Buy and Sell Stocks All Variants in DSA Javascript - 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 changes as the number of days grows.

We ask: How does the number of price checks and decisions grow with more days?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function maxProfit(prices) {
  let minPrice = Infinity;
  let maxProfit = 0;
  for (let 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 one buy and one sell by scanning prices once.

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 price (n times).

How Execution Grows With Input

As the number of days (prices) grows, the code checks each price once.

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

Pattern observation: The operations grow directly with the number of prices.

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 single pass approach keeps track of the lowest price and max profit, so no need to check all pairs.

Interview Connect

Understanding this linear time solution shows you can find efficient answers by smartly scanning data once, a key skill in interviews.

Self-Check

"What if we wanted to allow multiple buy and sell transactions? How would the time complexity change?"

Practice

1. What is the main goal when solving the Buy and Sell Stocks problem variants?

easy

A. Maximize the profit by choosing the best days to buy and sell stocks

B. Minimize the number of transactions regardless of profit

C. Find the day with the highest stock price only

D. Calculate the average stock price over all days

Solution

1. Step 1: Understand the problem goal

   The problem asks to find the best days to buy and sell stocks to get the highest profit.

2. Step 2: Identify the objective

   Maximizing profit means buying low and selling high at the right times.

3. Final Answer:

   Maximize the profit by choosing the best days to buy and sell stocks -> Option A

4. Quick Check:

   Goal = Maximize profit [OK]

Hint: Focus on profit maximization, not just prices or counts [OK]

Common Mistakes:

• Thinking the goal is to minimize transactions
• Only looking for highest price day
• Calculating average price instead of profit

2. Which of the following is the correct syntax to initialize a variable maxProfit to 0 in JavaScript?

easy

A. var maxProfit == 0;

B. let maxProfit = 0;

C. maxProfit := 0;

D. int maxProfit = 0;

Solution

1. Step 1: Identify correct JavaScript variable declaration

   JavaScript uses let or const for variable declaration and single equals = for assignment.

2. Step 2: Check each option

   let maxProfit = 0; uses let maxProfit = 0; which is correct syntax.

3. Final Answer:

   let maxProfit = 0; -> Option B

4. Quick Check:

   Variable declaration with let and = assignment [OK]

Hint: Use let or const with single = for assignment in JavaScript [OK]

Common Mistakes:

• Using double equals == for assignment
• Using := which is not JavaScript syntax
• Using int which is not a JavaScript keyword

3. What will be the output of the following code snippet?

const prices = [7,1,5,3,6,4];
let minPrice = Infinity;
let maxProfit = 0;
for (let price of prices) {
  if (price < minPrice) minPrice = price;
  else if (price - minPrice > maxProfit) maxProfit = price - minPrice;
}
console.log(maxProfit);

medium

A. 7

B. 6

C. 4

D. 5

Solution

1. Step 1: Track minimum price and maximum profit

   Iterate prices: minPrice updates to 7, then 1; maxProfit updates when price - minPrice is larger.

2. Step 2: Calculate maxProfit step-by-step

   Prices: 7 (minPrice=7), 1 (minPrice=1), 5 (profit=4), 3 (profit=2), 6 (profit=5), 4 (profit=3). Max profit is 5.

3. Final Answer:

   5 -> Option D

4. Quick Check:

   Max profit = 5 [OK]

Hint: Track min price and update max profit on the fly [OK]

Common Mistakes:

• Confusing minPrice update logic
• Calculating profit incorrectly
• Using max price instead of max profit

4. The following code is intended to find the maximum profit from one transaction but has a bug. What is the bug?

function maxProfit(prices) {
  let minPrice = 0;
  let maxProfit = 0;
  for (let price of prices) {
    if (price < minPrice) minPrice = price;
    else if (price - minPrice > maxProfit) maxProfit = price - minPrice;
  }
  return maxProfit;
}

medium

A. minPrice should be initialized to Infinity, not 0

B. maxProfit should be initialized to -Infinity

C. The for loop should use index instead of for-of

D. The return statement is missing

Solution

1. Step 1: Check initialization of minPrice

   minPrice is set to 0, but prices can be greater than 0, so minPrice never updates correctly if all prices are above 0.

2. Step 2: Correct initialization

   minPrice should start at Infinity to ensure it updates to the first price correctly.

3. Final Answer:

   minPrice should be initialized to Infinity, not 0 -> Option A

4. Quick Check:

   minPrice init = Infinity [OK]

Hint: Initialize minPrice to Infinity to track minimum correctly [OK]

Common Mistakes:

• Initializing minPrice to 0 causes wrong min tracking
• Thinking maxProfit needs -Infinity initialization
• Confusing loop types for this problem

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 twice to maximize profit, but transactions cannot overlap.

2. Step 2: Calculate profits for two transactions

   First transaction: buy at 3, sell at 5 (profit 2). Second transaction: buy at 0, sell at 4 (profit 5). Total profit = 2 + 5 = 7.

3. Final Answer:

   7 -> Option C

4. Quick Check:

   Two transactions profit = 7 [OK]

Hint: Split array, find max profit in two parts, sum them [OK]

Common Mistakes:

• Trying to do more than two transactions
• Adding profits from overlapping transactions
• Missing the best buy/sell days for each transaction