Stock Span Problem Using Stack in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize an empty stack for the stock span problem.

DSA Python

stack = [1]

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

A[]

B()

C{}

Dset()

Attempts:

3 left

💡 Hint

Common Mistakes

Using {} which creates a dictionary, not a stack.

Using () which creates a tuple, which is immutable.

Using set() which is unordered and not suitable for stack.

We use a list as a stack in Python, so initializing with [] creates an empty stack.

2fill in blank

medium

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

DSA Python

stack.[1](i)

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

Ainsert

Bpop

Cremove

Dappend

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop() which removes an element instead of adding.

Using remove() which removes a specific value.

Using insert() which inserts at a specific position.

To add an element to the top of the stack (list), we use append().

3fill in blank

hard

Fix the error in the while loop condition to check if stack is not empty.

DSA Python

while stack and prices[stack[-1]] <= prices[[1]]:

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

Ai

Bprices

Cstack

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Using stack instead of i, which is incorrect index.

Using prices which is the list, not an index.

Using -1 which is used to access last element of stack, not current index.

We compare the price at the top index of the stack with the current price at index i.

4fill in blank

hard

Fill both blanks to calculate the span for the current day.

DSA Python

span[i] = i - [1] if stack else [2]

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

Astack[-1]

B0

C-1

Dlen(stack)

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 instead of -1 for empty stack case.

Using len(stack) which is unrelated to index.

Using stack instead of stack[-1].

If stack is not empty, span is difference between current index and top of stack index; else span is i - (-1) which is i + 1.

5fill in blank

hard

Fill all three blanks to complete the stock span function.

DSA Python

def calculate_span(prices):
    n = len(prices)
    span = [0] * n
    stack = []
    for [1] in range(n):
        while stack and prices[stack[-1]] <= prices[[2]]:
            stack.pop()
        span[[3]] = [2] - stack[-1] if stack else [2] + 1
        stack.append([2])
    return span

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

Ai

Bj

Ck

Dm

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names causing index errors.

Using j or k inconsistently.

The loop variable and all index references should be consistent as i for correct indexing.

Practice

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

easy

A. The maximum stock price in the entire list

B. The number of consecutive days before the current day with price less than or equal to current day's price

C. The difference between current day's price and previous day's price

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

Solution

1. Step 1: Understand the problem goal

   The Stock Span Problem finds how many days before today had prices less than or equal to today's price consecutively.

2. Step 2: Clarify what is counted

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

3. Final Answer:

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

4. Quick Check:

   Span counts consecutive lower/equal prices [OK]

Hint: Span counts consecutive days with price ≤ today's price [OK]

Common Mistakes:

• Thinking it counts all previous days regardless of price
• Confusing span with price difference
• Assuming it finds max price instead

2. Which of the following Python code snippets correctly initializes an empty stack for solving the Stock Span Problem?

easy

A. stack = None

B. stack = {}

C. stack = []

D. stack = ()

Solution

1. Step 1: Identify correct stack data structure

   In Python, a stack is commonly implemented using a list, so an empty stack is an empty list: stack = []

2. Step 2: Check other options

   {} is a dictionary, () is a tuple (immutable), None is no value, so they are incorrect for stack.

3. Final Answer:

   stack = [] -> Option C

4. Quick Check:

   Empty list is stack [OK]

Hint: Use empty list [] to start a stack in Python [OK]

Common Mistakes:

• Using {} which is a dictionary, not stack
• Using () which is immutable tuple
• Assigning None instead of a list

3. What is the output of the following code snippet for prices = [100, 80, 60, 70, 60, 75, 85]?

def calculate_span(prices):
    stack = []
    span = [0]*len(prices)
    for i in range(len(prices)):
        while stack and prices[stack[-1]] <= prices[i]:
            stack.pop()
        span[i] = i + 1 if not stack else i - stack[-1]
        stack.append(i)
    return span

print(calculate_span(prices))

medium

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

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

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

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

Solution

1. Step 1: Trace the loop and stack operations

   For each price, pop indices with prices ≤ current price, then calculate span as distance from last higher price index or i+1 if none.

2. Step 2: Calculate spans for each day

   Day 0: stack empty, span=1; Day 1: price 80 ≤ 100? No pop, span=1; Day 2: 60 ≤ 80? No pop, span=1; Day 3: 70 > 60 pop index 2, span=2; Day 4: 60 ≤ 70 no pop, span=1; Day 5: 75 > 60 pop index 4, 75 > 70 pop index 3, span=4; Day 6: 85 > 75 pop index 5, 85 > 100 no pop, span=6.

3. Final Answer:

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

4. Quick Check:

   Span array matches expected output [OK]

Hint: Pop smaller prices, span = current index - last higher index [OK]

Common Mistakes:

• Not popping all smaller or equal prices
• Calculating span as i instead of difference
• Confusing span with price values

4. Identify the error in this Stock Span code snippet:

def stock_span(prices):
    stack = []
    span = []
    for i in range(len(prices)):
        while stack and prices[stack[-1]] < prices[i]:
            stack.pop()
        if not stack:
            span.append(i)
        else:
            span.append(i - stack[-1])
        stack.append(i)
    return span

print(stock_span([10, 4, 5, 90, 120, 80]))

medium

A. Stack is not initialized correctly

B. The function returns None instead of span list

C. Comparison operator should be <= instead of <

D. Span for first day should be 1, but code appends 0

Solution

1. Step 1: Check span calculation for first day

   When stack is empty, span should be i+1 (1 for first day), but code appends i (0 for first day), causing wrong span.

2. Step 2: Confirm other parts are correct

   Stack initialization and popping logic are fine; comparison operator < is acceptable; function returns span list correctly.

3. Final Answer:

   Span for first day should be 1, but code appends 0 -> Option D

4. Quick Check:

   Span for day 0 must be 1 [OK]

Hint: Span on empty stack = i+1, not i [OK]

Common Mistakes:

• Appending i instead of i+1 when stack empty
• Using wrong comparison operator causing wrong pops
• Not initializing span list properly

5. Given a list of stock prices, how would you modify the Stock Span algorithm to handle the case where prices can be equal and you want the span to include days with strictly less price only (exclude equal prices)?

hard

A. Change the comparison in while loop to prices[stack[-1]] < prices[i]

B. Change the comparison in while loop to prices[stack[-1]] <= prices[i]

C. Add an extra condition to skip equal prices after popping

D. No change needed; current algorithm already excludes equal prices

Solution

1. Step 1: Understand current comparison

   Current algorithm uses <= to pop indices with prices less than or equal to current price, including equal prices in span.

2. Step 2: Modify to exclude equal prices

   To exclude equal prices, change comparison to strictly less than (<) so equal prices stop popping and are not included in span.

3. Final Answer:

   Change the comparison in while loop to prices[stack[-1]] < prices[i] -> Option A

4. Quick Check:

   Use < to exclude equal prices [OK]

Hint: Use < instead of <= to exclude equal prices in span [OK]

Common Mistakes:

• Using <= keeps equal prices included
• Adding extra conditions unnecessarily
• Assuming no change is needed