Think of it like...

Imagine a line of people holding balloons with numbers. Each person looks back to see how many people before them have balloons with smaller or equal numbers. The stack helps keep track of these people efficiently.

Day:    1   2   3   4   5   6   7
Price:  100 80  60  70  60  75  85
Span:   1   1   1   2   1   4   6

Stack stores indices of days with prices higher than current day to calculate spans.

Build-Up - 6 Steps

1

Foundation

Understanding Stock Span Concept

Concept: Learn what stock span means and how to calculate it by comparing prices day by day.

Stock span for a day is the count of consecutive days before it where the price was less or equal. For example, if prices are [100, 80, 60], spans are [1, 1, 1] because each day has no previous day with lower or equal price except itself.

Result

You understand that span counts how many days back the price stayed lower or equal.

Understanding the problem definition is key before optimizing with data structures.

2

Foundation

Introduction to Stack Data Structure

Concept: Learn what a stack is and how it works with push and pop operations.

A stack is like a pile of plates: you add (push) on top and remove (pop) from the top. It helps keep track of recent items in order.

Result

You can use a stack to remember recent days' prices or indices.

Knowing stack basics prepares you to use it for efficient span calculation.

3

Intermediate

Naive Span Calculation Approach

🤔Before reading on: Do you think checking all previous days for each day is efficient or slow? Commit to your answer.

Concept: Calculate span by checking all previous days one by one for each day.

For each day, look back at previous days until you find a day with price higher than current. Count how many days had price less or equal. This is simple but slow for many days.

Result

For prices [100, 80, 60, 70, 60, 75, 85], spans are [1,1,1,2,1,4,6] but calculation takes O(n^2) time.

This shows why a better method is needed for large data.

4

Intermediate

Using Stack to Optimize Span Calculation

🤔Before reading on: Do you think a stack can help avoid checking all previous days? Commit yes or no.

Concept: Use a stack to keep indices of days with prices higher than current to skip unnecessary checks.

We push day indices onto stack if their price is higher. For current day, pop stack while top price is less or equal. The span is difference between current day and top of stack after popping. Push current day index after.

Result

Span calculation becomes O(n) because each day is pushed and popped once.

Using stack cleverly reduces repeated comparisons and speeds up the process.

5

Advanced

Implementing Stock Span in Rust

🤔Before reading on: Can you predict how Rust's vector and stack operations will be used here? Commit your guess.

