Just No more free questions left!

Upgrade Now

I want to learn some big words so people think I'm smart.

I opened up a dictionary to a page in the middle and started flipping through, looking for words I didn't know. I put each word I didn't know at increasing indices in a huge array I created in memory. When I reached the end of the dictionary, I started from the beginning and did the same thing until I reached the page I started at.

Now I have an array of words that are mostly alphabetical, except they start somewhere in the middle of the alphabet, reach the end, and then start from the beginning of the alphabet. In other words, this is an alphabetically ordered array that has been "rotated." For example:

String[] words = new String[]{ "ptolemaic", "retrograde", "supplant", "undulate", "xenoepist", "asymptote", // <-- rotates here! "babka", "banoffee", "engender", "karpatka", "othellolagkage", };

Write a function for finding the index of the "rotation point," which is where I started working from the beginning of the dictionary. This array is huge (there are lots of words I don't know) so we want to be efficient here.

We can get time.

You must log in with one click to view the rest.

Once you're logged in, you'll get free full access to this and 4 other questions.

You must log in with one click to view the rest.

Once you're logged in, you'll get free full access to this and 4 other questions.

Each time we go through the while loop, we cut our range of indices in half, just like binary search. So we have loop iterations.

In each loop iteration, we do some arithmetic and a string comparison. The arithmetic is constant time, but the string comparison requires looking at characters in both words—every character in the worst case. Here, we'll assume our word lengths are bounded by some constant so we'll say the string comparison takes constant time.

The longest word in English is pneumonoultramicroscopicsilicovolcanoconiosis, a medical term. It's 45 letters long.

Putting everything together, we do iterations, and each iteration is time. So our time complexity is .

Some languages—like German, Russian, and Dutch—can have arbitrarily long words, so we might want to factor the length of the words into our runtime. We could say the length of the words is \ell, each string comparison takes time, and the whole algorithm takes time.

We use space to store the first word and the floor and ceiling indices.

This function assumes that the array is rotated. If it isn't, what index will it return? How can we fix our function to return 0 for an unrotated array?

You must log in with one click to view the rest.

Once you're logged in, you'll get free full access to this and 4 other questions.

What's next?

RUN
Code execution powered by Qualified.io

. . .