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India - Water #21
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India - Water #21
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Original file line number | Diff line number | Diff line change |
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@@ -1,19 +1,46 @@ | ||
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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!" | ||
anagrams = {} | ||
strings.each do |string| | ||
sorted_string = string.split('').sort.join | ||
if anagrams[sorted_string] | ||
anagrams[sorted_string] << string | ||
else | ||
anagrams[sorted_string] = [string] | ||
end | ||
end | ||
return anagrams.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(nlog(n)) | ||
# Space Complexity: O(n) | ||
def top_k_frequent_elements(list, k) | ||
Comment on lines
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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. 👍 |
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raise NotImplementedError, "Method hasn't been implemented yet!" | ||
return list if list.length == k | ||
return [] if list.length == 0 | ||
frequency = {} | ||
list.each do |element| | ||
if frequency[element] | ||
frequency[element] += 1 | ||
else | ||
frequency[element] = 1 | ||
end | ||
end | ||
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sorted_hash = frequency.sort_by{|k, v| -v} | ||
result = [] | ||
index = 0 | ||
k.times do | ||
result << sorted_hash[index][0] | ||
index += 1 | ||
end | ||
return result | ||
end | ||
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👍 Since you're sorting the strings, the time complexity is O(n + m log m) where m is the length of the words, but if you assume the strings are limited in size, you can say it's O(n).