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[haset-19]

anagram, frequency functions

Hash Table Practice

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Comprehension Questions

Question	Answer
Why is a good Hash Function Important?	to avoid big time complexity for search, deletion and insert
How can you judge if a hash function is good or not?	it has uniform distribution of values, fast insertion, close keys have values far apart
Is there a perfect hash function? If so what is it?	no, depends on use case
Describe a strategy to handle collisions in a hash table	linear probing, separate chaining, double probing, quadratic hashing
Describe a situation where a hash table wouldn't be as useful as a binary search tree	when we have duplicate keys, ordered list
What is one thing that is more clear to you on hash tables now	we can use it to solve many problem

[chimerror]

Grabbing this to grade!

[chimerror]

Pretty good...

I feel that your anagrams solution has a lot of issues (see the comments) that I'm a bit confused by as you seem to understand how to use dictionaries to solve the other problems. As such, I am marking this Yellow to encourage you to take another look at that part.

Additionally there's some very vague names as well as very repetitive code in your sudoku solution that should usually be seen as a sign that you can refactor it to avoid the repetition. There are also missing complexity calculations for the anagrams and sudoku solutions.

Nonetheless, your top-K solution looks fine.

Regardless, I do encourage you to think about the anagrams solution more so this is a Yellow.

[chimerror]

These are rather generic names that more describe what the data type is rather than what it is for. You don't even need list1 because you can just do [[strings[0]]], and something like return_value or the like would be more informative than listOfLists.

[chimerror]

Likewise, this could be better named something like matching_list_found rather than flag.

[chimerror]

This solution redoes a great deal of work as it recalculates the frequency map through anagram_helper again and again for each word in the list.

Additionally going back over every list makes this algorithm balloon to O(n^2) in time complexity since in the worst case, every word is in its own list and there are none that share anagrams.

I think you can achieve a much better solution that is O(n) through using a dictionary, because with this solution you don't gain advantage of the O(1) lookup time of dictionaries.

[chimerror]

Calling out the missing complexity calculations here, but I haven't been dinging people for it.

[chimerror]

This works, but usually when you repeat almost the same code so many times, it's usually worth thinking about how you can cut down on duplication through perhaps defining a helper function.

Code repetition like this tends to lead to errors and higher maintenance as the different versions of almost the same code fall out of sync with each other.

[chimerror]

Likewise, complexity calculations are missing here.