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

Practice - 5 Tasks

1fill in blank

easy

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

DSA Typescript

function maxProfit(prices: number[]): number {
  let minPrice = Infinity;
  let maxProfit = 0;
  for (let i = 0; i < prices.length; i++) {
    if (prices[i] < [1]) {
      minPrice = prices[i];
    } else if (prices[i] - minPrice > maxProfit) {
      maxProfit = prices[i] - minPrice;
    }
  }
  return maxProfit;
}

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

AminPrice

Bprices[i]

CmaxProfit

DInfinity

Attempts:

3 left

💡 Hint

Common Mistakes

Using maxProfit instead of minPrice in the comparison.

Comparing prices[i] with Infinity directly.

We update minPrice when we find a price lower than the current minimum. So the condition checks if prices[i] is less than minPrice.

2fill in blank

medium

Complete the code to calculate maximum profit with unlimited transactions (buy and sell multiple times).

DSA Typescript

function maxProfit(prices: number[]): number {
  let profit = 0;
  for (let 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]

C0

Dprices[i + 1]

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing prices[i] with prices[i] (always false).

Using prices[i + 1] which may be out of bounds.

We buy at prices[i - 1] and sell at prices[i] if the price increased, so we check if prices[i] is greater than prices[i - 1].

3fill in blank

hard

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

DSA Typescript

function maxProfit(prices: number[]): number {
  let firstBuy = Infinity, firstProfit = 0;
  let secondBuy = Infinity, secondProfit = 0;
  for (const price of prices) {
    firstBuy = Math.min(firstBuy, price);
    firstProfit = Math.max(firstProfit, price - firstBuy);
    secondBuy = Math.min(secondBuy, price - [1]);
    secondProfit = Math.max(secondProfit, price - secondBuy);
  }
  return secondProfit;
}

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

AsecondProfit

BsecondBuy

CfirstBuy

DfirstProfit

Attempts:

3 left

💡 Hint

Common Mistakes

Subtracting firstBuy instead of firstProfit.

Using secondProfit or secondBuy incorrectly.

When calculating secondBuy, we subtract the profit from the first transaction (firstProfit) to simulate reinvesting that profit.

4fill in blank

hard

Fill both blanks to implement max profit with cooldown (cannot buy immediately after selling).

DSA Typescript

function maxProfit(prices: number[]): number {
  let sell = 0, prevSell = 0, buy = -Infinity, prevBuy;
  for (const price of prices) {
    prevBuy = buy;
    buy = Math.max(prevBuy, [1] - price);
    prevSell = sell;
    sell = Math.max(prevSell, [2] + price);
  }
  return sell;
}

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

AprevSell

Bsell

CprevBuy

Dbuy

Attempts:

3 left

💡 Hint

Common Mistakes

Using sell instead of prevSell for buying.

Using buy instead of prevBuy for selling.

For buying, we compare previous buy and previous sell minus current price (prevSell - price), so buy = Math.max(prevBuy, prevSell - price).

For selling, we compare previous sell and previous buy plus current price (prevBuy + price), so sell = Math.max(prevSell, prevBuy + price).

5fill in blank

hard

Fill both blanks to implement max profit with transaction fee.

DSA Typescript

function maxProfit(prices: number[], fee: number): number {
  let cash = 0, hold = -prices[0];
  for (let i = 1; i < prices.length; i++) {
    cash = Math.max(cash, hold + prices[i] - [1]);
    hold = Math.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, we pay the transaction fee, so subtract fee from the price.

When buying, we subtract the current price from cash.

Practice

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

easy

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

B. Minimize the number of transactions

C. Find the lowest price in the array

D. Calculate the average stock price

Solution

1. Step 1: Understand the problem goal

   The problem asks to find the maximum profit by buying and selling once, so we want the biggest positive difference between a sell price after a buy price.

2. Step 2: Identify the correct approach

   Maximizing the difference between a later selling price and an earlier buying price directly solves the problem.

