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Nara #14
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Nara #14
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# This method will return an array of arrays. | ||
# Each subarray will have strings which are anagrams of each other | ||
# Time Complexity: ? | ||
# Space Complexity: ? | ||
# Time Complexity: O(n) | ||
# Space Complexity: O(n) | ||
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def grouped_anagrams(strings) | ||
raise NotImplementedError, "Method hasn't been implemented yet!" | ||
return [] if strings.empty? | ||
hash = {} | ||
strings.each do |string| | ||
sorted_string = string.split("").sort.join | ||
if hash[sorted_string] | ||
hash[sorted_string] << string | ||
else | ||
hash[sorted_string] = [string] | ||
end | ||
end | ||
return hash.values | ||
end | ||
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# 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: ? | ||
# Time Complexity: O(n) | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. Look a bit down, you're doing a sort... how is that impacting things? |
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# Space Complexity: 0(n) | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. Zero of n? |
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def top_k_frequent_elements(list, k) | ||
raise NotImplementedError, "Method hasn't been implemented yet!" | ||
hash = {} | ||
frequent_elements =[] | ||
return [] if k == 0 || list.empty? | ||
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list.each do |element| | ||
if hash[element] | ||
hash[element] += 1 | ||
else | ||
hash[element] = 1 | ||
end | ||
end | ||
sorted_arrays = hash.sort | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. You're doing a sort here, how does this impact time complexity? |
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k.times do |i| | ||
frequent_elements << sorted_arrays[i][0] | ||
end | ||
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return frequent_elements | ||
end | ||
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If the words are expected to be small, this is true.