Buy and Sell Stocks All Variants in DSA Go - Build from Scratch

Buy and Sell Stocks All Variants

📖 Scenario: You are helping a friend who wants to analyze stock prices over several days to find the best times to buy and sell stocks for maximum profit.They have a list of daily stock prices and want to understand how to find the best single buy-sell pair to maximize profit.

🎯 Goal: Build a Go program that finds the maximum profit possible from buying and selling a stock once, given daily prices.This will teach you how to track minimum prices and calculate profits step-by-step.

📋 What You'll Learn

Create a slice called prices with the exact values: 7, 1, 5, 3, 6, 4

Create an integer variable called minPrice and set it to the first price in prices

Create an integer variable called maxProfit and set it to 0

Use a for loop with index i starting from 1 to iterate over prices

Inside the loop, update minPrice if the current price is lower

Calculate the profit by subtracting minPrice from the current price

Update maxProfit if the calculated profit is higher

Print the maxProfit at the end

💡 Why This Matters

🌍 Real World

Stock traders and investors use similar logic to decide the best times to buy and sell stocks to maximize profit.

💼 Career

Understanding how to analyze sequences of data and find optimal points is useful in finance, data analysis, and software development.

1

Create the stock prices slice

Create a slice called prices with the exact values 7, 1, 5, 3, 6, 4

DSA Go

# Create the prices slice with values 7, 1, 5, 3, 6, 4
# Your code here

Hint

Use prices := []int{7, 1, 5, 3, 6, 4} to create the slice.

2

Initialize minPrice and maxProfit variables

Create an integer variable called minPrice and set it to the first element of prices. Then create an integer variable called maxProfit and set it to 0

DSA Go

package main

func main() {
    prices := []int{7, 1, 5, 3, 6, 4}
    // Initialize minPrice and maxProfit
    // Your code here
}

Hint

Set minPrice to prices[0] and maxProfit to 0.

3

Loop through prices to find max profit

Use a for loop with index i starting from 1 to iterate over prices. Inside the loop, update minPrice if prices[i] is less than minPrice. Calculate the profit by subtracting minPrice from prices[i]. Update maxProfit if the profit is greater than maxProfit

DSA Go

package main

func main() {
    prices := []int{7, 1, 5, 3, 6, 4}
    minPrice := prices[0]
    maxProfit := 0

    // Loop through prices starting from index 1
    // Your code here
}

Hint

Use a for loop from 1 to len(prices). Update minPrice and maxProfit inside the loop.

4

Print the maximum profit

Print the value of maxProfit using fmt.Println(maxProfit)

DSA Go

package main

import "fmt"

func main() {
    prices := []int{7, 1, 5, 3, 6, 4}
    minPrice := prices[0]
    maxProfit := 0

    for i := 1; i  maxProfit {
            maxProfit = profit
        }
    }

    // Print the maxProfit
    // Your code here
}

Hint

Use fmt.Println(maxProfit) to print the result.

Practice

1. What is the main goal when solving the Buy and Sell Stocks problem variants?

easy

A. Find the day with the highest stock price only

B. Minimize the number of transactions regardless of profit

C. Maximize the profit by choosing the best days to buy and sell stocks

D. Calculate the average stock price over all days

Solution

1. Step 1: Understand the problem goal

   The problem asks to maximize profit by buying and selling stocks on certain days.

2. Step 2: Identify the correct objective

   Maximizing profit means choosing days to buy low and sell high, not just counting transactions or prices.

3. Final Answer:

   Maximize the profit by choosing the best days to buy and sell stocks -> Option C

4. Quick Check:

   Goal = Maximize profit [OK]

Hint: Focus on profit maximization, not just prices or counts [OK]

Common Mistakes:

• Confusing profit with number of transactions
• Thinking only about highest price day
• Ignoring the buy day in calculations

2. Which of the following Go code snippets correctly initializes a slice to store stock prices for 5 days?

easy

A. prices := []int{7, 1, 5, 3, 6}

B. prices := [5]int{7, 1, 5, 3, 6}

C. prices := make([]int, 5, 5)

D. prices := new([]int{7, 1, 5, 3, 6})

Solution

1. Step 1: Understand slice initialization in Go

   To create a slice with values, use the literal syntax: []int{values}.

