Buy and Sell Stocks All Variants in DSA Typescript - Execution Trace

Concept Flow - Buy and Sell Stocks All Variants

Start with price array

↓

Choose variant: single, multiple, cooldown, fee

↓

Initialize variables: buy, sell, cooldown, fee

↓

Iterate over prices

↓

Update buy/sell states based on rules

↓

Track max profit

↓

End iteration

↓

Return max profit

The flow starts with the price list, selects the variant rule, iterates over prices updating buy/sell states, and finally returns the max profit.

Execution Sample

DSA Typescript

const prices = [7,1,5,3,6,4];
let buy = -Infinity;
let sell = 0;
for (const price of prices) {
  buy = Math.max(buy, -price);
  sell = Math.max(sell, buy + price);
}
console.log(sell);

This code finds max profit for single buy-sell transaction by tracking best buy and sell states.

Execution Table

Step	Operation	Price	buy (max profit after buy)	sell (max profit after sell)	Visual State
1	Initialize	-	-Infinity	0	buy=-∞, sell=0
2	Price=7	7	max(-∞, -7) = -7	max(0, -7+7) = 0	buy=-7, sell=0
3	Price=1	1	max(-7, -1) = -1	max(0, -1+1) = 0	buy=-1, sell=0
4	Price=5	5	max(-1, -5) = -1	max(0, -1+5) = 4	buy=-1, sell=4
5	Price=3	3	max(-1, -3) = -1	max(4, -1+3) = 4	buy=-1, sell=4
6	Price=6	6	max(-1, -6) = -1	max(4, -1+6) = 5	buy=-1, sell=5
7	Price=4	4	max(-1, -4) = -1	max(5, -1+4) = 5	buy=-1, sell=5
8	End	-	-	-	Max profit = 5

💡 All prices processed, max profit is sell = 5

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 4	After Step 5	After Step 6	After Step 7	Final
buy	-Infinity	-7	-1	-1	-1	-1	-1	-1
sell	0	0	0	4	4	5	5	5

Key Moments - 3 Insights

Why do we initialize buy with -Infinity?

We start with buy = -Infinity to ensure any price will update buy to a valid negative value representing cost. See execution_table row 1 and 2 where buy changes from -Infinity to -7.

Why does sell never decrease?

sell tracks max profit after selling, so it only updates if a better profit is found. It never decreases because we use max() function. See execution_table rows 3 to 7 where sell stays or increases.

Why do we update buy with max(buy, -price)?

buy represents the best cost to buy so far (negative price). max(buy, -price) means we keep the best (highest) buy value, which is the least cost. See execution_table rows 2 to 7.

Visual Quiz - 3 Questions

Look at the execution_table at Step 4, what is the value of sell?

A 0

B -1

C 4

D 5

💡 Hint

Check the 'sell' column in execution_table row 4.

At which step does buy first update from -Infinity?

A Step 1

B Step 2

C Step 3

D Step 4

💡 Hint

Look at the 'buy' column in execution_table rows 1 and 2.

If prices were all increasing, how would sell change?

A sell would increase steadily

B sell would decrease

C sell would stay 0

D sell would fluctuate

💡 Hint

Refer to how sell updates in execution_table when price increases.

Concept Snapshot

Buy and Sell Stocks Variants:
- Track buy (max profit after buy) and sell (max profit after sell)
- For single transaction: buy = max(buy, -price), sell = max(sell, buy + price)
- For multiple transactions, cooldown, or fee, update states accordingly
- Return sell as max profit
- Use iteration over prices to update states

Full Transcript

This visualization shows how to find the maximum profit from buying and selling stocks in different variants. We start with a price list and initialize buy to negative infinity and sell to zero. We iterate over each price, updating buy to the maximum of current buy or negative current price, representing the best buying cost. We update sell to the maximum of current sell or buy plus current price, representing the best profit after selling. The execution table tracks these values step-by-step. Key moments clarify why buy starts at negative infinity, why sell never decreases, and why buy updates with max(buy, -price). The quiz tests understanding of these updates. The concept snapshot summarizes the approach for all variants.

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