nice job using only one counter instead of two.

brilliant

very nice best practice and clever problem that has the blazingly fast speed of O(N^2)

welcome to codewars

Assigning size_t to int isn't safe.

There's an issue with empty vectors, easy to fix by moving "-1" to the left in the condition.

This first if condition is not needed here. If arr1 is empty, arr1.size() returns 0 and the second if condition is never executed. If arr2 is empty, for each statement is not executed and outArray is returned as an empty collection anyway.

Why is an O(n*2) solution considered a "best practice"?

Can you explain what you did changing n to a constant char? How is this used as a condition in the second parameter in the for loop

that is clever

Not sure why some are suggesting O(log n). Binary search only works on an ordered set. These numbers are not ordered. The fastest you can possibly achieve this is O(n). You must iterate through all the digits to ensure even the very last digit is within the rules.

I've actually tried to do this, but it made the code slower lol.

This is less efficient than the second solution. The complexity is O(n^2) for this one, while the complexity of the the second one is O(n). (count has a complexity of O(n)).