Buy and Sell Stocks All Variants in DSA Java - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main goal in the 'Buy and Sell Stocks' problem?

To maximize profit by choosing the best days to buy and sell stocks given their prices over time.

beginner

Explain the difference between 'Buy and Sell Stocks I' and 'Buy and Sell Stocks II'.

'Buy and Sell Stocks I' allows only one transaction (one buy and one sell). 'Buy and Sell Stocks II' allows multiple transactions but you must sell before you buy again.

intermediate

What does the 'cooldown' mean in the 'Buy and Sell Stocks with Cooldown' variant?

After selling a stock, you must wait one day before buying again. This enforces a cooldown period to prevent immediate re-buying.

intermediate

How does the 'Buy and Sell Stocks with Transaction Fee' variant affect the profit calculation?

Each transaction (buy and sell pair) incurs a fixed fee, which reduces the profit from that transaction.

intermediate

What is the key idea behind using dynamic programming in 'Buy and Sell Stocks' problems?

To keep track of the best profit at each day considering whether you hold a stock or not, and use previous results to build the solution efficiently.

In 'Buy and Sell Stocks I', what is the maximum number of transactions allowed?

A One transaction

B Multiple transactions

C Unlimited transactions with cooldown

D No transactions

Only one buy and one sell are allowed in 'Buy and Sell Stocks I'.

What does the cooldown period prevent in the 'Buy and Sell Stocks with Cooldown' problem?

A Selling multiple stocks at once

B Buying immediately after selling

C Buying multiple stocks at once

D Selling before buying

Cooldown means you cannot buy a stock immediately after selling one; you must wait one day.

In the 'Buy and Sell Stocks with Transaction Fee' variant, what happens to profit when you pay the fee?

A Profit increases by the fee amount

B Profit stays the same

C Profit decreases by the fee amount

D Fee is added to profit

The transaction fee reduces the profit earned from each transaction.

Which approach is commonly used to solve all variants of the 'Buy and Sell Stocks' problem efficiently?

A Dynamic Programming

B Brute Force

C Sorting

D Greedy without state tracking

Dynamic programming tracks states like holding or not holding stock to optimize profit.

In 'Buy and Sell Stocks II', what is allowed that is not in 'Buy and Sell Stocks I'?

A No selling

B No transactions

C Only one transaction

D Multiple transactions

'Buy and Sell Stocks II' allows multiple buy and sell transactions.

Describe how dynamic programming helps solve the 'Buy and Sell Stocks with Cooldown' problem.

Think about how you track profit day by day with cooldown rules.

Explain the impact of transaction fees on the strategy to maximize profit in stock trading problems.

Consider how fees change when it is worth to sell.

Practice

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

easy

A. Minimize the number of transactions made

B. Find the day with the highest stock price

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 low and selling high on different days.

2. Step 2: Identify the correct objective

   Maximizing profit means choosing the best buy and sell days, not minimizing transactions or finding averages.

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 difference between buy and sell days [OK]

Common Mistakes:

• Confusing profit maximization with minimizing transactions
• Thinking the goal is to find highest price day only
• Ignoring the buy day before sell day order

2. Which of the following Java code snippets correctly initializes an array of stock prices for 5 days?

easy

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

B. int prices = [7, 1, 5, 3, 6];

C. int[] prices = (7, 1, 5, 3, 6);

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

Solution

1. Step 1: Check Java array initialization syntax

   Java arrays are initialized with curly braces {} and type declaration with [] after type.

2. Step 2: Validate each option

   Parentheses () are invalid, square brackets [] for values are invalid, missing braces {} and improper assignment are invalid. int[] prices = {7, 1, 5, 3, 6}; uses correct syntax.

3. Final Answer:

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

4. Quick Check:

   Correct array syntax = int[] prices = {7, 1, 5, 3, 6}; [OK]

Hint: Use curly braces {} for array values in Java [OK]

Common Mistakes:

• Using square brackets [] for values instead of curly braces
• Missing type[] before variable name
• Using parentheses () instead of braces

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

int[] prices = {7, 1, 5, 3, 6, 4};
int minPrice = Integer.MAX_VALUE;
int maxProfit = 0;
for (int price : prices) {
    if (price < minPrice) minPrice = price;
    else if (price - minPrice > maxProfit) maxProfit = price - minPrice;
}
System.out.println(maxProfit);

medium

A. 5

B. 6

C. 4

D. 7

Solution

1. Step 1: Track minimum price and max profit during iteration

   Prices: 7,1,5,3,6,4. minPrice updates to 7 then 1. maxProfit updates when price - minPrice is greater.

2. Step 2: Calculate max profit step-by-step

   At price=5, profit=5-1=4; maxProfit=4. At price=6, profit=6-1=5; maxProfit=5. At price=4, profit=4-1=3; maxProfit stays 5.

3. Final Answer:

   5 -> Option A

4. Quick Check:

   Max profit = 6 - 1 = 5 [OK]

Hint: Track min price and update max profit on the fly [OK]

Common Mistakes:

• Confusing max profit with max price
• Not updating minPrice correctly
• Calculating profit without checking minPrice

4. Identify the error in this Java code snippet that attempts to calculate max profit with multiple transactions:

int maxProfit = 0;
for (int i = 1; i < prices.length; i++) {
    if (prices[i] > prices[i - 1])
        maxProfit += prices[i] - prices[i - 1];
    else
        i++;
}
System.out.println(maxProfit);

medium

A. Loop should start from i = 0

B. Missing initialization of maxProfit

C. Using prices[i] instead of prices[i-1] in calculation

D. Incorrect increment of loop variable inside else block

Solution

1. Step 1: Analyze loop increment and conditions

   The loop increments i each iteration, but inside else block it increments i again causing skipping elements.

2. Step 2: Identify why extra increment causes error

   Extra i++ skips days, missing some profit opportunities and causing incorrect result.

3. Final Answer:

   Incorrect increment of loop variable inside else block -> Option D

4. Quick Check:

   Loop variable increment only once per iteration [OK]

Hint: Avoid changing loop variable inside loop body [OK]

Common Mistakes:

• Incrementing loop variable twice causing skips
• Starting loop from wrong index
• Confusing price indices in calculation

5. Given an array of stock prices, how would you modify the algorithm to find the maximum profit with at most two transactions? Choose the best approach.

hard

A. Buy on the lowest price day and sell on the highest price day only.

B. Use two passes: first pass calculates max profit up to each day, second pass calculates max profit from each day to end, then combine.

C. Use a single pass and sum all positive differences between consecutive days.

D. Sort the prices array and pick the two largest differences.

Solution

1. Step 1: Understand the two transaction constraint

   At most two transactions means we can buy and sell twice, but transactions cannot overlap.

2. Step 2: Use two-pass approach for max profit calculation

   First pass calculates max profit from day 0 to i, second pass calculates max profit from i to end. Combining these gives max profit with two transactions.

3. Final Answer:

   Use two passes: first pass calculates max profit up to each day, second pass calculates max profit from each day to end, then combine. -> Option B

4. Quick Check:

   Two-pass method = Use two passes: first pass calculates max profit up to each day, second pass calculates max profit from each day to end, then combine. [OK]

Hint: Split problem into two single-transaction profits [OK]

Common Mistakes:

• Sorting prices loses day order information
• Summing all positive differences allows unlimited transactions
• Ignoring transaction count limit