Understanding Big O notation - Cracking the Coding Interview

Question

I need help understanding how the author got the answer of problem 11 in the Big O chapter.

The problem goes like this:

The following code prints

Answer 1

In short, no. The correct answer is O(kc^k) as in the book.

After you went over a string to check if its characters were ordered, which would take O(k), printing it would add only O(k) - which does not change your complexity.

Suppose testing whether a string is ordered takes a*k operations, and printing it takes b*k. Then the total number of operations for each string is at most (a+b)*k which is still O(k).

Edit: Regarding the second part of your question, going over all words with some fixed length will result in an exponential runtime complexity, since there are c^k such words where c is the size of the alphabet and k is the length of the word.

Answer 2

In general the printing of a constant length string is considered constant as well, but if we want to be precise let's consider the print of a single character as the basic operation: this means that to print a k length string we have O(k).

Since we have O(c^k) possible strings and for each of them we have to check if it is sorted (with O(k)) and to print them (another O(k)), the total complexity became O(c^k(k + k)) = O(2c^kk).

But multiplying a function for a constant factor doesn't change it's complexity, and therefore the answer remains O(c^kk).

Answer 3

Printing the string is only an extra addition to the k time.

Checking whether each string is sorted is O(k) and say that printing it is O(dk) for some integer d (a constant). Adding the two we get O(k + dk), which can be re written as O(k(1 + d)). Because this just a scalar we know O(k + dk) = O(k) so the answer does not change.