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

Stock Span Problem Using Stack

📖 Scenario: You are working as a stock analyst. You want to find out how many consecutive days before today the stock price was less than or equal to today's price. This helps in understanding the stock's performance trend.

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

📋 What You'll Learn

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

Create an empty list called span to store the span values

Use a stack implemented as a list called stack to help calculate spans

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 top of the stack points to a price less than or equal to prices[i]

Calculate the span for each day and append it to span

Print the span list at the end

💡 Why This Matters

🌍 Real World

Stock analysts use the stock span problem to understand how long a stock price has been rising or stable, helping in making investment decisions.

💼 Career

This problem teaches stack usage, which is important in many software engineering tasks like parsing, undo operations, and managing function calls.

1

Create the list of stock prices

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

DSA Python

# Create the list of stock prices
# Your code here

Hint

Use square brackets to create a list and separate values with commas.

2

Create the span list and stack

Create an empty list called span and an empty list called stack to use as a stack

DSA Python

prices = [100, 80, 60, 70, 60, 75, 85]
# Create empty lists called span and stack
# Your code here

Hint

Use [] to create empty lists.

3

Calculate the stock span using a 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 stack is not empty and prices[stack[-1]] <= prices[i]. Then calculate the span for day i as i + 1 if stack is empty, else i - stack[-1]. Append the span to span and push i onto stack.

DSA Python

prices = [100, 80, 60, 70, 60, 75, 85]
span = []
stack = []
# Calculate the stock span using a stack
# Your code here

Hint

Use range(len(prices)) to loop over indices. Use stack[-1] to get the top of the stack.

4

Print the span list

Print the span list to display the stock span for each day

DSA Python

prices = [100, 80, 60, 70, 60, 75, 85]
span = []
stack = []
for i in range(len(prices)):
    while stack and prices[stack[-1]] <= prices[i]:
        stack.pop()
    if not stack:
        span.append(i + 1)
    else:
        span.append(i - stack[-1])
    stack.append(i)
# Print the span list
# Your code here

Hint

Use print(span) to show the result.

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