# Best Time to Buy and Sell Stock III

Updated: Mar 24, 2021

You are given an array prices where prices[i] is the price of a given stock on the ith 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:**

**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:**

**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:**

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

**Example 4:**

**Input****:** prices = [1]
**Output****:** 0

**Constraints:**

1 <= prices.length <= 105

0 <= prices[i] <= 105

**Solution:**

```
class Solution {
public int maxProfit(int[] prices) {
if(prices.length<2) return 0;
int k=2;
int havingstock;
int[][] nothavingstock = new int[k+1][prices.length+1];
for(int t=1;t<=k;t++)
{
havingstock = -prices[0];
for(int i=1;i<prices.length;i++)
{
havingstock = Math.max(havingstock,nothavingstock[t-1][i-1]-prices[i]);
nothavingstock[t][i] = Math.max(nothavingstock[t][i-1],havingstock+prices[i]);
}
}
return nothavingstock[k][prices.length-1];
}
}
```