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)~