3. Final Answer:

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

4. Quick Check:

   Goal = Maximize profit from one buy and one sell [OK]

Hint: Look for max difference with sell after buy [OK]

Common Mistakes:

• Thinking to minimize transactions instead of maximizing profit
• Confusing lowest price with max profit
• Ignoring the order of buy before sell

2. Which of the following TypeScript function signatures correctly represents a function to find max profit from multiple stock transactions?

easy

A. function maxProfit(prices: number[]): number

B. function maxProfit(prices: string): number[]

C. function maxProfit(prices: number): string

D. function maxProfit(prices: any): void

Solution

1. Step 1: Analyze the input and output types

   The input should be an array of numbers representing prices, so prices: number[]. The output is the maximum profit, a number.

2. Step 2: Match the correct function signature

   function maxProfit(prices: number[]): number matches input as number array and output as number, which is correct.

3. Final Answer:

   function maxProfit(prices: number[]): number -> Option A

4. Quick Check:

   Input array, output number [OK]

Hint: Input array of numbers, output single number [OK]

Common Mistakes:

• Using string type for prices
• Returning array instead of number
• Using void return type

3. What is the output of the following TypeScript code for max profit with unlimited transactions?

function maxProfit(prices: number[]): number {
  let profit = 0;
  for (let i = 1; i < prices.length; i++) {
    if (prices[i] > prices[i - 1]) {
      profit += prices[i] - prices[i - 1];
    }
  }
  return profit;
}

console.log(maxProfit([7,1,5,3,6,4]));

medium

A. 5

B. 7

C. 8

D. 6

Solution

1. Step 1: Trace the loop and calculate profit increments

   Prices: [7,1,5,3,6,4] - i=1: 1 > 7? No, profit=0 - i=2: 5 > 1? Yes, profit += 5-1=4 - i=3: 3 > 5? No, profit=4 - i=4: 6 > 3? Yes, profit += 6-3=3, total=7 - i=5: 4 > 6? No, profit=7

2. Step 2: Return total profit

   Total profit after loop is 7.

3. Final Answer:

   7 -> Option B

4. Quick Check:

   Sum of positive diffs = 7 [OK]

Hint: Add all positive price differences [OK]

Common Mistakes:

• Adding negative differences
• Only considering one transaction
• Misreading array indices

4. Identify the bug in this TypeScript code for max profit with one transaction:

function maxProfit(prices: number[]): number {
  let minPrice = prices[0];
  let maxProfit = 0;
  for (let i = 1; i <= prices.length; i++) {
    if (prices[i] < minPrice) {
      minPrice = prices[i];
    } else if (prices[i] - minPrice > maxProfit) {
      maxProfit = prices[i] - minPrice;
    }
  }
  return maxProfit;
}

medium

A. prices array should be sorted first

B. minPrice should be initialized to 0

C. maxProfit should start at -Infinity

D. Loop condition should be < instead of <=

Solution

1. Step 1: Check loop boundary

   The loop runs with i from 1 to prices.length inclusive, but array indices go from 0 to prices.length - 1. Using <= causes out-of-bounds access at prices[prices.length].

2. Step 2: Correct the loop condition

   Change loop condition to i < prices.length to avoid accessing undefined elements.

3. Final Answer:

   Loop condition should be < instead of <= -> Option D

4. Quick Check:

   Array index out-of-bounds fixed by < condition [OK]

Hint: Use < for array loops, not <= [OK]

Common Mistakes:

• Using <= causing runtime errors
• Initializing minPrice incorrectly
• Sorting prices unnecessarily

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

hard

A. 5

B. 6

C. 7

D. 4

Solution

1. Step 1: Understand two transactions limit

   We can buy and sell at most twice. We want to split the array into two parts and maximize profit from each.

2. Step 2: Calculate max profit for each split

   Best first transaction: buy at 3, sell at 5 = 2 profit Best second transaction: buy at 0, sell at 4 = 4 profit Total max profit = 2 + 4 = 6

3. Step 3: Check for better splits

   Splitting at index 5: first part max profit = 4 (buy at 0 sell at 3), second part max profit = 3 (buy at 1 sell at 4), total = 7

4. Final Answer:

   7 -> Option C

5. Quick Check:

   Sum of two best transactions = 7 [OK]

Hint: Split array, sum max profits from two parts [OK]

Common Mistakes:

• Counting more than two transactions
• Not splitting array correctly
• Choosing wrong buy/sell days