Buy and Sell Stocks All Variants in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to find the maximum profit from a single buy and sell.

DSA Java

public int maxProfit(int[] prices) {
    int minPrice = Integer.MAX_VALUE;
    int maxProfit = 0;
    for (int price : prices) {
        if (price < [1]) {
            minPrice = price;
        } else if (price - minPrice > maxProfit) {
            maxProfit = price - minPrice;
        }
    }
    return maxProfit;
}

Drag options to blanks, or click blank then click option'

Aprice

BmaxProfit

CminPrice

DInteger.MAX_VALUE

Attempts:

3 left

💡 Hint

Common Mistakes

Using maxProfit instead of minPrice in the comparison.

Not updating minPrice correctly.

We update minPrice when the current price is less than the minimum price seen so far.

2fill in blank

medium

Complete the code to calculate the maximum profit with unlimited transactions.

DSA Java

public int maxProfit(int[] prices) {
    int profit = 0;
    for (int i = 1; i < prices.length; i++) {
        if (prices[i] > [1]) {
            profit += prices[i] - prices[i - 1];
        }
    }
    return profit;
}

Drag options to blanks, or click blank then click option'

Aprices[i - 1]

Bprices[i]

Cprices[i + 1]

Dprices[0]

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing with prices[i] instead of prices[i - 1].

Using prices[i + 1] which causes index errors.

We compare the current price with the previous day's price to decide if we can make a profit.

3fill in blank

hard

Fix the error in the code to find max profit with at most two transactions.

DSA Java

public int maxProfit(int[] prices) {
    int n = prices.length;
    int[] leftProfits = new int[n];
    int minPrice = prices[0];
    for (int i = 1; i < n; i++) {
        minPrice = Math.min(minPrice, prices[i]);
        leftProfits[i] = Math.max(leftProfits[i - 1], prices[i] - [1]);
    }
    int maxPrice = prices[n - 1];
    int maxProfit = leftProfits[n - 1];
    int rightProfit = 0;
    for (int i = n - 2; i >= 0; i--) {
        maxPrice = Math.max(maxPrice, prices[i]);
        rightProfit = Math.max(rightProfit, maxPrice - prices[i]);
        maxProfit = Math.max(maxProfit, rightProfit + leftProfits[i]);
    }
    return maxProfit;
}

Drag options to blanks, or click blank then click option'

AminPrice

BmaxPrice

Cprices[i - 1]

Dprices[i]

Attempts:

3 left

💡 Hint

Common Mistakes

Using maxPrice instead of minPrice in leftProfits calculation.

Using prices[i] instead of minPrice.

We calculate profit by subtracting the minimum price seen so far from the current price.

4fill in blank

hard

Fill both blanks to implement max profit with cooldown period.

DSA Java

public int maxProfit(int[] prices) {
    int n = prices.length;
    if (n <= 1) return 0;
    int[] sell = new int[n];
    int[] buy = new int[n];
    buy[0] = -prices[0];
    sell[0] = 0;
    buy[1] = Math.max(-prices[0], -prices[1]);
    sell[1] = Math.max(0, prices[1] - prices[0]);
    for (int i = 2; i < n; i++) {
        buy[i] = Math.max(buy[i - 1], sell[i - 2] - [1]);
        sell[i] = Math.max(sell[i - 1], buy[i - 1] + [2]);
    }
    return sell[n - 1];
}

Drag options to blanks, or click blank then click option'

Aprices[i]

Bprices[i - 1]

Cprices[i + 1]

Dprices[0]

Attempts:

3 left

💡 Hint

Common Mistakes

Using prices[i - 1] or prices[i + 1] causing index errors.

Mixing buy and sell prices incorrectly.

We subtract the current price when buying and add the current price when selling.

5fill in blank

hard

Fill all three blanks to implement max profit with at most k transactions.

DSA Java

public int maxProfit(int k, int[] prices) {
    int n = prices.length;
    if (n == 0) return 0;
    if (k >= n / 2) {
        int profit = 0;
        for (int i = 1; i < n; i++) {
            if (prices[i] > prices[i - 1]) {
                profit += prices[i] - prices[i - 1];
            }
        }
        return profit;
    }
    int[][] dp = new int[k + 1][n];
    for (int i = 1; i <= k; i++) {
        int maxDiff = -prices[0];
        for (int j = 1; j < n; j++) {
            dp[i][j] = Math.max(dp[i][j - 1], prices[j] + [1]);
            maxDiff = Math.max(maxDiff, dp[i - 1][j - 1] - [2]);
        }
    }
    return dp[k][n - 1];
}

Drag options to blanks, or click blank then click option'

AmaxDiff

Bprices[j]

Cdp[i - 1][j - 1]

Ddp[i][j - 1]

Attempts:

3 left

💡 Hint

Common Mistakes

Using dp[i][j - 1] instead of maxDiff in dp update.

Mixing indices in maxDiff update.

We use maxDiff to track the best balance and prices[j] for current price in calculations.

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