Stock Span Problem Using Stack in DSA Go - Build from Scratch

Stock Span Problem Using Stack

📖 Scenario: You are working as a stock analyst. You want to find out how many consecutive days before today the stock price was less than or equal to today's price. This helps you understand the stock's recent performance trend.

🎯 Goal: Build a Go program that calculates the stock span for each day using a stack. The stock span is the number of consecutive days the price was less than or equal to the current day's price.

📋 What You'll Learn

Create a slice called prices with the exact values: 100, 80, 60, 70, 60, 75, 85

Create an empty stack using a slice of integers called stack

Create a slice called span to store the span values for each day

Use a for loop with variable i to iterate over prices

Use a for loop with variable len(stack) > 0 and compare prices[stack[len(stack)-1]] with prices[i]

Calculate the span for each day and store it in span[i]

Print the span slice at the end

💡 Why This Matters

🌍 Real World

Stock analysts use the stock span problem to understand how long a stock price has been rising or stable, helping in making investment decisions.

💼 Career

This problem teaches stack usage and array manipulation, which are common in technical interviews and real-world software development involving data processing.

1

Create the stock prices slice

Create a slice called prices with these exact values: 100, 80, 60, 70, 60, 75, 85

DSA Go

# Create the prices slice with the given values
# Your code here

Hint

Use prices := []int{100, 80, 60, 70, 60, 75, 85} to create the slice.

2

Create the stack and span slices

Create an empty slice of integers called stack to use as a stack, and create a slice called span with the same length as prices to store the span values

DSA Go

package main

func main() {
    prices := []int{100, 80, 60, 70, 60, 75, 85}
    // Create an empty stack slice
    // Create a span slice with the same length as prices
    // Your code here
}

Hint

Use stack := []int{} and span := make([]int, len(prices)).

3

Calculate the stock span using stack

Use a for loop with variable i to iterate over prices. Inside it, use a for loop to pop from stack while len(stack) > 0 and prices[stack[len(stack)-1]] <= prices[i]. Calculate the span for day i as i + 1 if stack is empty, else i - stack[len(stack)-1]. Append i to stack.

DSA Go

package main

func main() {
    prices := []int{100, 80, 60, 70, 60, 75, 85}
    stack := []int{}
    span := make([]int, len(prices))

    // Calculate span for each day
    // Your code here
}

Hint

Use nested loops and stack operations to calculate the span.

4

Print the span slice

Print the span slice using fmt.Println(span). Remember to import fmt at the top.

DSA Go

package main

import "fmt"

func main() {
    prices := []int{100, 80, 60, 70, 60, 75, 85}
    stack := []int{}
    span := make([]int, len(prices))

    for i := 0; i  0 && prices[stack[len(stack)-1]] <= prices[i] {
            stack = stack[:len(stack)-1]
        }
        if len(stack) == 0 {
            span[i] = i + 1
        } else {
            span[i] = i - stack[len(stack)-1]
        }
        stack = append(stack, i)
    }

    // Print the span slice
    // Your code here
}

Hint

Use fmt.Println(span) to print the span slice.

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 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 for a day is the count of consecutive previous days with prices less than or equal to that day's price.

2. Step 2: Clarify what is being calculated

   It is not about maximum price or sums, but about consecutive days with lower or equal prices.

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 A

4. Quick Check:

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

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

Common Mistakes:

• Confusing span with maximum price
• Thinking span is total sum of prices
• Ignoring consecutive days condition

2. Which Go data structure is best suited to efficiently solve the stock span problem?

easy

A. Stack

B. Array

C. Queue

D. Map

Solution

1. Step 1: Identify the data structure for tracking previous prices

   The stack helps keep track of indices of days with prices to compare efficiently.

2. Step 2: Understand why stack is efficient

   Stack allows popping smaller prices and quickly finding the last higher price day, enabling O(n) time complexity.

