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Here's my code to find the middle of the linked list:

I used a HashMap to store the occurrences of each element, and then iterated over the hash map to get duplicated element, but something doesn't feel right about this solution.

Problem statement in Firecode.io:

Write a method duplicate to find the repeated or duplicate elements in an array. This method should return a list of repeated integers in a string with the elements sorted in ascending order (as illustrated below).

duplicate({1,3,4,2,1}) --> "[1]"

duplicate({1,3,4,2,1,2,4}) --> "[1, 2, 4]"

Note: You may use toString() method to return the standard string representation of most data structures, and Arrays.sort() to sort your result.*

Here is my code:

I suppose this is not easy. I have a simple update query that works fine, the problem is that 7-8% of the records are not updated. It seems the LEFT JOIN cannot always match the ON clause.

I have checked the not matched fields to see if there are some white spaces but there are none. I also checked for duplicates in the LEFT table, none also here. Of course I've also checked that the records exist in both tables.

I'm using MySQL 8.0 under Windows 10

This is my code:

I'm solving a puzzle that needs to check if two strings are permutation of each other in O(N) with O(1) Space.

So my idea is to concatenate the two strings, and xor the result if its zero then it's a permutation.

But it fails in just one test case

I know this is a very basic question and I found sample codes online but I cannot figure out why it works.

If we need to traverse a binary tree in preorder fashion, one of the way to do so (quoted here http://interactivepython.org/courselib/static/pythonds/Trees/TreeTraversals.html) is by using something like: