Why Stock Span Problem Using Stack in DSA Java?

The Big Idea

Discover how a simple stack can turn a slow, repetitive task into a fast, smart solution for stock analysis!

The Scenario

Imagine you are tracking daily stock prices on paper. For each day, you want to know how many consecutive previous days had prices less than or equal to today's price.

Doing this by checking every previous day manually for each new price is tiring and slow.

The Problem

Manually comparing each day's price with all previous days means repeating many checks again and again.

This takes a lot of time and can easily cause mistakes, especially with many days of data.

The Solution

The Stock Span Problem uses a stack to remember important previous days' prices efficiently.

This way, you only check necessary days once, making the process fast and simple.

Before vs After

✗ Before

for (int i = 0; i < prices.length; i++) {
  int span = 1;
  for (int j = i - 1; j >= 0 && prices[j] <= prices[i]; j--) {
    span++;
  }
  spans[i] = span;
}

✓ After

Stack<Integer> stack = new Stack<>();
for (int i = 0; i < prices.length; i++) {
  while (!stack.isEmpty() && prices[stack.peek()] <= prices[i]) {
    stack.pop();
  }
  spans[i] = stack.isEmpty() ? i + 1 : i - stack.peek();
  stack.push(i);
}

What It Enables

This method lets you quickly find how many days in a row the stock price was lower or equal, helping with fast stock trend analysis.

Real Life Example

Traders use this to quickly understand how strong a stock's price rise is by seeing how many days it has been increasing or stable.

Key Takeaways

Manual checking repeats many comparisons and is slow.

Using a stack remembers key days and skips unnecessary checks.

Stack-based solution is efficient and easy to implement.

Practice

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

easy

A. The number of consecutive days before and including the current day with a price less than or equal to the current day's price

B. The maximum stock price in the entire list

C. The difference between the highest and lowest stock price

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

Solution

1. Step 1: Understand the problem definition

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

2. Step 2: Clarify what is being counted

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

3. Final Answer:

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

4. Quick Check:

   Span counts consecutive lower/equal prices [OK]

Hint: Span counts consecutive days with prices ≤ current day price including current day [OK]

Common Mistakes:

• Confusing span with maximum price
• Counting all previous days instead of consecutive ones
• Calculating price differences instead of counts

2. Which Java data structure is most suitable to efficiently solve the Stock Span Problem?

easy

A. Queue

B. ArrayList

C. HashMap

D. Stack

Solution

1. Step 1: Identify the need for tracking previous prices

   The problem requires checking previous days' prices to find the span efficiently.

2. Step 2: Choose a data structure that supports last-in-first-out access

   A stack allows pushing and popping prices or indices to quickly find the last higher price.

3. Final Answer:

   Stack -> Option D

4. Quick Check:

   Stack supports efficient previous day price checks [OK]

Hint: Use stack to track previous prices efficiently [OK]

Common Mistakes:

• Using ArrayList which is inefficient for this problem
• Choosing HashMap which doesn't maintain order
• Using Queue which is FIFO, not suitable here

3. What is the output span array for the stock prices [100, 80, 60, 70, 60, 75, 85] using the Stock Span algorithm?

medium

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

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

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

D. [1, 1, 1, 1, 1, 1, 1]

Solution

1. Step 1: Calculate span for each day

   Day 0: 100, no previous higher price, span=1
   Day 1: 80 ≤ 100? Yes, but previous higher price is 100, so span=1
   Day 2: 60 ≤ 80? Yes, but previous higher price is 80, so span=1
   Day 3: 70 > 60, span=2 (days 2 and 3)
   Day 4: 60 ≤ 70? Yes, but previous higher price is 70, so span=1
   Day 5: 75 > 60 and 70, span=4 (days 2,3,4,5)
   Day 6: 85 > 75, 70, 60, span=6 (days 1 to 6)

2. Step 2: Confirm final span array

   The span array is [1, 1, 1, 2, 1, 4, 6].

3. Final Answer:

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

4. Quick Check:

   Span matches consecutive lower/equal days [OK]

Hint: Count consecutive days with price ≤ current day [OK]

Common Mistakes:

• Counting all previous days instead of consecutive
• Misapplying stack logic causing wrong spans
• Confusing span with price values

4. Identify the error in this Java snippet for Stock Span calculation:

for (int i = 0; i < prices.length; i++) {
  while (!stack.isEmpty() && prices[stack.peek()] < prices[i]) {
    stack.pop();
  }
  span[i] = stack.isEmpty() ? i + 1 : i - stack.peek();
  stack.push(i);
}

medium

A. The stack should store prices, not indices

B. The span calculation should use i + stack.peek() instead of i - stack.peek()

C. The comparison operator should be <= instead of <

D. The loop should start from i = 1 instead of 0

Solution

1. Step 1: Analyze the while condition for popping stack

   The condition uses < to pop indices with prices less than current price, but it should pop prices less than or equal to current price to correctly count span.

2. Step 2: Confirm correct comparison operator

   Using <= ensures all previous days with price ≤ current are popped, giving correct span.

3. Final Answer:

   The comparison operator should be <= instead of < -> Option C

4. Quick Check:

   Use <= to pop all lower or equal prices [OK]

Hint: Use <= in while to pop all lower or equal prices [OK]

Common Mistakes:

• Using < instead of <= causing wrong spans
• Storing prices instead of indices in stack
• Starting loop at wrong index

5. Given stock prices [10, 4, 5, 90, 120, 80], what is the span array if the stack is implemented to store prices instead of indices?

hard

A. [1, 1, 2, 4, 5, 1]

B. [1, 1, 1, 1, 1, 1]

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

D. [1, 1, 2, 1, 2, 1]

Solution

1. Step 1: Understand stack storing prices instead of indices

   Without indices, we cannot calculate span as difference of indices; span calculation fails.

2. Step 2: Resulting span values

   Each day is treated as isolated, so span defaults to 1 for all days.

3. Final Answer:

   [1, 1, 1, 1, 1, 1] -> Option B

4. Quick Check:

   Stack must store indices for correct span [OK]

Hint: Stack must store indices, not prices, for correct span [OK]

Common Mistakes:

• Storing prices causes incorrect span calculation
• Assuming span grows without index info
• Ignoring index difference in span formula