TRIVIA: If you know what this is, you are a OG: "Long ago in a distant land, I, Aku, the Shapeshifting Master of Darkness, unleashed an unspeakable evil. But a foolish samurai warrior wielding a magic sword stepped forth to oppose me. Before the final blow was struck, I tore open a portal in time, and flung him into the future where my evil is law. Now the fool seeks to return to the past and undo the future THAT IS AKUU!"

What is word processing and word wrap?

Some of your friends have gotten into the burgeoning field of time-series data mining, in which one looks for patterns in sequences of events that occur over time. purchases at stock exchanges--what’s being bought-- are one source of data with a natural ordering in time. given a long sequence s of such events, your friends want an efficient way to detect certain "patterns" in them--for example, they may want to know if the four events buy yahoo, buy ebay, buy yahoo, buy oracle occur in this sequence s, in order but not necessarily consecutively. they begin with a collection of possible events (e.g., the possible’ transactions) and a sequence s of n of these events. a given event may occur multiple times in s (e.g., yahoo stock may be bought many times in a single sequence s). we will say that a sequence s’ is a subsequence of s if there is a way to delete certain of the events from s so that the remaining events, in order, are equal to the sequence s’. so, for example, the sequence of four events above is a subsequence of the sequence buy amazon, buy yahoo, buy ebay, buy yahoo, buy yahoo, buy oracle their goal is to be able to dream up short sequences and quickly detect whether they are subsequences of s. so this is the problem they pose to you: give an algorithm that takes two sequences of even~s--s’ of length m and s of length n, each possibly containing an event more than once--and decides in time o(m n) whether s’ is a subsequence of s

The following function returns a string of length n whose characters are all 'x'. give the order of growth (as a function of n) of the running time. recall that concatenating two strings in java takes time proportional to the sum of their lengths. public static string f(int n) { if (n == 0) return ""; if (n == 1) return "x"; return f(n/2) + f(n - n/2); } options: a) constant b) logarithmic c) linear d) linearithmic e)quadratic f)cubic g) exponential