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space kate #27
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Nice work Kate, you hit the learning goals here. Well done.
# Time Complexity: O(n log n) | ||
# Space Complexity: O(n) | ||
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def grouped_anagrams(strings) |
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Since you're sorting the words and English words are small this method actually depends on the number of words, not the length of them.
So it's O(n * m log m) where m is the length of the words, or if you assume the words are short O(n)
key = word.chars.sort.join | ||
#check if key is in the hash, push the word into the hash | ||
#else, set the key as the word | ||
if hash.key?(key) |
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Just a bit more compact.
if hash.key?(key) | |
if hash[key] |
# Time Complexity: O(n log n) | ||
# Space Complexity: O(n) | ||
def top_k_frequent_elements(list, k) |
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👍
Hash Table Practice
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Comprehension Questions