Solution for E:

Editorial (don't look at the sample code, it's broken rn)
Basically, you convert the strings into a graph, where a node connects $$$i$$$ and $$$j$$$ if $$$s_j$$$ starts with the same $$$3$$$ characters that $$$s_i$$$ ends with. Then you do topological sort to get the order you do dp updates, and then do dp on the graph.

Yes, I was also going on that path, from a node we need to see whether there is a dead-end node (from where no further words exist such that they start with the last 3 characters) at an odd distance, but I was not able to optimize this processes.

I also had the intuition to form a graph then construct the edges in 52^5(forgot to remove the useless letters in-between) instead of 52^3, such a silly mistake

The search results depends of your country and of your own search history, I got results in french, what do most people see?

Lol, try searching broken profile dp

It's a literal translation from Japanese term 「挿入DP」, so there's no doubt the term is not known. I'm curious, what is this kind of DP called in English?

Suppose A, C are two sequences such that 1 <= A_i <= C_i, A_i != A_j (the two conditions listed in the bullets in the problem statement)

Then if you apply the same permutation to A and C, they continue to satisfy the two conditions.

let P be a permutation which gets from C to sorted C, let $$$l_1,...,l_n$$$ a solution to the original problem, a permutation of l according to P will give you a solution to sorted C and conversely you can get back a solution to C from solution to sorted C.

it's a one-to-one function between the two set of solutions so they have same size.

Suppose at some i array C has value C_i. Now let there be x elements in the array C (other than i) whose value is less than or equal to C_i. This implies the number of possible values for i will be reduced by x which is (C_i — x).

Notice when we sort x did not change, so our answer won't change at all.

Plus we'll also get the value of x (which is i-1) effortlessly as a bonus ;)

One thing that I faced is, In this approach total no of edges in your graph in worst case is around N^2 which will not fit in time limits. Refer the editorial for optimal way of graph construction.