Kristina - Maple

[k0axaca]

Hash Table Practice

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Comprehension Questions

Question	Answer
Why is a good Hash Function Important?	Good Hash Functions minimize collisions.
How can you judge if a hash function is good or not?	A hash function is good if it maps different keys to different values, executes in constant time, and appears random.
Is there a perfect hash function? If so what is it?	No.
Describe a strategy to handle collisions in a hash table	You could utilize linear probing to minimize collisions.
Describe a situation where a hash table wouldn't be as useful as a binary search tree	When there are not unique keys available.
What is one thing that is more clear to you on hash tables now	Time complexity in relation to hash tables is cleared now, for example O(1) lookup time.

[chimerror]

Grabbing this to grade!

[chimerror]

Good work!

I added some corrections to your time complexity calculations, but overall this demonstrates sufficient exercise of the information presented and thus is being marked Green.

[chimerror]

Just a minor efficiency thing, if you find yourself calling the same function multiple times with the exact same arguments like you do with create_anagram_key(ele) here, it's usually worth saving the result into a local variable, since every call will recalculate the same data that doesn't change.

[chimerror]

I'm going to assume that n is the number of words in the list and that by m you mean the number of letters in each word. That being said, note that the call to sorted will actually take O(m * log(m)) time, making this more accurately O(n * m * log(m)).

However, we can make a simplifying assumption since we know that the input list is made up of English words. English words tend not to get long, only about 5 letters per word on average, so the effect of m is going to be dwarfed by the effect of n, since there can easily be hundreds or thousands of words in the list. With that assumption, we can just say the time complexity is just O(n).

[chimerror]

Don't forget to account for the call to sort on line 42, which will take O(n * log(n)) time and thus be the dominant part of the algorithm for the final time complexity.

[kyra-patton]

✨🌟 Nice work, Kristina. I left a couple comments. Let me know what questions you have.

🟢

[kyra-patton]

Instead of calling your helper function three times, consider calling it just once, at the top of the for loop body.

Suggested change

	if grouped_words.get(create_anagram_key(ele)) == None:
	grouped_words[create_anagram_key(ele)] = [ele]
	else:
	grouped_words[create_anagram_key(ele)].append(ele)
	anagram_key = create_anagram_key(ele)
	if grouped_words.get(anagram_key) == None:
	grouped_words[anagram_key] = [ele]
	else:
	grouped_words[anagram_key].append(ele)

[kyra-patton]

Yes! Where n is the length of strings and m is the length of the longest word in strings. However, if we assume each word in strings is a valid English word, since those are all of a finite length, we can reduce this to say the solution is O(n)

[kyra-patton]

✨