# Best Time To Buy And Sell Stock Iii Problem

## Description

LeetCode Problem 123.

You are given an array prices where prices[i] is the price of a given stock on the i^th day. Find the maximum profit you can achieve. You may complete at most two transactions.

Note: You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).

Example 1:

``````1
2
3
4
Input: prices = [3,3,5,0,0,3,1,4]
Output: 6
Explanation: Buy on day 4 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.
Then buy on day 7 (price = 1) and sell on day 8 (price = 4), profit = 4-1 = 3.
``````

Example 2:

``````1
2
3
4
Input: prices = [1,2,3,4,5]
Output: 4
Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are engaging multiple transactions at the same time. You must sell before buying again.
``````

Example 3:

``````1
2
3
Input: prices = [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.
``````

Example 4:

``````1
2
Input: prices = [1]
Output: 0
``````

Constraints:

• 1 <= prices.length <= 10^5
• 0 <= prices[i] <= 10^5

## Sample C++ Code

``````1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
class Solution {
public:
int MaxProfitDp(int[] prices) {
// DP formula:
// dp[k, i] = max(dp[k, i-1], prices[i] - prices[j] + dp[k-1, j-1]), j=[0..i-1]
if (prices.Length == 0) return 0;
var dp = new int[3, prices.Length];
for (int k = 1; k <= 2; k++) {
int min = prices[0];
for (int i = 1; i < prices.Length; i++) {
min = Math.Min(min, prices[i] - dp[k-1, i-1]);
dp[k, i] = Math.Max(dp[k, i-1], prices[i] - min);
}
}

return dp[2, prices.Length - 1];
}
};
``````