Buy and Sell Stocks All Variants in DSA Typescript - Time & Space Complexity

Time Complexity: Buy and Sell Stocks All Variants

O(n)

Understanding Time Complexity

When solving buy and sell stock problems, we want to know how the time needed grows as the number of days increases.

We ask: How does the number of price checks and decisions change when we have more days?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function maxProfit(prices: number[]): number {
  let minPrice = Infinity;
  let maxProfit = 0;
  for (const price of prices) {
    if (price < minPrice) minPrice = price;
    else if (price - minPrice > maxProfit) maxProfit = price - minPrice;
  }
  return maxProfit;
}
    

This code finds the best profit from buying and selling once by checking each day's price.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: One loop through the prices array.
• How many times: Exactly once for each day (n times).

How Execution Grows With Input

As the number of days increases, the code checks each price once.

Input Size (n)	Approx. Operations
10	10 checks
100	100 checks
1000	1000 checks

Pattern observation: The work grows directly with the number of days.

Final Time Complexity

Time Complexity: O(n)

This means the time needed grows in a straight line with the number of days.

Common Mistake

[X] Wrong: "Checking every pair of days is needed to find max profit."

[OK] Correct: The code smartly tracks the lowest price so far and only compares current price to it, avoiding extra checks.

Interview Connect

Understanding this time complexity shows you can write efficient code that scales well, a key skill interviewers look for.

Self-Check

"What if we allowed unlimited transactions instead of just one? How would the time complexity change?"

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