Why Stock Span Problem Using Stack in DSA C?

The Big Idea

Discover how a simple stack can save you hours of tedious price comparisons!

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 today's price with all previous days for every new price 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

Using a stack helps keep track of previous days with higher prices efficiently.

This way, you only compare with relevant days, skipping unnecessary checks and speeding up the process.

Before vs After

✗ Before

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

✓ After

stack<int> s;
for (int i = 0; i < n; i++) {
  while (!s.empty() && prices[s.top()] <= prices[i]) {
    s.pop();
  }
  span[i] = s.empty() ? (i + 1) : (i - s.top());
  s.push(i);
}

What It Enables

This method allows quick calculation of stock spans for large data sets, enabling fast financial analysis.

Real Life Example

Traders use this to quickly understand how many days in a row a stock price has been rising or stable, helping them decide when to buy or sell.

Key Takeaways

Manual checking repeats many comparisons and is slow.

Stack stores indices of days with higher prices to skip unnecessary checks.

Efficiently calculates spans in one pass through the data.

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 a price less than or equal to the current day's price

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 how many consecutive previous days had stock prices less than or equal to the current day's price.

2. Step 2: Identify what is calculated for each day

   For each day, it counts the number of consecutive days before it with prices not greater than the current day's price.

3. Final Answer:

   The number of consecutive days with price less or equal to current day -> Option B

4. Quick Check:

   Span counts consecutive smaller/equal prices [OK]

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

Common Mistakes:

• Confusing span with maximum price
• Calculating total sum instead of span
• Counting days with price greater than current

2. Which of the following C code snippets correctly initializes an empty stack for solving the Stock Span Problem?

easy

A. int stack[100]; int top = -1;

B. int stack[] = {0}; int top = 0;

C. int *stack; top = NULL;

D. stack = new int[100]; top = -1;

Solution

1. Step 1: Identify correct stack initialization in C

   In C, stack is usually an array with an integer top initialized to -1 to indicate empty stack.

2. Step 2: Check each option

   int stack[100]; int top = -1; correctly declares an int array and sets top to -1. int stack[] = {0}; int top = 0; initializes top to 0 which is incorrect for empty stack. int *stack; top = NULL; uses pointer and NULL incorrectly. stack = new int[100]; top = -1; uses 'new' which is C++ syntax, not C.

3. Final Answer:

   int stack[100]; int top = -1; -> Option A

4. Quick Check:

   Array + top = -1 initializes empty stack in C [OK]

Hint: Use array and top = -1 for empty stack in C [OK]

Common Mistakes:

• Using C++ 'new' keyword in C code
• Setting top to 0 instead of -1
• Using uninitialized pointers for stack

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: price=100, no previous, span=1
   Day 1: price=80, previous 100 > 80, span=1
   Day 2: price=60, previous 80 > 60, span=1
   Day 3: price=70, previous 60 ≤ 70, span=1+1=2
   Day 4: price=60, previous 70 > 60, span=1
   Day 5: price=75, previous 60 ≤ 75, count 1 (day4), previous 70 ≤ 75, count 2 (day3), previous 60 ≤ 75, count 3 (day2), previous 80 > 75 stop, span=4
   Day 6: price=85, previous 75 ≤ 85, count 1 (day5), previous 60 ≤ 85, count 2 (day4), previous 70 ≤ 85, count 3 (day3), previous 60 ≤ 85, count 4 (day2), previous 80 ≤ 85, count 5 (day1), previous 100 > 85 stop, span=6

2. Step 2: Write 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 counts consecutive smaller/equal prices [OK]

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

Common Mistakes:

• Counting all previous days instead of consecutive
• Ignoring equal prices in span
• Miscounting span for days with fluctuating prices

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

for (int i = 0; i < n; i++) {
  while (top != -1 && prices[stack[top]] <= prices[i]) {
    top--;
  }
  span[i] = (top == -1) ? (i + 1) : (i - stack[top]);
  stack[++top] = i;
}

medium

A. The condition 'top != -1' should be 'top == -1'

B. The stack push operation should be before calculating span

C. No error; code correctly calculates the stock span

D. The comparison operator in while loop should be '<' instead of '<='

Solution

1. Step 1: Analyze the while loop condition

   The while loop pops indices from stack while the price at those indices is less than or equal to current price, which is correct for span calculation.

2. Step 2: Check span calculation and stack push

   Span is set to i+1 if stack empty, else difference between current index and top of stack index. Then current index is pushed. This matches the standard algorithm.

3. Final Answer:

   No error; code correctly calculates the stock span -> Option C

4. Quick Check:

   Standard stack-based span algorithm used [OK]

Hint: Check while condition and span formula carefully [OK]

Common Mistakes:

• Changing <= to < breaks span logic
• Pushing before span calculation causes wrong results
• Incorrect stack empty condition

5. Given stock prices [10, 4, 5, 90, 120, 80], how would the Stock Span algorithm change if we want to count days with strictly less price (not less or equal)? What is the resulting span array?

hard

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

B. [1, 1, 1, 4, 5, 1]

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

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

Solution

1. Step 1: Understand change from ≤ to < in comparison

   Changing from less or equal to strictly less means while popping from stack, we pop only if previous price is strictly less than current price, not equal.

2. Step 2: Calculate span with strict less condition

   Day 0: 10, span=1
   Day 1: 4, previous 10 > 4, span=1
   Day 2: 5, previous 4 < 5, pop 4, span=2
   Day 3: 90, previous 5 < 90, pop 5; previous 10 < 90, pop 10; stack empty, span=4
   Day 4: 120, previous 90 < 120, pop 90; stack empty, span=5
   Day 5: 80, previous 120 > 80, span=1 (no pop)
   But since we pop only if strictly less, equal prices would not pop. Here no equal prices affect.

3. Final Answer:

   [1, 1, 2, 4, 5, 1] -> Option D

4. Quick Check:

   Strict less changes popping condition, span updated accordingly [OK]

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

Common Mistakes:

• Not updating while condition correctly
• Miscounting span for last day
• Assuming equal prices pop stack