3. Final Answer:

   Stack -> Option A

4. Quick Check:

   Stack is used for efficient previous price tracking [OK]

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

Common Mistakes:

• Using array leads to O(n²) time
• Queue does not support backward checks
• Map is not suitable for order-based checks

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, 2, 2, 3, 3, 4]

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

C. [1, 1, 1, 1, 1, 1, 1]

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

Solution

1. Step 1: Calculate span for each day

   Day 0: 100 -> span=1 (no previous)
   Day 1: 80 ≤ 100 -> span=1
   Day 2: 60 ≤ 80 -> span=1
   Day 3: 70 > 60 -> span=2 (includes day 2)
   Day 4: 60 ≤ 70 -> span=1
   Day 5: 75 > 60, 70 -> span=4 (includes days 4,3,2)
   Day 6: 85 > 75, 70, 60, 80 -> span=6 (includes days 5,4,3,2,1)

2. Step 2: Confirm final span array

   The spans are [1, 1, 1, 2, 1, 4, 6].

3. Final Answer:

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

4. Quick Check:

   Span array matches calculated values [OK]

Hint: Pop smaller prices from stack to count span [OK]

Common Mistakes:

• Counting all previous days instead of consecutive
• Not popping smaller prices from stack
• Confusing span with price values

4. Identify the bug in this Go code snippet for stock span calculation:

func calculateSpan(prices []int) []int {
    n := len(prices)
    span := make([]int, n)
    stack := []int{}
    for i := 0; i < n; i++ {
        for len(stack) > 0 && prices[stack[len(stack)-1]] <= prices[i] {
            stack = stack[:len(stack)-1]
        }
        if len(stack) == 0 {
            span[i] = i + 1
        } else {
            span[i] = i - stack[len(stack)-1]
        }
        stack = append(stack, i)
    }
    return span
}

medium

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

B. The comparison operator should be <= instead of < in the inner loop

C. Span array is not initialized with correct size

D. Stack should store prices, not indices

Solution

1. Step 1: Analyze the inner loop condition

   The inner loop pops indices while the price at stack top is less than current price. It should pop while less than or equal to handle equal prices correctly.

2. Step 2: Understand impact of using < instead of <=

   Using < misses popping equal prices, causing incorrect span counts for equal prices.

3. Final Answer:

   The comparison operator should be <= instead of < in the inner loop -> Option B

4. Quick Check:

   Use <= to correctly handle equal prices [OK]

Hint: Use <= in comparison to handle equal prices correctly [OK]

Common Mistakes:

• Stack storing prices instead of indices
• Incorrect span array size
• Starting loop from wrong index

5. Given stock prices [10, 4, 5, 90, 120, 80], how would you modify the stock span algorithm to also return the day index of the last higher price before the current day? What is the output for the last higher price indices array?

hard

A. [ -1, 0, 1, -1, -1, 4 ]

B. [ -1, 0, 0, 2, 3, 3 ]

C. [-1, 0, 0, -1, -1, 4]

D. [ 0, -1, 1, 2, 3, 4 ]

Solution

1. Step 1: Modify algorithm to track last higher price index

   Use stack to store indices; for each day, pop while current price ≥ price at stack top. If stack empty, last higher index is -1; else top of stack is last higher index.

2. Step 2: Calculate last higher price indices for given prices

   Prices: [10,4,5,90,120,80]
   Day 0: empty -> -1
   Day 1: 10 > 4, no pop -> 0
   Day 2: 4 <= 5 pop 1, 10 > 5 -> 0
   Day 3: 5 <= 90 pop 2, 10 <= 90 pop 0, empty -> -1
   Day 4: 90 <= 120 pop 3, empty -> -1
   Day 5: 120 > 80, no pop -> 4

3. Final Answer:

   [-1, 0, 0, -1, -1, 4] -> Option C

4. Quick Check:

   Last higher indices track previous greater price day [OK]

Hint: Track last higher price index using stack top or -1 if empty [OK]

Common Mistakes:

• Returning span instead of last higher index
• Not handling empty stack case (-1)
• Confusing last higher index with span count