Stock Span Problem Using Stack in DSA Rust - Build from Scratch

Stock Span Problem Using Stack

📖 Scenario: You are working as a stock market assistant. You want to help traders understand how many consecutive days before today the stock price was less than or equal to today's price. This helps traders see the strength of the current price.

🎯 Goal: Build a Rust program that calculates the stock span for each day using a stack. The stock span for a day is the number of consecutive days before it with a price less than or equal to that day's price.

📋 What You'll Learn

Create a vector called prices with the exact values: [100, 80, 60, 70, 60, 75, 85]

Create a mutable stack using a vector called stack to store indices

Create a vector called span of the same length as prices initialized with zeros

Use a for loop with variable i to iterate over the indices of prices

Inside the loop, use a while loop to pop from stack while the current price is greater or equal to the price at the top index of stack

Calculate the span for each day as i + 1 if stack is empty, otherwise as i - *stack.last().unwrap()

Push the current index i onto stack

Print the span vector at the end

💡 Why This Matters

🌍 Real World

Stock traders use the stock span to understand price trends and make decisions about buying or selling stocks.

💼 Career

This problem teaches stack usage and algorithmic thinking, important for software roles involving data processing and financial applications.

1

Create the price list

Create a vector called prices with these exact values: [100, 80, 60, 70, 60, 75, 85]

DSA Rust

# Create the prices vector with given values
# Your code here

Hint

Use let prices = vec![100, 80, 60, 70, 60, 75, 85]; to create the vector.

2

Setup stack and span vector

Create a mutable vector called stack to store usize indices and a mutable vector called span with the same length as prices initialized with zeros.

DSA Rust

fn main() {
    let prices = vec![100, 80, 60, 70, 60, 75, 85];
    // Create a mutable stack vector to store indices
    // Create a mutable span vector with zeros of length prices.len()
    // Your code here
}

Hint

Use Vec::new() for stack and vec![0usize; prices.len()] for span.

3

Calculate the stock span using stack

Use a for loop with variable i to iterate over the indices of prices. Inside the loop, use a while loop to pop from stack while prices[i] is greater or equal to prices[*stack.last().unwrap()]. Then calculate span[i] as i + 1 if stack is empty, otherwise as i - *stack.last().unwrap(). Finally, push i onto stack.

DSA Rust

fn main() {
    let prices = vec![100, 80, 60, 70, 60, 75, 85];
    let mut stack: Vec = Vec::new();
    let mut span = vec![0usize; prices.len()];

    // Loop over prices indices
    // Use while loop to pop from stack while current price >= price at stack top
    // Calculate span[i]
    // Push i onto stack
    // Your code here
}

Hint

Use for i in 0..prices.len() and inside use while !stack.is_empty() && prices[i] >= prices[*stack.last().unwrap()] to pop. Calculate span with if stack.is_empty() condition.

4

Print the stock span

Print the span vector using println!("{:?}", span) to display the stock span for each day.

DSA Rust

fn main() {
    let prices = vec![100, 80, 60, 70, 60, 75, 85];
    let mut stack: Vec = Vec::new();
    let mut span = vec![0usize; prices.len()];

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

    // Print the span vector
    // Your code here
}

Hint

Use println!("{:?}", span) to print the vector.

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