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

Buy and Sell Stocks All Variants

📖 Scenario: You are helping a friend who wants to understand how to find the best time to buy and sell stocks to maximize profit. They have a list of daily stock prices and want to learn how to find the best single buy-sell pair.

🎯 Goal: Build a Rust program that finds the maximum profit from buying and selling a stock once, given a list of daily prices.

📋 What You'll Learn

Create a vector called prices with the exact values [7, 1, 5, 3, 6, 4]

Create a variable called min_price and set it to the first price in prices

Use a for loop with variable price to iterate over prices starting from the second element

Inside the loop, update min_price if price is lower

Inside the loop, calculate profit as price - min_price and update max_profit if profit is higher

Print the max_profit at the end

💡 Why This Matters

🌍 Real World

This project helps understand how to analyze stock price data to make smart buy and sell decisions for profit.

💼 Career

Knowing how to process and analyze data with loops and conditions is essential for software developers working with financial or time-series data.

1

Create the stock prices vector

Create a vector called prices with these exact values: [7, 1, 5, 3, 6, 4]

DSA Rust

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

Hint

Use let prices = vec![7, 1, 5, 3, 6, 4]; to create the vector.

2

Set up minimum price and maximum profit variables

Create a variable called min_price and set it to the first element of prices. Also create a variable called max_profit and set it to 0.

DSA Rust

fn main() {
    let prices = vec![7, 1, 5, 3, 6, 4];
    // Create min_price and max_profit variables
    // Your code here
}

Hint

Use let mut min_price = prices[0]; and let mut max_profit = 0;.

3

Loop through prices to find max profit

Use a for loop with variable price to iterate over prices[1..]. Inside the loop, update min_price if price is less than min_price. Calculate profit as price - min_price and update max_profit if profit is greater.

DSA Rust

fn main() {
    let prices = vec![7, 1, 5, 3, 6, 4];
    let mut min_price = prices[0];
    let mut max_profit = 0;
    // Loop through prices starting from index 1
    // Your code here
}

Hint

Use a for loop over &prices[1..]. Use *price to get the value. Update min_price and max_profit accordingly.

4

Print the maximum profit

Print the value of max_profit using println!.

DSA Rust

fn main() {
    let prices = vec![7, 1, 5, 3, 6, 4];
    let mut min_price = prices[0];
    let mut max_profit = 0;
    for price in &prices[1..] {
        if *price  max_profit {
                max_profit = profit;
            }
        }
    }
    // Print max_profit
    // Your code here
}

Hint

Use println!("{}", max_profit); to print the result.

Practice

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

easy

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

B. Minimize the number of transactions regardless of profit

C. Find the stock price on the last day

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 key objective

   Maximizing profit means buying low and selling high at the best possible times.

3. Final Answer:

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

4. Quick Check:

   Goal = Maximize profit [OK]

Hint: Focus on profit maximization, not just transactions [OK]

Common Mistakes:

• Confusing profit maximization with minimizing transactions
• Ignoring the buy before sell rule
• Thinking the goal is to find average price

2. Which of the following Rust code snippets correctly initializes a vector of stock prices for the Buy and Sell Stocks problem?

easy

A. let prices = {7, 1, 5, 3, 6, 4};

B. let prices = vec![7, 1, 5, 3, 6, 4];

C. let prices = (7, 1, 5, 3, 6, 4);

D. let prices = [7, 1, 5, 3, 6, 4];

Solution

1. Step 1: Identify Rust vector syntax

   In Rust, vectors are created using vec![...] macro.

2. Step 2: Check each option

   let prices = vec![7, 1, 5, 3, 6, 4]; uses vec![...], which is correct. let prices = [7, 1, 5, 3, 6, 4]; is an array, not a vector. let prices = (7, 1, 5, 3, 6, 4); is a tuple. let prices = {7, 1, 5, 3, 6, 4}; is invalid syntax.

3. Final Answer:

   let prices = vec![7, 1, 5, 3, 6, 4]; -> Option B

4. Quick Check:

   Vector = vec![...] syntax [OK]

Hint: Use vec![] to create vectors in Rust [OK]

Common Mistakes:

• Using array syntax instead of vector
• Using tuple instead of vector
• Invalid curly braces for initialization

3. What is the output of the following Rust code that solves the 'Best Time to Buy and Sell Stock I' problem?

fn max_profit(prices: Vec<i32>) -> i32 {
    let mut min_price = i32::MAX;
    let mut max_profit = 0;
    for price in prices {
        if price < min_price {
            min_price = price;
        } else if price - min_price > max_profit {
            max_profit = price - min_price;
        }
    }
    max_profit
}

fn main() {
    let prices = vec![7, 1, 5, 3, 6, 4];
    println!("{}", max_profit(prices));
}

medium

A. 7

B. 6

C. 4

D. 5

Solution

1. Step 1: Trace the loop to find min_price and max_profit

   Start with min_price = MAX, max_profit = 0. Prices: 7,1,5,3,6,4 - price=7: min_price=7, max_profit=0 - price=1: min_price=1, max_profit=0 - price=5: profit=5-1=4 > 0, max_profit=4 - price=3: profit=3-1=2 < 4, max_profit=4 - price=6: profit=6-1=5 > 4, max_profit=5 - price=4: profit=4-1=3 < 5, max_profit=5

2. Step 2: Return max_profit

   max_profit is 5 after loop ends.

3. Final Answer:

   5 -> Option D

4. Quick Check:

   Max profit = 5 [OK]

Hint: Track min price and update max profit on each day [OK]

Common Mistakes:

• Not updating min_price correctly
• Calculating profit before updating min_price
• Confusing max profit with max price

4. Identify the bug in this Rust function intended to solve the 'Best Time to Buy and Sell Stock II' (multiple transactions allowed):

fn max_profit(prices: Vec<i32>) -> i32 {
    let mut profit = 0;
    for i in 0..prices.len() {
        if prices[i] > prices[i - 1] {
            profit += prices[i] - prices[i - 1];
        }
    }
    profit
}

medium

A. Index out of bounds error when i = 0

B. Profit calculation is incorrect for multiple transactions

C. Loop should start from 1 instead of 0

D. Prices vector should be mutable

Solution

1. Step 1: Analyze loop 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 bounds.

2. Step 2: Identify fix

   Loop should start from i=1 to avoid negative index access.

3. Final Answer:

   Index out of bounds error when i = 0 -> Option A

4. Quick Check:

   Negative index access causes runtime error [OK]

Hint: Start loop from 1 to avoid negative index [OK]

Common Mistakes:

• Starting loop at 0 causing invalid index
• Not checking vector length before access
• Assuming negative indices are valid

5. Given the stock prices [3, 3, 5, 0, 0, 3, 1, 4], what is the maximum profit achievable with at most two transactions (buy-sell pairs)?

hard

A. 5

B. 7

C. 6

D. 8

Solution

1. Step 1: Understand two transactions limit

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

2. Step 2: Find best two transactions

   Prices: 3,3,5,0,0,3,1,4 - First transaction: buy at 3 (day 0), sell at 5 (day 2), profit = 2 - Second transaction: buy at 0 (day 3), sell at 4 (day 7), profit = 4 Total profit = 2 + 4 = 6

3. Final Answer:

   6 -> Option C

4. Quick Check:

   Sum of two best profits = 6 [OK]

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

Common Mistakes:

• Counting overlapping transactions
• Choosing wrong buy/sell days
• Ignoring the two transactions limit