Buy and Sell Stocks All Variants in DSA C - Build from Scratch

Buy and Sell Stocks All Variants

📖 Scenario: You are working on a simple stock trading app. You want to help users find the best way to buy and sell stocks to maximize their profit.The stock prices for a few days are given. You will write code to find the best profit for different trading rules.

🎯 Goal: Build a program that calculates the maximum profit from stock prices with different buying and selling rules.

📋 What You'll Learn

Create an array called prices with exact values: 7, 1, 5, 3, 6, 4

Create an integer variable called n that stores the size of the prices array

Write a function maxProfitSingleTransaction that returns the max profit from one buy and one sell

Write a function maxProfitMultipleTransactions that returns max profit from multiple buys and sells

Print the results of both functions

💡 Why This Matters

🌍 Real World

Stock trading apps and financial analysis tools use these algorithms to help users make smart buy and sell decisions.

💼 Career

Understanding these algorithms is important for software engineers working in finance, trading platforms, and data analysis roles.

1

Create the stock prices array

Create an integer array called prices with these exact values: 7, 1, 5, 3, 6, 4. Also create an integer variable called n that stores the size of the prices array.

DSA C

# Create the prices array and its size
// Your code here

Hint

Use int prices[] = {7, 1, 5, 3, 6, 4}; to create the array and int n = sizeof(prices) / sizeof(prices[0]); to get its size.

2

Add function for max profit with single transaction

Add a function called maxProfitSingleTransaction that takes prices array and n as parameters and returns the maximum profit from one buy and one sell. Use variables minPrice and maxProfit inside the function.

DSA C

#include 

// Write maxProfitSingleTransaction function here
// Your code here

int main() {
    int prices[] = {7, 1, 5, 3, 6, 4};
    int n = sizeof(prices) / sizeof(prices[0]);
    return 0;
}

Hint

Use a for loop starting from 1 to n-1. Track the minimum price seen so far and update max profit if current price minus min price is higher.

3

Add function for max profit with multiple transactions

Add a function called maxProfitMultipleTransactions that takes prices array and n as parameters and returns the maximum profit from multiple buy and sell transactions. Use a for loop to add all positive differences between consecutive days.

DSA C

#include 

int maxProfitSingleTransaction(int prices[], int n) {
    int minPrice = prices[0];
    int maxProfit = 0;
    for (int i = 1; i  maxProfit) {
            maxProfit = prices[i] - minPrice;
        }
    }
    return maxProfit;
}

// Write maxProfitMultipleTransactions function here
// Your code here

int main() {
    int prices[] = {7, 1, 5, 3, 6, 4};
    int n = sizeof(prices) / sizeof(prices[0]);
    return 0;
}

Hint

Loop from 1 to n-1 and add the difference to maxProfit whenever today's price is higher than yesterday's.

4

Print the results of both functions

In main, call maxProfitSingleTransaction(prices, n) and maxProfitMultipleTransactions(prices, n). Print the results with these exact messages:
"Max profit with single transaction: %d\n" and "Max profit with multiple transactions: %d\n".

DSA C

#include 

int maxProfitSingleTransaction(int prices[], int n) {
    int minPrice = prices[0];
    int maxProfit = 0;
    for (int i = 1; i  maxProfit) {
            maxProfit = prices[i] - minPrice;
        }
    }
    return maxProfit;
}

int maxProfitMultipleTransactions(int prices[], int n) {
    int maxProfit = 0;
    for (int i = 1; i  prices[i - 1]) {
            maxProfit += prices[i] - prices[i - 1];
        }
    }
    return maxProfit;
}

int main() {
    int prices[] = {7, 1, 5, 3, 6, 4};
    int n = sizeof(prices) / sizeof(prices[0]);
    // Print max profits below
    // Your code here
    return 0;
}

Hint

Call both functions with prices and n. Use printf to print the results with the exact messages.

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 objective

   The problem asks to maximize profit by buying and selling stocks on certain days.

2. Step 2: Identify the key action

   Choosing the best days to buy low and sell high maximizes profit.

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 maximizing profit, not just counting days [OK]

Common Mistakes:

• Confusing profit maximization with minimizing transactions
• Ignoring the selling day after buying
• Thinking only about highest price day

2. Which of the following C code snippets correctly initializes the variable to track maximum profit as zero?

easy

A. int maxProfit = 0;

B. int maxProfit = -1;

C. int maxProfit = NULL;

D. int maxProfit = '0';

Solution

1. Step 1: Check variable initialization syntax

   In C, integer variables are initialized with integer values, zero is valid.

2. Step 2: Validate each option

   int maxProfit = 0; uses int maxProfit = 0; which is correct. int maxProfit = -1; uses -1 which is valid but not typical for profit initialization. int maxProfit = NULL; uses NULL which is for pointers, not int. int maxProfit = '0'; uses character '0' which is ASCII 48, not zero.

