But what if the testcase is like 11110, the decreasing character is at index 3. Follow your solution then we have 11109 (not a tidy number, the correct answer is 9999). It seems that after decreasing the left part by 1, we also need to check those digits stand before the decreasing character again.

Btw, base on your idea then I found out a solution (edit a little bit):
We need to store the index of the first appearance of decreasing character, for example:

11110 ( the decreasing character is '1', but we need to store the index of the first time appearance of it in the number, the index here is number 0).

12344296 ( the decreasing character is '4' and we store the index of its first appearance, is 3)
Finally, we split the origin in 3 parts: ( assume our digits testcase right now is 12344296)

Firstly, the starting digits are digits.substring(0, indexAbove) -- 123

Secondly, only 1 digit in which value decreases by 1. -- 3

Finally, the latter part is 9999, you can concat the result to the number which equals to 10** (digits.length - indexAbove) - 1 , ** is power operator.
Thank you for reading my idea.

## Discussion (6)

But what if the testcase is like 11110, the decreasing character is at index 3. Follow your solution then we have 11109 (not a tidy number, the correct answer is 9999). It seems that after decreasing the left part by 1, we also need to check those digits stand before the decreasing character again.

You are perfectly right!

I just updated the gist with the fix (I forgot to make

`solve()`

recursive, that's the key!).Btw, base on your idea then I found out a solution (edit a little bit):

We need to store the index of the first appearance of decreasing character, for example:

Your recursive solution is coded beautifully and easy to understand, thanks a lot!

Especially for the qualification round, where you have more time, I like the idea of your brute force algorithm as a unit test.

Thanks!

Indeed this is not a good solution for the next rounds, even if it could help you understand the problem.