Best Time to Buy and Sell Stock - C++ Solution with Time Complexity Analysis

Input: [2,4,1], k = 2
Output: 2
Explanation: Buy on day 1 (price = 2) and sell on day 2 (price = 4), profit = 4-2 = 2.

Input: [3,2,6,5,0,3], k = 2
Output: 7
Explanation: Buy on day 2 (price = 2) and sell on day 3 (price = 6), profit = 6-2 = 4.
             Then buy on day 5 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.

Understanding the Problem

The core challenge of this problem is to maximize the profit from at most k transactions. A transaction consists of buying and then selling one share of the stock. The constraint is that you cannot engage in multiple transactions simultaneously, meaning you must sell the stock before you can buy again.

This problem is significant in financial markets where traders aim to maximize their returns by strategically buying and selling stocks. A common pitfall is to overlook the constraint of non-overlapping transactions, which can lead to incorrect solutions.

Approach

To solve this problem, we can use dynamic programming. The idea is to maintain two arrays: one for the maximum profit up to the current day with at most j transactions, and another for the maximum profit up to the previous day with at most j-1 transactions.

We start with a naive solution that considers all possible pairs of buy and sell days, but this approach is not optimal due to its high time complexity. Instead, we use dynamic programming to optimize the solution.

Naive Solution

The naive solution involves iterating through all possible pairs of buy and sell days, which results in a time complexity of O(n^2). This is not feasible for large inputs.

Optimized Solution

We use dynamic programming to reduce the time complexity. The key idea is to use two arrays: dp and maxDiff. The dp array stores the maximum profit up to day i with at most j transactions, and the maxDiff array helps in calculating the maximum difference between the current price and the previous maximum profit.

Algorithm

1. Initialize a 2D array dp where dp[j][i] represents the maximum profit up to day i with at most j transactions.

2. Iterate through the number of transactions from 1 to k.

3. For each transaction, iterate through the days from 1 to n.

4. Use a variable maxDiff to keep track of the maximum difference between the current price and the previous maximum profit.

5. Update the dp array based on the maximum profit calculated using maxDiff.

Code Implementation

#include <vector>
#include <algorithm>
#include <climits>

using namespace std;

// Function to calculate the maximum profit with at most k transactions
int maxProfit(int k, vector<int>& prices) {
    int n = prices.size();
    if (n == 0) return 0;

    // If k is greater than n/2, it's equivalent to unlimited transactions
    if (k >= n / 2) {
        int maxProfit = 0;
        for (int i = 1; i < n; ++i) {
            if (prices[i] > prices[i - 1]) {
                maxProfit += prices[i] - prices[i - 1];
            }
        }
        return maxProfit;
    }

    // Initialize dp array
    vector<vector<int>> dp(k + 1, vector<int>(n, 0));

    for (int j = 1; j <= k; ++j) {
        int maxDiff = -prices[0];
        for (int i = 1; i < n; ++i) {
            dp[j][i] = max(dp[j][i - 1], prices[i] + maxDiff);
            maxDiff = max(maxDiff, dp[j - 1][i] - prices[i]);
        }
    }

    return dp[k][n - 1];
}

// Example usage
int main() {
    vector<int> prices1 = {2, 4, 1};
    int k1 = 2;
    cout << "Maximum profit for example 1: " << maxProfit(k1, prices1) << endl;

    vector<int> prices2 = {3, 2, 6, 5, 0, 3};
    int k2 = 2;
    cout << "Maximum profit for example 2: " << maxProfit(k2, prices2) << endl;

    return 0;
}

Complexity Analysis

The time complexity of the optimized solution is O(n*k), where n is the number of days and k is the number of transactions. The space complexity is also O(n*k) due to the use of the dp array.

Compared to the naive solution with a time complexity of O(n^2), the optimized solution is significantly more efficient, especially for large inputs.

Edge Cases

1. If the number of days n is 0, the maximum profit is 0.

2. If k is greater than or equal to n/2, it is equivalent to unlimited transactions.

3. If the prices array is strictly decreasing, the maximum profit is 0.

Testing

To test the solution comprehensively, consider the following test cases:

• Simple cases with small arrays.
• Cases with large arrays to test performance.
• Edge cases such as empty arrays or arrays with all elements the same.

Use testing frameworks like Google Test for C++ to automate the testing process.

Thinking and Problem-Solving Tips

1. Break down the problem into smaller subproblems and solve them using dynamic programming.

2. Understand the constraints and optimize the solution accordingly.

3. Practice solving similar problems to improve problem-solving skills.

Conclusion

In this blog post, we discussed the problem of finding the best time to buy and sell stock with at most k transactions. We explored a naive solution and an optimized solution using dynamic programming. We also analyzed the time and space complexity of the solutions and discussed edge cases and testing strategies.

Understanding and solving such problems is crucial for developing strong problem-solving skills and optimizing algorithms for real-world applications.

Additional Resources

• LeetCode Problem
• GeeksforGeeks Article
• TopCoder Dynamic Programming Tutorial