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45 changes: 39 additions & 6 deletions lib/exercises.rb
Original file line number Diff line number Diff line change
@@ -1,19 +1,52 @@

# 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(1)

def grouped_anagrams(strings)
Comment on lines +4 to 7

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👍 , however you have O(n) space complexity.

raise NotImplementedError, "Method hasn't been implemented yet!"
string_hashmap = {}
strings.each do |string|
if string_hashmap[grouped_anagrams_helper(string)]
string_hashmap[grouped_anagrams_helper(string)] << string
else
string_hashmap[grouped_anagrams_helper(string)] = []
string_hashmap[grouped_anagrams_helper(string)] << string
end
end
return string_hashmap.values
end

def grouped_anagrams_helper(string)
return string.chars.sort.join
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^2) in worst case because of the sorting to be done
# Space Complexity: O(1)
def top_k_frequent_elements(list, k)
Comment on lines +26 to 28

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You have an O(n log n) time complexity algorithm which is pretty good and O(n) space complexity because you're making a hash and an array both proportional to the size of the original list.

raise NotImplementedError, "Method hasn't been implemented yet!"
# build hashmap
num_hash = {}
list.each do |num|
if num_hash[num]
num_hash[num] += 1
else
num_hash[num] = 1
end
end

# convert hashmap to array
array_num_hash = num_hash.to_a

# sort by occurences
sorted_array_num_hash = array_num_hash.sort_by {|num| -num[1]}

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just noting this is an O(n log n) operation.


# sorted elements only
sorted_numbers = sorted_array_num_hash.map{|num| num[0]}

# slice sorted array by k elements
return sorted_numbers.slice(0, k)
end


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