Stock Span Problem Using Stack in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a new empty stack for indices.

DSA Rust

let mut stack: Vec<usize> = Vec::new();
stack.[1](0);

Drag options to blanks, or click blank then click option'

Apush

Bremove

Cinsert

Dpop

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop instead of push will remove an element instead of adding.

Using insert or remove requires an index and is not typical for stack push.

We use push to add an element to the top of the stack.

2fill in blank

medium

Complete the code to pop elements from the stack while the current price is higher than the price at the top index.

DSA Rust

while !stack.is_empty() && prices[i] > prices[stack.[1]()] {
    stack.pop();
}

Drag options to blanks, or click blank then click option'

Afirst

Blen

Clast

Dpop

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop() removes the element, which is done later, not for comparison.

Using len() or first() does not give the top element.

last() returns the last element of the vector without removing it, which is needed to compare prices.

3fill in blank

hard

Fix the error in calculating the span when the stack is empty.

DSA Rust

let span = if stack.is_empty() {
    [1]
} else {
    i - stack.last().unwrap()[2]
};

Drag options to blanks, or click blank then click option'

Ai + 1

Bi - 1

C+ 1

D- 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using i - 1 when stack is empty gives wrong span.

Adding 1 instead of subtracting 1 in else case causes off-by-one error.

If the stack is empty, the span is i + 1. Otherwise, subtract the top index and subtract 1 to get the span.

4fill in blank

hard

Fill both blanks to correctly push the current index and update the span vector.

DSA Rust

spans.push(i - stack.last().unwrap()[1]);
stack.[2](i);

Drag options to blanks, or click blank then click option'

A- 1

B+ 1

Cpush

Dpop

Attempts:

3 left

💡 Hint

Common Mistakes

Using + 1 causes incorrect span calculation.

Using pop instead of push removes elements incorrectly.

We subtract 1 to get the correct span length and push the current index onto the stack.

5fill in blank

hard

Fill all three blanks to complete the main loop for calculating stock spans.

DSA Rust

for i in 0..prices.len() {
    while !stack.is_empty() && prices[i] > prices[stack.[1]()] {
        stack.[2]();
    }
    let span = if stack.is_empty() { i [3] 1 } else { i - stack.last().unwrap() - 1 };
    spans.push(span);
    stack.push(i);
}

Drag options to blanks, or click blank then click option'

Alast

Bpop

C+

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop() instead of last() for peeking causes errors.

Subtracting 1 instead of adding 1 when stack is empty causes off-by-one error.

We use last() to peek, pop() to remove, and add 1 to i when stack is empty for span calculation.

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