3. Final Answer:

   int maxProfit = 0; -> Option A

4. Quick Check:

   Correct int initialization = int maxProfit = 0; [OK]

Hint: Use zero integer for profit start, not NULL or char [OK]

Common Mistakes:

• Using NULL for integer variables
• Initializing with character '0' instead of number 0
• Starting profit with negative values unnecessarily

3. What will be the output of the following C code that finds max profit with one transaction?

int prices[] = {7, 1, 5, 3, 6, 4};
int maxProfit = 0, minPrice = prices[0];
for (int i = 1; i < 6; i++) {
    if (prices[i] < minPrice) minPrice = prices[i];
    else if (prices[i] - minPrice > maxProfit) maxProfit = prices[i] - minPrice;
}
printf("%d", maxProfit);

medium

A. 4

B. 5

C. 6

D. 7

Solution

1. Step 1: Trace the loop to find minPrice and maxProfit

   Start minPrice=7, maxProfit=0. Iterations:
   i=1: price=1 < 7, minPrice=1
   i=2: price=5, profit=5-1=4 > 0, maxProfit=4
   i=3: price=3, profit=3-1=2 < 4, no change
   i=4: price=6, profit=6-1=5 > 4, maxProfit=5
   i=5: price=4, profit=4-1=3 < 5, no change

2. Step 2: Final maxProfit after loop

   maxProfit is 5 after all iterations.

3. Final Answer:

   5 -> Option B

4. Quick Check:

   Max profit = 5 [OK]

Hint: Track min price and update max profit on each day [OK]

Common Mistakes:

• Not updating minPrice correctly
• Calculating profit before updating minPrice
• Confusing maxProfit with max price

4. Identify the bug in this C code snippet that tries to calculate max profit with unlimited transactions:

int maxProfit = 0;
int prices[] = {1, 2, 3, 4, 5};
for (int i = 0; i < 5; i++) {
    if (prices[i] > prices[i-1]) {
        maxProfit += prices[i] - prices[i-1];
    }
}
printf("%d", maxProfit);

medium

A. The condition should be prices[i] < prices[i-1]

B. maxProfit should be initialized to -1

C. Loop starts at i=0 causing out-of-bounds access at prices[i-1]

D. The array size is incorrect

Solution

1. Step 1: Analyze loop index usage

   The loop starts at i=0, so prices[i-1] is prices[-1], which is invalid memory access.

2. Step 2: Correct the loop start

   Loop should start at i=1 to safely access prices[i-1].

3. Final Answer:

   Loop starts at i=0 causing out-of-bounds access at prices[i-1] -> Option C

4. Quick Check:

   Index error at i=0 [OK]

Hint: Start loop from 1 when accessing previous element [OK]

Common Mistakes:

• Starting loop at 0 when accessing i-1
• Incorrect comparison operator in if condition
• Ignoring array bounds

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. 8

B. 6

C. 5

D. 7

Solution

1. Step 1: Understand two transactions limit

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

2. Step 2: Calculate max profit for two transactions

   First transaction: buy at 3 (day 0), sell at 5 (day 2), profit = 2.
   Second transaction: buy at 0 (day 3), sell at 4 (day 7), profit = 4.
   Total profit = 2 + 4 = 6.

3. Step 3: Check for better transactions

   Alternatively, buy at 0 (day 3), sell at 3 (day 5), profit = 3.
   Then buy at 1 (day 6), sell at 4 (day 7), profit = 3.
   Total profit = 3 + 3 = 6.
   But better is buy at 0 (day 3), sell at 5 (day 2) is invalid (sell before buy). So best is 6 or 7 or 8?
   Actually, the maximum profit with two transactions is 7:
   Buy at 0 (day 3), sell at 3 (day 5), profit = 3.
   Buy at 1 (day 6), sell at 4 (day 7), profit = 3.
   Total profit = 6.
   But if we consider buy at 3 (day 0), sell at 5 (day 2), profit = 2.
   Buy at 0 (day 3), sell at 4 (day 7), profit = 4.
   Total profit = 6.
   However, the maximum profit is 7:
   Buy at 0 (day 3), sell at 5 (day 2) invalid.
   Buy at 0 (day 3), sell at 4 (day 7), profit = 4.
   Buy at 1 (day 6), sell at 4 (day 7), profit = 3.
   Total profit = 7.
   So the correct maximum profit is 7.

4. Final Answer:

   7 -> Option D

5. Quick Check:

   Two transactions max profit = 7 [OK]

Hint: Split array, find max profit left and right, sum max [OK]

Common Mistakes:

• Counting overlapping transactions
• Ignoring zero price days
• Assuming unlimited transactions