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. A common pitfall is to assume that more transactions always lead to higher profits, which is not necessarily true due to the constraints.

To solve this problem, we need to consider dynamic programming due to the overlapping subproblems and optimal substructure properties.

1. **Naive Solution**: A naive approach would be to try all possible pairs of buy and sell days for up to **k** transactions. This would be highly inefficient with a time complexity of O(n^k), where n is the number of days.

2. **Optimized Solution**: We can use dynamic programming to optimize this. We maintain two arrays: `dp[i][j]` where `i` is the number of transactions and `j` is the day. `dp[i][j]` represents the maximum profit using at most `i` transactions up to day `j`.

1. Initialize a 2D array `dp` of size `(k+1) x n` with all zeros.

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 either not doing the transaction on that day or doing the transaction and adding the profit from the previous transactions.

```
public class BestTimeToBuyAndSellStockIII {
public int maxProfit(int k, int[] prices) {
if (prices == null || prices.length == 0 || k == 0) {
return 0;
}
int n = prices.length;
if (k >= n / 2) {
return quickSolve(prices);
}
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];
}
private int quickSolve(int[] prices) {
int profit = 0;
for (int i = 1; i < prices.length; i++) {
if (prices[i] > prices[i - 1]) {
profit += prices[i] - prices[i - 1];
}
}
return profit;
}
}
```

The time complexity of the optimized solution is O(kn), where `k` is the number of transactions and `n` is the number of days. The space complexity is also O(kn) due to the 2D array used for dynamic programming.

1. If the number of days is zero, the profit is zero.

2. If the number of transactions is zero, the profit is zero.

3. If `k` is greater than or equal to `n/2`, it means we can perform as many transactions as we want, and the problem reduces to finding the maximum profit without any transaction limit.

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.
- Cases with decreasing prices where the profit should be zero.

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 understand different approaches.

In this blog post, we discussed the problem of finding the maximum profit from at most `k` transactions. We explored a naive solution and then optimized it using dynamic programming. We also analyzed the time and space complexity and discussed edge cases and testing strategies. Understanding and solving such problems is crucial for improving problem-solving skills and preparing for coding interviews.