Say you have an array for which the *i-*th element is the price of a given stock on day *i*.

Design an algorithm to find the maximum profit. You may complete at most **k** transactions.

**Note:**

You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

**Example 1:**

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

**Example 2:**

Input:[3,2,6,5,0,3], k = 2Output:7Explanation: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.

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. Potential pitfalls include misunderstanding the constraint of non-overlapping transactions and not considering the optimal substructure of the problem.

To solve this problem, we can use dynamic programming. The idea is to maintain a table where the entry `dp[i][j]`

represents the maximum profit achievable using at most `i`

transactions up to day `j`

.

1. **Naive Solution**: A naive solution would involve trying all possible pairs of buy and sell days for each transaction, which would be computationally expensive and inefficient.

2. **Optimized Solution**: We can optimize this using dynamic programming. The key observation is that the maximum profit up to day `j`

with `i`

transactions can be derived from the maximum profit up to day `j-1`

with `i`

transactions and the maximum profit up to day `j`

with `i-1`

transactions.

1. Initialize a 2D array `dp`

where `dp[i][j]`

represents the maximum profit using at most `i`

transactions up to day `j`

.

2. Iterate over the number of transactions from 1 to `k`

.

3. For each transaction, iterate over the days from 1 to `n-1`

.

4. For each day, calculate the maximum profit by considering whether to sell on that day or not.

5. Update the `dp`

table accordingly.

```
public class BestTimeToBuyAndSellStock {
public int maxProfit(int k, int[] prices) {
int n = prices.length;
if (n == 0) return 0;
// If k is greater than n/2, then 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;
}
// dp[i][j] represents the max profit up to day j with at most i transactions
int[][] dp = new int[k + 1][n];
for (int i = 1; i <= k; i++) {
int maxDiff = -prices[0];
for (int j = 1; j < n; j++) {
dp[i][j] = Math.max(dp[i][j - 1], prices[j] + maxDiff);
maxDiff = Math.max(maxDiff, dp[i - 1][j] - prices[j]);
}
}
return dp[k][n - 1];
}
public static void main(String[] args) {
BestTimeToBuyAndSellStock solution = new BestTimeToBuyAndSellStock();
int[] prices1 = {2, 4, 1};
int k1 = 2;
System.out.println("Max Profit: " + solution.maxProfit(k1, prices1)); // Output: 2
int[] prices2 = {3, 2, 6, 5, 0, 3};
int k2 = 2;
System.out.println("Max Profit: " + solution.maxProfit(k2, prices2)); // Output: 7
}
}
```

The time complexity of this approach 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 2D array used for dynamic programming.

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.

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

- Simple cases with small arrays.
- Cases where
`k`

is greater than or equal to`n/2`

. - Edge cases with empty arrays or arrays with one element.
- Arrays with strictly increasing or decreasing prices.

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 similar problems to improve problem-solving skills and gain a deeper understanding of dynamic programming techniques.

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 dynamic programming approach to solve the problem efficiently and provided a detailed explanation of the algorithm and its implementation in Java. By understanding and practicing such problems, you can improve your problem-solving skills and gain a deeper understanding of dynamic programming techniques.