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Earth - Anya #25

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34 changes: 28 additions & 6 deletions lib/exercises.rb
Original file line number Diff line number Diff line change
@@ -1,19 +1,41 @@

# 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 * m) n is the length of the array, m is the max length of string within the array.
# Could pull chars method to run first on each element of strings, for O(n) n being the longest string or array.
# Space Complexity: O(n)

def grouped_anagrams(strings)
Comment on lines +4 to 8

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👍 But the time complexity is O(n) if the strings are limited in size (so you can ignore the sort because it's limited in size), or O(n * m log m) where m is the length of the strings, if the string is not limited in size.

raise NotImplementedError, "Method hasn't been implemented yet!"
return [] if strings.nil?
return [] if strings.empty?

counts_hash = Hash.new([])
strings.each do |string|
counts_hash[string.chars.sort] += [string]
end
return counts_hash.map {|k, v| v}
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(n) n is list length
# Space Complexity: O(n^2 * k) is that right?
def top_k_frequent_elements(list, k)
Comment on lines +21 to 23

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👍 The time complexity is O(n log n) because you have an O(n) loop (the list.each loop) then a sort (O(n log n)), and followed by another loop which goes k times where k <= n.

So the time complexity is O(n + n log n + k) ==> O(n log n) because n log n is the dominate term. It's larger than n and k.

In the space complexity you're creating a hash O(n), a sorted hash O(n), and k_most_common_ele. So O(n + n + k) and so the final space complexity is O(n).

raise NotImplementedError, "Method hasn't been implemented yet!"
return [] if list.nil?
return [] if list.empty?

counts_hash = Hash.new(0) # space = O(n)
list.each do |num|
counts_hash[num] += 1
end

sorted_values_2d = counts_hash.sort_by {|k, v| -v} # space = O(n)

k_most_common_ele = [] # space = O(k)
k.times do |i|
k_most_common_ele << sorted_values_2d[i][0]
end
return k_most_common_ele
end


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