Defines f[i] = max profit we can get at the end of day i

In day i+1, we can

- do nothing, f[i+1] = f[i]
- sell the stock buy before, f[i+1] = max{f[k] + a[i] - a[k+2]} (buy the stock in day k+2)

Then we have:

```
f[i+1] = max{ f[i], f[k] + a[i] - a[k+2] }, k+2 < i+1
```

We only need to store the max value of f[k] - a[k+2] (k+2<i+1) in the calculating process.

```
f[i+1] = max{ f[i], g[i-1] + a[i] }
g[i+1] = max{ g[i], f[i-2] - a[i] }
```

Notice that k may less than 0, and f[i<0] = 0;

Finally, eliminate the redundant space usage:

```
int maxProfit(int* a, int n) {
int f = 0, f1 = 0, f2 = 0, g = -a[0];
for (int i=1; i<n; i++) {
g = max(g, f2 - a[i]);
f = max(f1, g + a[i]);
// for next iteration
f2 = f1;
f1 = f;
}
return f;
}
```