## Recent solutions

Below is the list of recent solutions.
This solution is complete and solves the problem. The approach is straightforward and easy to follow.
Jan 31

The candidate's solution is correct and demonstrates a level of completeness. The candidate's approach is also correct, as it uses a dictionary to store the number of times each element appears in the list. This approach is optimal because it only iterates through the list once and then iterates through the dictionary once. Therefore, the time complexity is O(n) where n is the size of the list.
Jan 31

The candidate's solution is complete and solves the problem. The candidate's approach is efficient, using the modulo operator to check if a number is even.
Jan 31

This solution is correct and demonstrates a good understanding of list comprehensions.
Jan 30

This solution correctly sums a list of integers. However, it is not very robust - for example, it would not work if the input was not a list of integers. Additionally, the function does not have any documentation to explain what it does or how to use it.
Jan 30

The candidate's solution correctly implements the function and returns the expected output. The candidate uses the asyncio library to create asynchronous tasks and wait for them to complete. This is a good approach to solving the problem.
Jan 30

This solution is complete and solves the problem. The approach is straightforward and easy to understand.
Jan 29

The candidate's solution correctly returns the sum of the odd numbers in the list. The approach is straightforward and easy to follow.
Jan 29