They're in a numbered list, so minimizing i should take precedence over minimizing j.

Chrono79, as you say i must be as small as possible. But it also says the same about j with no statement about which should take precedence. In the case of 0, 1 it could also be done as 1, 0 with the same effect (and this would be the simpler way too). I might have misunderstood but it seems for the 0, 1 case to make sense another explicit rule would need to be in the text.

No problem, but starting from 269045 and using i = 0 and j = 3 would be 690245, don't forget the 0 will be still there ;)

I get it now, [26945, 0, 3] will give 69245, wich is > 26945.
Thank you very much :D!

The input is 269045 right?
The smallest number you get following the kata instructions is 26945
i is the index of the digit you move, you don't get 26945 moving the 2 (index 0), you get that number moving the 0 (index 3)

2. the index i of the digit d you took, i as small as possible
3. the index j (as small as possible) where you insert this digit d to have the smallest number.

Maybe you'll see it better if you write it as 026945 that as a number is 26945. You don't swap two digits either, you take a single digit and move it.

The solution I dont understand is not this one [29917, 0, 1].
It it this one [26945, 3, 0], if you want i as small as possible it should be [26945, 0, 3].

No, they're not, read what I wrote again, first case you move the 2 and second case you move the 0, why? Because you want i as small as possible, so [29917, 1, 0] is not a solution and [29917, 0, 1] is. In the second case you have to move the 0, not the 2 to get the smallest possible number.

Sorry I'll go for the question next time, but I still dont get it.
Both cases are 2 (index 0) inserted at the index of 0 (index 1 and 3).
What am I missing?

Next time use Question, not issue if you don't understand something.
Finding the smallest digit and moving it is not the first step in both cases.
In the first case, you take the 2 (index 0) and insert it back at index 1.
In the second case, you take the 0 (index 3) and insert it back at index 0.

Can someone explain me why both of these case doesnt work the same way?
testing(209917, "[29917, 0, 1]");
testing(269045, "[26945, 3, 0]");
The way I understand it:
1): You find the smallest number (0)
2): You find the index of the digit i as small as possible (0 and 0)
3): You find the index of j, both the zeros (1 and 3)
In both cases you need to swap a Zero with the First element, why are the results [0,1] and [3,0] not in the same order?
Thanks :)

Test.assertDeepEquals should be used in JavaScript.

Nobody's going to change kata specifications after 6k+ people solved it.

A plain dictionary should be returned.

Bumping an issue from 2 years ago - Python 3 should be enabled.