This was a really nice problem I remember solving at the SEERC. But it would be helpful if you mentioned which part is unclear to you.

In general this is the solution:

• notice that you can't have $$$p_i \lt p_{i+1}$$$ except when $$$p_i \leq 2$$$. This means that we will have a pseudo-decreasing array, which actually consists of 1, 2, decreasing array between 1-2, decreasing array between 2-1

• if we can find the ways to separate the elements into two decreasing arrays like this, we can easily calculate the solution

• notice that elements $$$a$$$ and $$$a-1$$$ can always go after another, so we can place entire segments of consecutive elements into one segment

• therefore, we can have $$$dp_i$$$ which is the number of ways to construct these arrays, such that one ends on $$$i$$$ and another ends on $$$i+1$$$

• we can "select" the previous element $$$x$$$ which is the array which ends with $$$i$$$

• as these two arrays are symmetrical, this adds $$$dp[x-1]$$$ to $$$dp[i]$$$

• there are about $$$3\cdot \frac{n}{i}$$$ of this options which sums into $$$O(NlogN)$$$ (google harmonic sum if you don't know this)

I don't have code as I participated onsite