Stock Span Problem Using Stack in DSA C# - 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 stock spans 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.

public int[] CalculateSpan(int[] prices) {
    int n = prices.Length;
    int[] span = new int[n];
    Stack<int> stack = new Stack<int>();
    for (int i = 0; i < n; i++) {
        while (stack.Count > 0 && prices[stack.Peek()] <= prices[i]) {
            stack.Pop();
        }
        span[i] = (stack.Count == 0) ? (i + 1) : (i - stack.Peek());
        stack.Push(i);
    }
    return span;
}
    

This code calculates the stock span for each day using a stack to track previous higher prices.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The for-loop runs once for each price (n times).
• 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 over all iterations.

How Execution Grows With Input

As the number of days (n) increases, the total operations grow roughly in a straight line.

Input Size (n)	Approx. Operations
10	About 20 (each element pushed and popped once)
100	About 200
1000	About 2000

Pattern observation: Operations increase roughly in direct proportion to input size.

Final Time Complexity

Time Complexity: O(n)

This means the time to calculate spans grows linearly with the number of prices.

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 price index is pushed and popped only once, so total while-loop operations across all iterations add up to n, not n².

Interview Connect

Understanding this linear time solution shows you how to use stacks smartly to avoid repeated work, a skill valuable 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 difference between the current day's price and the previous day's price

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

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

Solution

1. Step 1: Understand the problem definition

   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 calculated for each day

   For each day, it counts consecutive previous days with prices not higher than the current day.

3. Final Answer:

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

4. Quick Check:

   Stock span = consecutive days with price ≤ current day price [OK]

Hint: Span counts consecutive previous days with price ≤ today [OK]

Common Mistakes:

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

2. Which of the following C# code snippets correctly declares a stack to be used for the Stock Span Problem?

easy

A. Stack<int> stack = new Stack<int>();

B. List<int> stack = new List<int>();

C. int[] stack = new int[10];

D. Queue<int> stack = new Queue<int>();

Solution

1. Step 1: Identify the correct data structure for stack

   In C#, Stack<T> is the built-in class for stack operations.

2. Step 2: Check each option's data structure

   Stack<int> stack = new Stack<int>(); uses Stack<int>, which is correct. int[] stack = new int[10]; uses array, List<int> stack = new List<int>(); uses List, which are not stacks. Queue<int> stack = new Queue<int>(); uses Queue, which is FIFO, not LIFO.

3. Final Answer:

   Stack<int> stack = new Stack<int>(); -> Option A

4. Quick Check:

   Stack<T> is the correct stack type in C# [OK]

Hint: Use Stack<int> for stack, not List or Queue [OK]

Common Mistakes:

• Using List instead of Stack
• Using Queue which is FIFO, not stack
• Using fixed-size array without stack methods

3. What is the output of the following C# code snippet for stock prices [100, 80, 60, 70, 60, 75, 85]?

int[] prices = {100, 80, 60, 70, 60, 75, 85};
int[] span = new int[prices.Length];
Stack<int> stack = new Stack<int>();

for (int i = 0; i < prices.Length; i++) {
    while (stack.Count > 0 && prices[stack.Peek()] <= prices[i]) {
        stack.Pop();
    }
    span[i] = (stack.Count == 0) ? (i + 1) : (i - stack.Peek());
    stack.Push(i);
}
Console.WriteLine(string.Join(", ", span));

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 stack and span calculation for each price

   Day 0: stack empty, span[0] = 1, push 0
   Day 1: prices[0]=100 > 80, no pop, span[1] = 1 - 0 = 1, push 1
   Day 2: prices[1]=80 > 60, no pop, span[2] = 2 - 1 = 1, push 2
   Day 3: prices[2]=60 <= 70, pop 2; prices[1]=80 > 70, span[3] = 3 - 1 = 2, push 3
   Day 4: prices[3]=70 > 60, no pop, span[4] = 4 - 3 = 1, push 4
   Day 5: prices[4]=60 <= 75 pop 4; prices[3]=70 <= 75 pop 3; prices[1]=80 > 75 no pop, span[5] = 5 - 1 = 4, push 5
   Day 6: prices[5]=75 <= 85 pop 5; prices[1]=80 <= 85 pop 1; prices[0]=100 > 85 no pop, span[6] = 6 - 0 = 6, push 6

2. Step 2: Collect the 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 days with price ≤ current day [OK]

Hint: Pop smaller prices, span = current index - top stack index [OK]

Common Mistakes:

• Not popping all smaller or equal prices
• Using wrong index difference for span
• Confusing span with price values

4. Identify the error in the following C# code snippet for the Stock Span Problem:

int[] prices = {10, 4, 5, 90, 120, 80};
int[] span = new int[prices.Length];
Stack<int> stack = new Stack<int>();

for (int i = 0; i < prices.Length; i++) {
    while (stack.Count > 0 && prices[stack.Peek()] <= prices[i]) {
        stack.Pop();
    }
    span[i] = (stack.Count == 0) ? i : (i - stack.Peek());
    stack.Push(i);
}
Console.WriteLine(string.Join(", ", span));

medium

A. The loop should start from i = 1, not 0

B. The comparison operator in while should be <= instead of <

C. Stack should store prices, not indices

D. The span calculation should use i + 1 when stack is empty

Solution

1. Step 1: Check the span calculation when stack is empty

   When stack is empty, span should be i + 1 because all previous days are counted.

2. Step 2: Identify the error in span assignment

   Code uses span[i] = i instead of i + 1, which undercounts span for first element and others when stack is empty.

3. Final Answer:

   The span calculation should use i + 1 when stack is empty -> Option D

4. Quick Check:

   Empty stack span = i + 1, not i [OK]

Hint: Use i+1 for span if stack empty, not i [OK]

Common Mistakes:

• Using i instead of i+1 for empty stack span
• Storing prices in stack instead of indices
• Wrong comparison operator causing incorrect pops

5. Given a list of daily stock prices, how would you modify the Stock Span Problem algorithm to find the span of days where the price was strictly less than the current day's price (not less than or equal)?

hard

A. Calculate span by counting all previous days regardless of price

B. Change the comparison in the while loop from prices[stack.Peek()] <= prices[i] to prices[stack.Peek()] < prices[i]

C. Use a queue instead of a stack to track indices

D. Change the comparison in the while loop from prices[stack.Peek()] <= prices[i] to prices[stack.Peek()] >= prices[i]

Solution

1. Step 1: Understand the original comparison

   The original algorithm pops indices while prices at stack top are less than or equal to current price.

2. Step 2: Modify comparison for strictly less

   To count only days with price strictly less, change <= to < in the while condition.

3. Final Answer:

   Change the comparison in the while loop from prices[stack.Peek()] <= prices[i] to prices[stack.Peek()] < prices[i] -> Option B

4. Quick Check:

   Use < instead of ≤ for strict less span [OK]

Hint: Use < instead of <= in while condition for strict less [OK]

Common Mistakes:

• Using >= instead of < causing wrong pops
• Switching to queue which breaks logic
• Ignoring price comparison and counting all days