Stock Span Problem Using Stack in DSA Python - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the Stock Span Problem?

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

intermediate

Why do we use a stack to solve the Stock Span Problem?

A stack helps keep track of previous days with higher stock prices efficiently, allowing us to find spans in O(n) time by popping smaller prices until a higher price is found.

intermediate

What does each element in the stack represent in the Stock Span Problem?

Each element in the stack stores a pair: (index of the day, stock price on that day). This helps calculate the span by comparing prices and indices.

intermediate

How is the span for the current day calculated using the stack?

Span = current day index - index of the last higher price day found on the stack. If no higher price day exists, span = current day index + 1.

intermediate

What is the time complexity of the Stock Span Problem solution using a stack?

The time complexity is O(n) because each element is pushed and popped at most once from the stack.

What does the stack store in the Stock Span Problem?

A Pairs of (day index, stock price)

B Only stock prices

C Only day indices

D Pairs of (stock price, span)

The stack stores pairs of (day index, stock price) to help calculate spans efficiently.

What is the span for the first day in the stock span problem?

A Depends on price

B 0

C 1

D Same as second day

The first day always has a span of 1 because there are no previous days.

If the current day's price is less than the price at the top of the stack, what do we do?

A Pop from stack

B Push current day onto stack

C Ignore current day

D Clear the stack

We push the current day onto the stack because it is a new higher price boundary.

What is the time complexity of the stack-based solution for the Stock Span Problem?

A O(n)

B O(n log n)

C O(log n)

D O(n^2)

Each element is pushed and popped at most once, so the solution runs in O(n) time.

What happens when the stack is empty while calculating the span for a day?

A Span is zero

B Span is previous day's span

C Span is 1

D Span is current day index + 1

If the stack is empty, it means no previous higher price exists, so span = current day index + 1.

Explain how a stack helps solve the Stock Span Problem efficiently.

Think about how you find the last day with a higher price.

Describe the step-by-step process to calculate the span for each day using a stack.

Imagine you are checking previous days one by one but faster.

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