Why Stock Span Problem Using Stack in DSA Rust?

The Big Idea

Discover how a simple stack can turn a slow, tiring task into a fast, smart solution for stock analysis!

The Scenario

Imagine you are tracking daily stock prices on paper. For each day, you want to know how many consecutive previous days had prices less than or equal to today's price.

Doing this manually means checking every previous day one by one, which takes a lot of time as days go by.

The Problem

Manually checking each previous day's price for every new day is slow and tiring. As the number of days grows, it becomes harder to keep track without mistakes.

You might lose count or forget some days, making your results wrong.

The Solution

The Stock Span Problem uses a stack to remember important days' prices efficiently.

Instead of checking all previous days, the stack helps jump over days that are not needed, making the process fast and easy.

Before vs After

✗ Before

for i in 0..n {
    span[i] = 1;
    for j in (0..i).rev() {
        if prices[j] <= prices[i] {
            span[i] += 1;
        } else {
            break;
        }
    }
}

✓ After

let mut stack = Vec::new();
for i in 0..n {
    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);
}

What It Enables

This method lets you quickly find how many days in a row the stock price was lower or equal, helping you analyze trends instantly.

Real Life Example

Traders use this to understand how strong a stock's price is compared to recent days, helping them decide when to buy or sell.

Key Takeaways

Manual checking is slow and error-prone for large data.

Stack helps skip unnecessary checks, speeding up the process.

Stock span shows how many days prices stayed below or equal to today's price.

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