Buy and Sell Stocks All Variants in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA C++

int maxProfit(vector<int>& prices) {
    int minPrice = INT_MAX;
    int maxProfit = 0;
    for (int price : prices) {
        minPrice = min(minPrice, [1]);
        maxProfit = max(maxProfit, price - minPrice);
    }
    return maxProfit;
}

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

Aprice

BmaxProfit

Cprices[0]

DminPrice

Attempts:

3 left

💡 Hint

Common Mistakes

Using maxProfit instead of price to update minPrice.

Using minPrice instead of price in min function.

We update the minimum price seen so far by comparing it with the current price.

2fill in blank

medium

Complete the code to calculate maximum profit with unlimited transactions.

DSA C++

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

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

A>

B<

C==

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' instead of '>' causing no profit calculation.

Using '==' which ignores profitable days.

We add profit only when today's price is greater than yesterday's price.

3fill in blank

hard

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

DSA C++

int maxProfit(vector<int>& prices) {
    int n = prices.size();
    if (n == 0) return 0;
    vector<int> leftProfits(n, 0);
    int minPrice = prices[0];
    for (int i = 1; i < n; i++) {
        minPrice = min(minPrice, prices[i]);
        leftProfits[i] = max(leftProfits[i - 1], prices[i] - [1]);
    }
    vector<int> rightProfits(n + 1, 0);
    int maxPrice = prices[n - 1];
    for (int i = n - 2; i >= 0; i--) {
        maxPrice = max(maxPrice, prices[i]);
        rightProfits[i] = max(rightProfits[i + 1], maxPrice - prices[i]);
    }
    int maxProfit = 0;
    for (int i = 0; i < n; i++) {
        maxProfit = max(maxProfit, leftProfits[i] + rightProfits[i + 1]);
    }
    return maxProfit;
}

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

Aprices[i - 1]

BmaxPrice

CminPrice

DleftProfits[i]

Attempts:

3 left

💡 Hint

Common Mistakes

Using maxPrice instead of minPrice in leftProfits calculation.

Using prices[i - 1] which is incorrect for profit calculation.

We subtract the minimum price seen so far to calculate profit for the left side.

4fill in blank

hard

Fill both blanks to implement max profit with cooldown period after selling.

DSA C++

int maxProfit(vector<int>& prices) {
    int n = prices.size();
    if (n <= 1) return 0;
    vector<int> sell(n, 0), buy(n, 0), cooldown(n, 0);
    buy[0] = -prices[0];
    for (int i = 1; i < n; i++) {
        buy[i] = max(buy[i - 1], cooldown[i - 1] - [1]);
        sell[i] = max(sell[i - 1], buy[i - 1] + prices[i]);
        cooldown[i] = max(cooldown[i - 1], sell[i - 1]);
    }
    return max(sell[n - 1], cooldown[n - 1]);
}

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

Aprices[i]

Bprices[i - 1]

Csell[i - 1]

Dbuy[i - 1]

Attempts:

3 left

💡 Hint

Common Mistakes

Using prices[i - 1] instead of prices[i] for buying cost.

Using buy[i - 1] instead of sell[i - 1] for cooldown.

We buy by subtracting today's price from cooldown state, and cooldown is max of previous cooldown or sell.

5fill in blank

hard

Fill all three blanks to implement max profit with transaction fee.

DSA C++

int maxProfit(vector<int>& prices, int fee) {
    int n = prices.size();
    int cash = 0, hold = -prices[0];
    for (int i = 1; i < n; i++) {
        cash = max(cash, hold + prices[i] - [1]);
        hold = max(hold, cash - [2]);
    }
    return cash;
}

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

Afee

Bprices[i]

Attempts:

3 left

💡 Hint

Common Mistakes

Not subtracting fee when selling.

Subtracting fee when buying instead of price.

When selling, subtract the fee from the profit. When buying, subtract the current price.

Practice

1. What is the main goal when solving the Buy and Sell Stocks problem with only one transaction allowed?

easy

A. Sell before buying to maximize profit

B. Minimize the buying price regardless of selling price

C. Buy and sell on the same day to avoid loss

D. Maximize the difference between a later selling price and an earlier buying price

Solution

1. Step 1: Understand the problem constraints

   The problem allows only one buy and one sell, so the goal is to find the maximum profit from one pair of days where buying day is before selling day.

2. Step 2: Identify the profit calculation

   Profit = selling price - buying price. We want to maximize this difference with the buying day before selling day.