Concept: Write Rust code using Vec as stack to compute spans efficiently.

```rust fn calculate_span(prices: &[i32]) -> Vec { let mut stack: Vec = Vec::new(); let mut spans = vec![0; prices.len()]; for i in 0..prices.len() { while let Some(&top) = stack.last() { if prices[top] <= prices[i] { stack.pop(); } else { break; } } spans[i] = if let Some(&top) = stack.last() { i - top } else { i + 1 }; stack.push(i); } spans } fn main() { let prices = vec![100, 80, 60, 70, 60, 75, 85]; let spans = calculate_span(&prices); println!("{:?}", spans); } ```

Result

[1, 1, 1, 2, 1, 4, 6]

Seeing the full Rust implementation clarifies how stack operations map to span calculation.

6

Expert

Analyzing Time and Space Complexity

🤔Before reading on: Do you think each element is pushed and popped multiple times or just once? Commit your answer.

Concept: Understand why the stack approach runs in linear time and uses linear space.

Each price index is pushed once and popped at most once, so total operations are proportional to number of days. Space is used to store indices in stack and spans array, both O(n). This makes the solution efficient for large inputs.

Result

Time complexity: O(n), Space complexity: O(n)

Knowing complexity helps choose this method for performance-critical applications.

Under the Hood

The stack stores indices of days with strictly higher prices. When a new price comes, it pops all indices with prices less or equal, effectively skipping them. This way, the top of the stack always points to the nearest previous day with a higher price, allowing quick span calculation.

Why designed this way?

This design avoids repeated comparisons by remembering only relevant days. Alternatives like checking all previous days are slower. The stack approach balances simplicity and efficiency, making it suitable for real-time stock analysis.

Prices: 100  80  60  70  60  75  85
Stack:  [0] [0,1] [0,1,2] [0,1,3] [0,1,3,4] [0,1,5] [0,6]

At each step:
- Pop indices with price <= current
- Calculate span = current_index - top_of_stack
- Push current_index

Myth Busters - 3 Common Misconceptions

Quick: Does the stack store prices or indices? Commit your answer.

Common Belief: The stack stores the actual stock prices to compare.

Reality: The stack stores indices of days, not prices, to efficiently calculate spans using indices.

Why it matters: Storing prices would make it harder to calculate spans and increase complexity.

Quick: Is the time complexity of the stack solution O(n^2)? Commit yes or no.

Common Belief: Since there is a loop inside a loop, the time complexity is quadratic O(n^2).

Reality: Each element is pushed and popped at most once, so the time complexity is linear O(n).

Why it matters: Misunderstanding complexity can lead to rejecting an efficient solution.

Quick: Does the stack always contain all previous days? Commit your answer.

Common Belief: The stack keeps all previous days' indices at all times.

Reality: The stack only keeps indices of days with prices higher than the current day, popping others.

Why it matters: This selective storage is key to efficiency; misunderstanding it leads to wrong implementations.

Expert Zone

1

The stack approach can be adapted to find spans for other comparison criteria, like strictly less or strictly greater, by adjusting the comparison operator.

2

In real-time streaming data, the stack can be maintained incrementally without recomputing spans from scratch.

3

Rust's ownership and borrowing rules ensure safe stack manipulation without runtime errors, which is crucial for production reliability.

When NOT to use

If you need spans for non-consecutive days or complex queries like maximum spans in a range, segment trees or binary indexed trees are better. Also, if memory is extremely limited, the O(n) space of the stack might be too large.

Production Patterns

Used in financial software to quickly analyze stock trends, in trading algorithms to detect breakout points, and in interview coding tests to demonstrate stack mastery and optimization skills.

Connections

Next Greater Element Problem

Builds-on and similar pattern using stack to find next larger values.

Understanding stock span helps grasp next greater element, as both use stacks to skip unnecessary comparisons.

Monotonic Stack

Stock span is a classic example of monotonic stack usage where stack elements maintain order.

Knowing monotonic stacks generalizes the approach to many problems involving ordered sequences.

Real-time Event Processing

Stack-based span calculation resembles event windowing where recent events are tracked efficiently.

This connection shows how data structures help in streaming data analysis beyond stock prices.

Common Pitfalls

#1 Pushing prices instead of indices on the stack.

Wrong approach:

stack.push(prices[i]); // pushing price values

Correct approach:

stack.push(i); // pushing indices for span calculation

Root cause:

Confusing what data is needed to calculate spans efficiently.

#2 Not popping all smaller or equal prices before calculating span.

Wrong approach:

if prices[stack.last()] <= prices[i] { /* do nothing */ }

Correct approach:

while let Some(&top) = stack.last() { if prices[top] <= prices[i] { stack.pop(); } else { break; } }

Root cause:

Missing the loop to remove all irrelevant days leads to incorrect spans.

#3 Calculating span as current index minus stack top without handling empty stack.

Wrong approach:

span = i - stack.last().unwrap(); // unwrap without check

Correct approach:

span = if let Some(&top) = stack.last() { i - top } else { i + 1 };

Root cause:

Ignoring empty stack case causes runtime errors or wrong spans.

Key Takeaways

The stock span problem measures how many consecutive previous days had prices less or equal to today.

Using a stack to store indices of days with higher prices allows efficient O(n) span calculation.

Each price index is pushed and popped at most once, ensuring linear time complexity.

