hanoi | Pair Programming TDD - Write tests | Unit Testing library

If the problem is that you simply need to deal with the case where element.genre_ids is not defined in the API result, I think you could simply change the assignment of objectResults to be:

If you just want to print letters instead of numbers, cast your value to char

String.valueOf((char)(tower[1].get(i) + 97))

You might want to use 96 instead, since I believe you are using 1 as the first index.

There is the ability to select better xpaths. You can use a relative xpaths using the label of associated select

Let's define 𝑛 as the number of discs.

The time complexity is O(2^𝑛), because that is the number of iterations done in the only loops present in the code, while all other code runs in constant time.

The space complexity can be split up in two parts:

1. The "towers" themselves (stacks) have a O(𝑛) space complexity
2. The auxiliary space has a O(1) space complexity as there are no other vectors, and the call stack has a fixed size (no dynamic recursion)

So combined this has a O(𝑛) space complexity.

It seems the comments are too short to explain it clearly.

Each function call starts the function from its first line. And after the function ends, the code goes back to where it got invoked.

A friend recommended me to write down the code on paper in order to understand it

That's what we're gonna do here. Analyse the code line-by-line and see when it goes deeper and when it leaves.

• hanoi(3, "source", "target", "spare") This will go to the else. And run hanoi with n=2

  • So n=2, right? So we enter else again. And run hanoi again but with n=1
    • we go deeper! Now n=1, so we finally hit print! That's where Move disk 1 from to target came from! The function exits.
  • We get back to where it was called. We did first hanoi in else. Now we hit another print. That's where Move disk 2 from source to spare comes from.
  • We get to the next line. We run hanoi with n=1 again.
    • We're inside again. n=1, so we get into if and print Move disk 1 from target to spare and exit
  • We got to where we invoked it (so we're in n=2 level) but we don't have any more lines to run. We exit

• We're back to the top level. Now we hit the print Move disk 3 from source to target

• Next line. Another call to hanoi, with n=2

  • n=2, again: else, go deeper
    • n=1, print, exit
  • we went back, so we print
  • and go deeper again
    • n=1, print, exit
  • exit

• exit

To understand recursion, you can actually do debug printing when entering and leaving.

The for loop version of the list comprehension would be

Try this:

Works as expected to me https://pl.kotl.in/otINdg8E3:

The degrees are initialized with 0 and never changed due to no invocation of the setter, which lacks a value for cities that are not expected (maybe that's why you initialized the degrees?).

You could do what you want way shorter: