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              | Original file line number | Diff line number | Diff line change | 
|---|---|---|
| @@ -1,19 +1,49 @@ | ||
|  | ||
| # 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) | ||
|  | ||
| def grouped_anagrams(strings) | ||
| raise NotImplementedError, "Method hasn't been implemented yet!" | ||
| # raise NotImplementedError, "Method hasn't been implemented yet!" | ||
| return [] if strings.empty? | ||
|  | ||
| hash = Hash.new | ||
|  | ||
| strings.each do |word| | ||
| array_letters = word.split("").sort | ||
|  | ||
| if hash[array_letters] | ||
| hash[array_letters] << word | ||
| else | ||
| hash[array_letters] = [word] | ||
| end | ||
| end | ||
|  | ||
| return hash.values | ||
|  | ||
|  | ||
| end | ||
|  | ||
| # 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(nlogn) | ||
| # Space Complexity: O(n) | ||
| def top_k_frequent_elements(list, k) | ||
| 
      Comment on lines
    
      +30
     to 
      32
    
   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. 👍 | ||
| raise NotImplementedError, "Method hasn't been implemented yet!" | ||
| # raise NotImplementedError, "Method hasn't been implemented yet!" | ||
| return [] if list.empty? | ||
|  | ||
| hash = Hash.new(0) | ||
|  | ||
| list.each do |num| | ||
| hash[num] += 1 | ||
| end | ||
|  | ||
| sorted_hash = hash.sort_by {|key, value| -value} | ||
|  | ||
| return sorted_hash[0..k-1].map { |num, count| num} | ||
|  | ||
|  | ||
| end | ||
|  | ||
|  | ||
|  | ||
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👍 However since you're sorting the words, the time complexity is O(n + m log m) where m is the length of the words. If you assume the words are limited in size, then it would be O(n).