Rust's vector can be used as a stack with push and pop to implement this algorithm safely and efficiently.

Understanding this problem builds a foundation for many other stack-based sequence problems.

Practice

1. What does the Stock Span Problem calculate for each day in a list of stock prices?

easy

A. The number of consecutive days ending with the current day with a price less than or equal to the current day's price

B. The maximum stock price in the entire list

C. The minimum stock price in the entire list

D. The difference between the current day's price and the previous day's price

Solution

1. Step 1: Understand the problem definition

   The Stock Span Problem finds, for each day, the length of consecutive days ending on that day where prices are less than or equal to the current day's price.

2. Step 2: Clarify what is calculated per day

   It counts days backward until a higher price is found, including the current day.

3. Final Answer:

   The number of consecutive days ending with the current day with a price less than or equal to the current day's price -> Option A

4. Quick Check:

   Span counts consecutive lower/equal prices [OK]

Hint: Span counts days with price ≤ current day price [OK]

Common Mistakes:

• Confusing span with max or min price
• Counting all previous days instead of consecutive ones
• Calculating price differences instead of counts

2. Which Rust data structure is most suitable to efficiently solve the Stock Span Problem?

easy

A. LinkedList

B. Vector (Vec) used as a stack

C. HashMap

D. BinaryHeap

Solution

1. Step 1: Identify the need for last-in-first-out access

   The Stock Span Problem requires checking previous days' prices in reverse order, which fits stack behavior.

2. Step 2: Choose Rust's stack implementation

   Rust's Vec can be used as a stack with push and pop operations efficiently.

3. Final Answer:

   Vector (Vec) used as a stack -> Option B

4. Quick Check:

   Vec as stack fits LIFO for span calculation [OK]

Hint: Use Vec as stack for efficient push/pop [OK]

Common Mistakes:

• Using HashMap which is unordered
• Using LinkedList which is less efficient
• Using BinaryHeap which is for priority, not order

3. What is the output span array for the stock prices [100, 80, 60, 70, 60, 75, 85] using the Stock Span algorithm?

medium

A. [1, 1, 1, 2, 1, 4, 6]

B. [1, 2, 3, 4, 5, 6, 7]

C. [1, 1, 2, 2, 3, 3, 4]

D. [7, 6, 5, 4, 3, 2, 1]

Solution

1. Step 1: Calculate span for each price

   Day 0: 100 -> span 1 (no previous)
   Day 1: 80 < 100 -> span 1
   Day 2: 60 < 80 -> span 1
   Day 3: 70 > 60 -> span 2 (includes day 2)
   Day 4: 60 < 70 -> span 1
   Day 5: 75 > 60, 70 -> span 4 (includes days 4,3,2)
   Day 6: 85 > 75, 70, 60, 80 -> span 6 (includes days 5,4,3,2,1)

2. Step 2: Confirm final span array

   The spans are [1, 1, 1, 2, 1, 4, 6].

3. Final Answer:

   [1, 1, 1, 2, 1, 4, 6] -> Option A

4. Quick Check:

   Span array matches calculated values [OK]

Hint: Count consecutive previous days with price ≤ current [OK]

Common Mistakes:

• Counting all previous days instead of consecutive
• Mixing up indices when calculating spans
• Assuming spans always increase

4. Given this Rust snippet for Stock Span, what is the bug?

let mut stack: Vec<usize> = Vec::new();
let mut span = vec![0; prices.len()];
for i in 0..prices.len() {
    while !stack.is_empty() && prices[*stack.last().unwrap()] < prices[i] {
        stack.pop();
    }
    span[i] = if stack.is_empty() { i + 1 } else { i - stack.last().unwrap() };
    stack.push(i);
}

medium

A. The comparison should be <= instead of < in the while condition

B. The stack should store prices, not indices

C. The span calculation subtracts without dereferencing the last stack element

D. The span vector is initialized with zeros instead of ones

Solution

1. Step 1: Analyze span calculation line

   span[i] = if stack.is_empty() { i + 1 } else { i - stack.last().unwrap() } uses stack.last().unwrap() without dereferencing, so it subtracts usize - &usize causing a type error or wrong calculation.

2. Step 2: Correct dereferencing needed

   It should be i - *stack.last().unwrap() to get the usize value from reference.

3. Final Answer:

   The span calculation subtracts without dereferencing the last stack element -> Option C

4. Quick Check:

   Dereference stack.last() for correct subtraction [OK]

Hint: Dereference stack.last() before subtraction [OK]

Common Mistakes:

• Using < instead of <= (not critical here)
• Not dereferencing stack.last() causing errors
• Initializing span with zeros is correct here
• Stack must store indices, not prices

5. How would you modify the Stock Span algorithm to handle a stream of stock prices arriving one by one, efficiently updating the span each time?

hard

A. Store all prices in a vector and recalculate spans from scratch each time

B. Use a hash map to store price frequencies and compute span from frequencies

C. Use a queue to store prices and calculate span by counting elements

D. Use a stack to store pairs of (price, span) and update span by accumulating popped spans

Solution

1. Step 1: Understand streaming requirement

   For streaming prices, we must update span without reprocessing all previous data.

2. Step 2: Use stack with (price, span) pairs

   Store each price with its span. When a new price arrives, pop all smaller prices and add their spans to current span, then push new pair.

3. Step 3: Efficient update

   This approach allows O(1) amortized time per new price, updating span incrementally.

4. Final Answer:

   Use a stack to store pairs of (price, span) and update span by accumulating popped spans -> Option D

5. Quick Check:

   Stack with (price, span) pairs for streaming [OK]

Hint: Accumulate spans while popping smaller prices in stack [OK]

Common Mistakes:

• Recalculating spans from scratch each time (inefficient)
• Using queue which doesn't support needed access pattern
• Using hash map which doesn't track order