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KS #23
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KS #23
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Thanks for turning in this PR Kate. Both methods work well are are efficient. Do take a look at my comments and consider adding the space/time complexities as well. Otherwise outstanding work.
# Each subarray will have strings which are anagrams of each other | ||
# Time Complexity: ? | ||
# Space Complexity: ? | ||
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def grouped_anagrams(strings) |
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👍 Space & Time Complexity?
deltas[max].each do |num| | ||
solution << num | ||
remains -= 1 | ||
break if remains == 0 |
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You don't need this break since your while loop runs while remains > 0
break if remains == 0 | ||
end | ||
end | ||
max -= 1 |
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indentation?
# This method will return the k most common elements | ||
# in the case of a tie it will select the first occuring element. | ||
# Time Complexity: ? | ||
# Space Complexity: ? | ||
def top_k_frequent_elements(list, k) |
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👍
Hash Table Practice
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Comprehension Questions