Hi,

I just come across this problem, and it's very frustating since I'm bad at DP.

So I just draw all the actions that can be done.

Here is the drawing (Feel like an elementary ...)

There are three states, according to the action that you can take.

Hence, from there, you can now the profit at a state at time i as:

```
s0[i] = max(s0[i - 1], s2[i - 1]); // Stay at s0, or rest from s2
s1[i] = max(s1[i - 1], s0[i - 1] - prices[i]); // Stay at s1, or buy from s0
s2[i] = s1[i - 1] + prices[i]; // Only one way from s1
```

Then, you just find the maximum of s0[n] and s2[n], since they will be the maximum profit we need (No one can buy stock and left with more profit that sell right :) )

Define base case:

```
s0[0] = 0; // At the start, you don't have any stock if you just rest
s1[0] = -prices[0]; // After buy, you should have -prices[0] profit. Be positive!
s2[0] = INT_MIN; // Lower base case
```

Here is the code :D

```
class Solution {
public:
int maxProfit(vector<int>& prices){
if (prices.size() <= 1) return 0;
vector<int> s0(prices.size(), 0);
vector<int> s1(prices.size(), 0);
vector<int> s2(prices.size(), 0);
s1[0] = -prices[0];
s0[0] = 0;
s2[0] = INT_MIN;
for (int i = 1; i < prices.size(); i++) {
s0[i] = max(s0[i - 1], s2[i - 1]);
s1[i] = max(s1[i - 1], s0[i - 1] - prices[i]);
s2[i] = s1[i - 1] + prices[i];
}
return max(s0[prices.size() - 1], s2[prices.size() - 1]);
}
};
```