Stock Span Problem Using Stack in DSA Python - Execution Trace

Concept Flow - Stock Span Problem Using Stack

Start with empty stack

↓

For each day's price

↓

While stack not empty and top price <= current price

↓

Pop stack top

↓

Calculate span: current index - top of stack index

↓

Push current day index onto stack

↓

Repeat for all days

↓

Return spans array

We use a stack to keep track of indices of days with higher prices. For each day, we pop days with lower or equal prices, then calculate the span as the difference between current day and last higher price day.

Execution Sample

DSA Python

prices = [100, 80, 60, 70, 60, 75, 85]
stack = []
spans = []
for i, price in enumerate(prices):
    while stack and prices[stack[-1]] <= price:
        stack.pop()
    span = i + 1 if not stack else i - stack[-1]
    spans.append(span)
    stack.append(i)
print(spans)

Calculates the stock span for each day using a stack to track previous higher prices.

Execution Table

Step	Operation	Current Day (i)	Current Price	Stack (Indices)	Span Calculated	Visual State (Prices with Spans)
1	Start	-	-	[]	-	[]
2	Process day 0	0	100	[] -> [0]	1	[100(1)]
3	Process day 1	1	80	[0]	1	[100(1), 80(1)]
4	Process day 2	2	60	[0,1]	1	[100(1), 80(1), 60(1)]
5	Process day 3	3	70	[0,1,2] -> pop 2	2	[100(1), 80(1), 60(1), 70(2)]
6	Process day 4	4	60	[0,1,3]	1	[100(1), 80(1), 60(1), 70(2), 60(1)]
7	Process day 5	5	75	[0,1,3,4] -> pop 4 -> pop 3	4	[100(1), 80(1), 60(1), 70(2), 60(1), 75(4)]
8	Process day 6	6	85	[0,1,5] -> pop 5 -> pop 1	6	[100(1), 80(1), 60(1), 70(2), 60(1), 75(4), 85(6)]
9	End	-	-	[0,6]	-	[100(1), 80(1), 60(1), 70(2), 60(1), 75(4), 85(6)]

💡 All days processed, spans calculated for each day.

Variable Tracker

Variable	Start	After Day 0	After Day 1	After Day 2	After Day 3	After Day 4	After Day 5	After Day 6	Final
stack	[]	[0]	[0,1]	[0,1,2]	[0,1,3]	[0,1,3,4]	[0,1,5]	[0,6]	[0,6]
spans	[]	[1]	[1,1]	[1,1,1]	[1,1,1,2]	[1,1,1,2,1]	[1,1,1,2,1,4]	[1,1,1,2,1,4,6]	[1,1,1,2,1,4,6]

Key Moments - 3 Insights

Why do we pop elements from the stack when the current price is higher or equal?

Because those days have prices less or equal to current, so their span is included in current day's span. See execution_table rows 5, 7, and 8 where popping happens before calculating span.

Why do we use 'i + 1' as span when stack is empty?

If stack is empty, it means no previous day has higher price, so span is all days from day 0 to current day (i+1). Check execution_table row 2 for day 0.

What does the stack store and why indices instead of prices?

Stack stores indices to calculate span easily by subtracting indices. Prices alone don't give position info. This is shown in variable_tracker where stack holds indices.

Visual Quiz - 3 Questions

Look at the execution_table at step 5. What is the stack after popping?

A [0,3]

B [0,1,2]

C [0,1,3]

D [1,3]

💡 Hint

Check the 'Stack (Indices)' column at step 5 in execution_table.

At which step does the span for day 5 become 4?

A Step 6

B Step 7

C Step 8

D Step 5

💡 Hint

Look at 'Span Calculated' column for day 5 in execution_table.

If the price on day 3 was 50 instead of 70, how would the span at step 5 change?

A Span would be 1

B Span would be 2

C Span would be 3

D Span would be 4

💡 Hint

Consider how popping depends on current price compared to stack top prices in execution_table.

Concept Snapshot

Stock Span Problem Using Stack:
- Use a stack to store indices of days with higher prices.
- For each day, pop stack while top price <= current price.
- Span = current index - top of stack index (or i+1 if stack empty).
- Push current day index onto stack.
- Result: spans array showing consecutive days with price <= current day.

Full Transcript

The Stock Span Problem calculates how many consecutive days before the current day had stock prices less than or equal to today's price. We use a stack to keep track of indices of days with higher prices. For each day, we pop from the stack all days with prices less or equal to the current price. If the stack is empty, it means no previous higher price exists, so the span is the current day index plus one. Otherwise, the span is the difference between the current day index and the index on top of the stack. We then push the current day index onto the stack. This process continues for all days, resulting in an array of spans. The execution table shows each step, the stack state, and the spans calculated. Key moments clarify why popping happens, why we use indices, and how spans are computed. The visual quiz tests understanding of stack changes and span calculations.

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