3. Final Answer:

   Maximize the difference between a later selling price and an earlier buying price -> Option D

4. Quick Check:

   Profit = sell price - buy price [OK]

Hint: Find max difference with buy day before sell day [OK]

Common Mistakes:

• Ignoring the order of buy and sell days
• Trying to minimize price without considering selling
• Assuming multiple transactions allowed

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

easy

A. std::vector prices = (7, 1, 5, 3, 6, 4);

B. std::vector<int> prices = {7, 1, 5, 3, 6, 4};

C. vector prices = [7, 1, 5, 3, 6, 4];

D. std::vector<int> prices = (7, 1, 5, 3, 6, 4);

Solution

1. Step 1: Recall correct vector initialization syntax

   In C++, a vector of integers is initialized with curly braces and type specified: std::vector<int> prices = {values};

2. Step 2: Check each option

   std::vector<int> prices = {7, 1, 5, 3, 6, 4}; uses correct syntax with std::vector<int> and curly braces. Options A, B, and D use incorrect syntax for vector initialization.

3. Final Answer:

   std::vector<int> prices = {7, 1, 5, 3, 6, 4}; -> Option B

4. Quick Check:

   Use std::vector<int> and braces {} [OK]

Hint: Use std::vector<int> and curly braces {} to initialize [OK]

Common Mistakes:

• Using square brackets [] instead of curly braces {}
• Omitting the type <int>
• Using parentheses () instead of braces

3. What is the output of the following C++ code that finds the maximum profit with unlimited transactions allowed?

std::vector<int> prices = {7, 1, 5, 3, 6, 4};
int maxProfit = 0;
for (int i = 1; i < prices.size(); i++) {
    if (prices[i] > prices[i - 1]) {
        maxProfit += prices[i] - prices[i - 1];
    }
}
std::cout << maxProfit << std::endl;

medium

A. 7

B. 5

C. 8

D. 6

Solution

1. Step 1: Understand the loop and profit calculation

   The code adds profit for every day where price increases from previous day, summing all positive differences.

2. Step 2: Calculate profits day by day

   Day 1 to 2: 1 - 7 = -6 (no profit)
   Day 2 to 3: 5 - 1 = 4 (add 4)
   Day 3 to 4: 3 - 5 = -2 (no profit)
   Day 4 to 5: 6 - 3 = 3 (add 3)
   Day 5 to 6: 4 - 6 = -2 (no profit)
   Total profit = 4 + 3 = 7

3. Final Answer:

   7 -> Option A

4. Quick Check:

   Sum of positive differences = 7 [OK]

Hint: Sum all positive price differences for unlimited transactions [OK]

Common Mistakes:

• Adding negative differences as profit
• Only considering one transaction
• Miscounting indices in loop

4. Identify the bug in this C++ code snippet that tries to find the maximum profit with only one transaction allowed:

std::vector<int> prices = {7, 1, 5, 3, 6, 4};
int minPrice = INT_MAX;
int maxProfit = 0;
for (int i = 0; i < prices.size(); i++) {
    if (prices[i] < minPrice) {
        minPrice = prices[i];
    } else if (prices[i] - minPrice > maxProfit) {
        maxProfit = prices[i] - minPrice;
    }
}
std::cout << maxProfit << std::endl;

medium

A. No bug, code works correctly

B. minPrice should be initialized to 0

C. maxProfit should be initialized to INT_MIN

D. Loop should start from i = 1 instead of 0

Solution

1. Step 1: Analyze initialization and loop

   minPrice is set to INT_MAX to find minimum price correctly. maxProfit starts at 0. Loop from 0 to end checks each price.

2. Step 2: Check logic correctness

   For each price, update minPrice if current price is lower. Else, calculate profit and update maxProfit if higher. This correctly finds max profit with one transaction.

3. Final Answer:

   No bug, code works correctly -> Option A

4. Quick Check:

   Correct minPrice and maxProfit updates [OK]

Hint: Initialize minPrice to INT_MAX and loop from 0 [OK]

Common Mistakes:

• Starting loop from 1 and missing first price
• Initializing minPrice to 0 causing wrong min
• Initializing maxProfit to INT_MIN causing negative profits

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

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. We must find two pairs of buy-sell days maximizing total profit without overlap.

2. Step 2: Calculate max profit with two transactions

   First transaction: Buy at 3 (day 1), sell at 5 (day 3) profit = 2
   Second transaction: Buy at 0 (day 4), sell at 4 (day 8) profit = 4
   Total profit = 2 + 4 = 6

3. Final Answer:

   6 -> Option C

4. Quick Check:

   Sum of two best transactions = 6 [OK]

Hint: Split prices into two parts, sum max profits [OK]

Common Mistakes:

• Counting overlapping transactions
• Only considering one transaction
• Misidentifying buy/sell days