Stock Span Problem Using Stack in DSA Java - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Stock Span Calculation

What is the output of the following Java code that calculates the stock span for given prices?

DSA Java

import java.util.*;
public class StockSpan {
    public static int[] calculateSpan(int[] prices) {
        int n = prices.length;
        int[] span = new int[n];
        Stack<Integer> stack = new Stack<>();
        for (int i = 0; i < n; i++) {
            while (!stack.isEmpty() && prices[stack.peek()] <= prices[i]) {
                stack.pop();
            }
            span[i] = stack.isEmpty() ? (i + 1) : (i - stack.peek());
            stack.push(i);
        }
        return span;
    }
    public static void main(String[] args) {
        int[] prices = {100, 80, 60, 70, 60, 75, 85};
        int[] result = calculateSpan(prices);
        for (int val : result) {
            System.out.print(val + " ");
        }
    }
}

A 1 1 2 2 3 3 4

B 1 1 1 2 1 4 6

C 1 2 3 4 5 6 7

D 1 1 1 1 1 1 1

💡 Hint

Think about how the stack helps track previous higher prices to calculate spans.

The stack stores indices of days with higher prices. For each day, span is the difference between current day and last higher price day. If none, span is current day index + 1.

🧠 Conceptual

intermediate

Understanding Stack Usage in Stock Span

Why is a stack used in the stock span problem instead of a simple loop checking all previous days?

A Because stack stores all prices in sorted order automatically.

B Because stack duplicates the array to avoid modifying original data.

C Because stack allows efficient tracking of previous higher prices, reducing time complexity from O(n^2) to O(n).

D Because stack is used to reverse the prices array before processing.

💡 Hint

Consider how many times each element is pushed and popped in the stack.

Stack helps skip unnecessary comparisons by storing indices of days with higher prices, so each element is pushed and popped at most once, making the algorithm O(n).

🔧 Debug

advanced

Identify the Error in Stock Span Code

What error will the following code produce when run?

DSA Java

import java.util.*;
public class StockSpan {
    public static int[] calculateSpan(int[] prices) {
        int n = prices.length;
        int[] span = new int[n];
        Stack<Integer> stack = new Stack<>();
        for (int i = 0; i <= n; i++) {
            while (!stack.isEmpty() && prices[stack.peek()] <= prices[i]) {
                stack.pop();
            }
            span[i] = stack.isEmpty() ? (i + 1) : (i - stack.peek());
            stack.push(i);
        }
        return span;
    }
}

A ArrayIndexOutOfBoundsException

B StackOverflowError

C No error, code runs correctly

D NullPointerException

💡 Hint

Check the loop boundary condition carefully.

The loop runs from i = 0 to i <= n, which causes i to reach n, an invalid index for the prices array, causing ArrayIndexOutOfBoundsException.

❓ Predict Output

advanced

Output of Stock Span with All Increasing Prices

What is the output of the stock span calculation for strictly increasing prices?

DSA Java

import java.util.*;
public class StockSpan {
    public static int[] calculateSpan(int[] prices) {
        int n = prices.length;
        int[] span = new int[n];
        Stack<Integer> stack = new Stack<>();
        for (int i = 0; i < n; i++) {
            while (!stack.isEmpty() && prices[stack.peek()] <= prices[i]) {
                stack.pop();
            }
            span[i] = stack.isEmpty() ? (i + 1) : (i - stack.peek());
            stack.push(i);
        }
        return span;
    }
    public static void main(String[] args) {
        int[] prices = {10, 20, 30, 40, 50};
        int[] result = calculateSpan(prices);
        for (int val : result) {
            System.out.print(val + " ");
        }
    }
}

A 1 2 3 4 5

B 5 4 3 2 1

C 1 1 1 1 1

D 1 2 1 2 1

💡 Hint

In increasing prices, each price is higher than all previous, so span grows.

Each new price is higher than all before, so span equals current day index + 1.

🧠 Conceptual

expert

Time Complexity Analysis of Stock Span Algorithm

What is the time complexity of the stock span algorithm using a stack and why?

A O(n log n) because of sorting inside the stack operations.

B O(n^2) because for each element, all previous elements are checked.

C O(log n) because the stack maintains a sorted order.

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

💡 Hint

Consider how many times each index is pushed and popped.

Each price index is pushed once and popped once, so total operations are proportional to n, making it O(n).

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