2. Step 2: Check each option

   prices := []int{7, 1, 5, 3, 6} correctly creates a slice with given values. prices := [5]int{7, 1, 5, 3, 6} creates an array, not a slice. prices := make([]int, 5, 5) creates an empty slice of length 5 but no values. prices := new([]int{7, 1, 5, 3, 6}) is invalid syntax.

3. Final Answer:

   prices := []int{7, 1, 5, 3, 6} -> Option A

4. Quick Check:

   Slice literal = []int{values} [OK]

Hint: Use []int{...} to create slices with values directly [OK]

Common Mistakes:

• Using array syntax instead of slice
• Using new() incorrectly for slices
• Creating empty slices without values

3. What is the output of the following Go code that finds the max profit with only one transaction?

func maxProfit(prices []int) int {
    minPrice := prices[0]
    maxProfit := 0
    for _, price := range prices {
        if price < minPrice {
            minPrice = price
        } else if price - minPrice > maxProfit {
            maxProfit = price - minPrice
        }
    }
    return maxProfit
}

fmt.Println(maxProfit([]int{7,1,5,3,6,4}))

medium

A. 5

B. 6

C. 4

D. 7

Solution

1. Step 1: Trace the loop to find minPrice and maxProfit

   Start minPrice=7, maxProfit=0. Iterate prices: - price=7: minPrice=7, maxProfit=0 - price=1: minPrice=1 (lower) - price=5: profit=5-1=4 >0, maxProfit=4 - price=3: profit=3-1=2 <4, maxProfit=4 - price=6: profit=6-1=5 >4, maxProfit=5 - price=4: profit=4-1=3 <5, maxProfit=5

2. Step 2: Return maxProfit

   maxProfit is 5 after loop ends.

3. Final Answer:

   5 -> Option A

4. Quick Check:

   Max profit = 5 [OK]

Hint: Track min price and update max profit during iteration [OK]

Common Mistakes:

• Not updating minPrice correctly
• Calculating profit before updating minPrice
• Returning wrong variable

4. Identify the bug in this Go function that tries to calculate max profit with unlimited transactions:

func maxProfit(prices []int) int {
    profit := 0
    for i := 0; i < len(prices); i++ {
        if prices[i] > prices[i-1] {
            profit += prices[i] - prices[i-1]
        }
    }
    return profit
}

medium

A. prices slice is not initialized

B. Incorrect profit calculation logic

C. Missing return statement

D. Index out of range error due to accessing prices[i-1] when i=0

Solution

1. Step 1: Check loop and index usage

   The loop starts at i=0, but code accesses prices[i-1], which is prices[-1] at i=0, causing index out of range error.

2. Step 2: Identify fix

   Start loop from i=1 to avoid negative index access.

3. Final Answer:

   Index out of range error due to accessing prices[i-1] when i=0 -> Option D

4. Quick Check:

   Loop index must start at 1 to avoid error [OK]

Hint: Avoid negative indices by starting loop at 1 [OK]

Common Mistakes:

• Starting loop at 0 but accessing i-1
• Ignoring runtime errors
• Not testing with small input arrays

5. Given the prices [3, 3, 5, 0, 0, 3, 1, 4], what is the maximum profit with at most two transactions?

hard

A. 7

B. 6

C. 5

D. 8

Solution

1. Step 1: Understand two transactions limit

   We can buy and sell stocks at most twice to maximize profit.

2. Step 2: Calculate max profit with two transactions

   First transaction: buy at 3, sell at 5 (profit 2) Second transaction: buy at 0, sell at 4 (profit 4) Total profit = 2 + 4 = 6

3. Final Answer:

   6 -> Option B

4. Quick Check:

   Two transactions max profit = 6 [OK]

Hint: Split array, find max profit left and right, sum max [OK]

Common Mistakes:

• Counting overlapping transactions
• Ignoring zero price days
• Calculating only one transaction profit