Buy and Sell Stocks All Variants in DSA Python - 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.

prices = [7,1,5,3,6,4]
max_profit = 0
min_price = float('inf')
for price in prices:
    if price < min_price:
        min_price = price
    elif price - min_price > max_profit:
        max_profit = price - min_price
print(max_profit)
    

This code finds the best profit from buying and selling once by scanning prices once.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

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

How Execution Grows With Input

As the number of days 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 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: We can find max profit by one pass, tracking the lowest price and best profit so far, without checking all pairs.

Interview Connect

Understanding how to scan prices efficiently shows you can find smart solutions that save time, a skill valued in real coding tasks.

Self-Check

"What if we could buy and sell multiple times? How would the time complexity change?"

Practice

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

easy

A. Find the day with the lowest stock price only

B. Minimize the number of transactions regardless of profit

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

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 C

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 finding lowest price day without selling
• Calculating averages instead of profit

2. Which of the following Python code snippets correctly initializes variables to track minimum price and maximum profit for the stock problem?

easy

A. min_price = float('inf'); max_profit = 0

B. min_price = 0; max_profit = float('-inf')

C. min_price = None; max_profit = None

D. min_price = 0; max_profit = 0

Solution

1. Step 1: Understand initialization needs

   We want min_price to start very high so any price is lower, and max_profit to start at zero.

2. Step 2: Check each option

   min_price = float('inf'); max_profit = 0 sets min_price to infinity and max_profit to zero, which is correct for tracking minimum and maximum values.

3. Final Answer:

   min_price = float('inf'); max_profit = 0 -> Option A

4. Quick Check:

   Initialization = min_price high, max_profit zero [OK]

Hint: Use infinity for min_price start, zero for max_profit [OK]

Common Mistakes:

• Setting min_price to zero causes wrong minimum tracking
• Using None without checks causes errors
• Setting max_profit to negative infinity can confuse logic

3. What is the output of the following code snippet for prices = [7,1,5,3,6,4]?

min_price = float('inf')
max_profit = 0
for price in prices:
    if price < min_price:
        min_price = price
    elif price - min_price > max_profit:
        max_profit = price - min_price
print(max_profit)

medium

A. 6

B. 7

C. 4

D. 5

Solution

1. Step 1: Trace the loop with given prices

   Prices: 7,1,5,3,6,4 - Start min_price=inf, max_profit=0 - price=7: min_price=7 - price=1: min_price=1 (lower) - price=5: profit=5-1=4 > 0, max_profit=4 - price=3: profit=3-1=2 < 4, no change - price=6: profit=6-1=5 > 4, max_profit=5 - price=4: profit=4-1=3 < 5, no change

2. Step 2: Final max_profit after loop

   max_profit is 5 after checking all prices.

3. Final Answer:

   5 -> Option D

4. Quick Check:

   Max profit = 5 [OK]

Hint: Track min price and update max profit on higher prices [OK]

Common Mistakes:

• Confusing max profit with max price
• Not updating min_price correctly
• Calculating profit incorrectly

4. Identify the error in this code snippet for the stock problem:

prices = [7,1,5,3,6,4]
min_price = 0
max_profit = 0
for price in prices:
    if price < min_price:
        min_price = price
    elif price - min_price > max_profit:
        max_profit = price - min_price
print(max_profit)

medium

A. Loop should start from second element

B. min_price initialized to 0 instead of a high value

C. max_profit should start at float('inf')

D. Using elif instead of if causes logic error

Solution

1. Step 1: Check variable initialization

   min_price is set to 0, but prices include values greater than 0, so min_price never updates to actual minimum (1).

2. Step 2: Effect of wrong min_price initialization

   Since min_price=0, price < min_price is never true, so min_price stays 0, causing incorrect profit calculations.

3. Final Answer:

   min_price initialized to 0 instead of a high value -> Option B

4. Quick Check:

   min_price must start high to track minimum [OK]

Hint: Initialize min_price to infinity, not zero [OK]

Common Mistakes:

• Starting min_price at zero causes wrong minimum tracking
• Confusing max_profit initialization
• Thinking elif causes logic error here

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. 6

B. 7

C. 5

D. 4

Solution

1. Step 1: Understand two transactions limit

   We can buy and sell stocks at most twice to maximize total profit.

2. Step 2: Calculate profits for two transactions

   First transaction: buy at 3, sell at 5 = 2 profit Second transaction: buy at 0, sell at 4 = 4 profit Total profit = 2 + 4 = 6

3. Final Answer:

   6 -> Option A

4. Quick Check:

   Two transactions max profit = 6 [OK]

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

Common Mistakes:

• Counting overlapping transactions
• Only considering one transaction
• Adding profits incorrectly