Stock Span Problem Using Stack in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to push the current index onto the stack.

DSA Java

stack.[1](i);

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

Apop

Bpush

Cpeek

Dempty

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop instead of push causes removal instead of addition.

Using peek does not add anything to the stack.

We use push() to add the current index i onto the stack.

2fill in blank

medium

Complete the while loop condition to pop indices with stock price less or equal to current price.

DSA Java

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

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

A<=

B>

C<

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using < misses equal prices.

Using > or >= reverses the logic.

We pop while the price at stack top is less than or equal to current price.

3fill in blank

hard

Fix the error in calculating span when stack is empty.

DSA Java

span[i] = stack.isEmpty() ? [1] : i - stack.peek();

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

Aprices[i]

Bi

C0

Di + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 results in zero span incorrectly.

Using i alone misses the +1 needed.

If stack is empty, span is i+1, but since i starts at 0, span is i+1 which equals i+1, so span[i] = i+1. Here, 1 is the base span for first day.

4fill in blank

hard

Fill both blanks to complete the for loop and initialize the stack.

DSA Java

Stack<Integer> stack = new Stack<>();
for (int [1] = 0; [2] < prices.length; [1]++) {
    // loop body
}

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

Ai

Bj

Ck

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names in initialization and condition causes errors.

Using j or k is valid but inconsistent here.

The loop variable is commonly named i and used consistently in initialization and condition.

5fill in blank

hard

Fill all three blanks to complete the span calculation and stack update.

DSA Java

span[i] = stack.isEmpty() ? [1] : i - stack.peek();
stack.[2](i);

return [3];

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

Ai + 1

Bpush

Cspan

Dpop

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop instead of push removes elements incorrectly.

Returning wrong variable causes errors.

Span is i+1 if stack empty, push current index i, and return the span array.

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 before and including 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 difference between the highest and lowest stock price

D. The total sum of stock prices up to the current day

Solution

1. Step 1: Understand the problem definition

   The Stock Span Problem finds, for each day, how many consecutive days before and including the current day had stock prices less than or equal to the current day's price.

2. Step 2: Clarify what is being counted

   It counts days backward until a higher price is found or the start is reached.

3. Final Answer:

   The number of consecutive days before and including 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 consecutive days with prices ≤ current day price including current day [OK]

Common Mistakes:

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

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

easy

A. Queue

B. ArrayList

C. HashMap

D. Stack

Solution

1. Step 1: Identify the need for tracking previous prices

   The problem requires checking previous days' prices to find the span efficiently.

2. Step 2: Choose a data structure that supports last-in-first-out access

   A stack allows pushing and popping prices or indices to quickly find the last higher price.

3. Final Answer:

   Stack -> Option D

4. Quick Check:

   Stack supports efficient previous day price checks [OK]

Hint: Use stack to track previous prices efficiently [OK]

Common Mistakes:

• Using ArrayList which is inefficient for this problem
• Choosing HashMap which doesn't maintain order
• Using Queue which is FIFO, not suitable here

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. [1, 1, 1, 1, 1, 1, 1]

Solution

1. Step 1: Calculate span for each day

   Day 0: 100, no previous higher price, span=1
   Day 1: 80 ≤ 100? Yes, but previous higher price is 100, so span=1
   Day 2: 60 ≤ 80? Yes, but previous higher price is 80, so span=1
   Day 3: 70 > 60, span=2 (days 2 and 3)
   Day 4: 60 ≤ 70? Yes, but previous higher price is 70, so span=1
   Day 5: 75 > 60 and 70, span=4 (days 2,3,4,5)
   Day 6: 85 > 75, 70, 60, span=6 (days 1 to 6)

2. Step 2: Confirm final span array

   The span array is [1, 1, 1, 2, 1, 4, 6].

3. Final Answer:

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

4. Quick Check:

   Span matches consecutive lower/equal days [OK]

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

Common Mistakes:

• Counting all previous days instead of consecutive
• Misapplying stack logic causing wrong spans
• Confusing span with price values

4. Identify the error in this Java snippet for Stock Span calculation:

for (int i = 0; i < prices.length; i++) {
  while (!stack.isEmpty() && prices[stack.peek()] < prices[i]) {
    stack.pop();
  }
  span[i] = stack.isEmpty() ? i + 1 : i - stack.peek();
  stack.push(i);
}

medium

A. The stack should store prices, not indices

B. The span calculation should use i + stack.peek() instead of i - stack.peek()

C. The comparison operator should be <= instead of <

D. The loop should start from i = 1 instead of 0

Solution

1. Step 1: Analyze the while condition for popping stack

   The condition uses < to pop indices with prices less than current price, but it should pop prices less than or equal to current price to correctly count span.

2. Step 2: Confirm correct comparison operator

   Using <= ensures all previous days with price ≤ current are popped, giving correct span.

3. Final Answer:

   The comparison operator should be <= instead of < -> Option C

4. Quick Check:

   Use <= to pop all lower or equal prices [OK]

Hint: Use <= in while to pop all lower or equal prices [OK]

Common Mistakes:

• Using < instead of <= causing wrong spans
• Storing prices instead of indices in stack
• Starting loop at wrong index

5. Given stock prices [10, 4, 5, 90, 120, 80], what is the span array if the stack is implemented to store prices instead of indices?

hard

A. [1, 1, 2, 4, 5, 1]

B. [1, 1, 1, 1, 1, 1]

C. [1, 2, 3, 4, 5, 6]

D. [1, 1, 2, 1, 2, 1]

Solution

1. Step 1: Understand stack storing prices instead of indices

   Without indices, we cannot calculate span as difference of indices; span calculation fails.

2. Step 2: Resulting span values

   Each day is treated as isolated, so span defaults to 1 for all days.

3. Final Answer:

   [1, 1, 1, 1, 1, 1] -> Option B

4. Quick Check:

   Stack must store indices for correct span [OK]

Hint: Stack must store indices, not prices, for correct span [OK]

Common Mistakes:

• Storing prices causes incorrect span calculation
• Assuming span grows without index info
• Ignoring index difference in span formula