For subtask 1, it suffices to print out the string achieved by appending all the strings together.

Time complexity: $\mathcal{O}(N)$

For subtask 2, we can use a bitmask-DP to create such a string. If we have $dp[i][mask]$ (where $mask$ is a bitmask for which strings were already used) represent whether or not it's possible to create a string with the chosen strings in $mask$, where the last string chosen was the $i^{\text{th}}$ one. Thus, we have $dp[i][mask]$ as $1$ if and only if at least one of $dp[j][mask-(2^i)]$ is $1$ where $s_j$ is a string that has a suffix equivalent to the prefix of $s_i$. This can be checked in $\mathcal{O}(N)$ time by looping through all the given strings.

Time complexity: $\mathcal{O}(N^2{2}^N)$

For subtasks 3 and 4, we can represent the given strings as a graph where each node represents a given $K-1$ length string, and each string is an edge connecting its $K-1$ length prefix to its $K-1$ length suffix. Thus, a solution string is represented by an Euler path or an Euler circuit in the graph. For subtask 3, we only need to determine the existence of an Euler path/circuit, while to get full points, we must determine a possible Euler circuit in the graph.

Time complexity: $\mathcal{O}(NK)$ or $\mathcal{O}({26}^{K-1}+NK)$