Stock Span Problem Using Stack in DSA Python - Time & Space Complexity

Time Complexity: Stock Span Problem Using Stack

O(n)

Understanding Time Complexity

We want to understand how the time taken grows when we calculate the stock span for many days.

How does the number of operations change as the list of stock prices gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

def calculate_span(prices):
    stack = []  # stores indices
    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
    

This code calculates how many consecutive days before today the stock price was less or equal.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The for-loop runs once for each price.
• Inside loop: The while-loop pops from the stack, but each element is pushed and popped at most once.
• Dominant operation: The combined push and pop operations on the stack.

How Execution Grows With Input

Each price is processed once, and each index is pushed and popped at most once.

Input Size (n)	Approx. Operations
10	About 20 (10 pushes + up to 10 pops)
100	About 200 (100 pushes + up to 100 pops)
1000	About 2000 (1000 pushes + up to 1000 pops)

Pattern observation: Operations grow roughly in a straight line with input size.

Final Time Complexity

Time Complexity: O(n)

This means the time taken grows directly in proportion to the number of days.

Common Mistake

[X] Wrong: "The while-loop inside the for-loop makes this O(n²) because it can run many times each iteration."

[OK] Correct: Each element is pushed and popped only once, so total while-loop operations across all iterations add up to n, not n².

Interview Connect

Understanding this problem shows you can use stacks smartly to reduce repeated work, a skill useful in many coding challenges.

Self-Check

"What if we used a simple nested loop without a stack? How would the time complexity change?"

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