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water - mackenzie #20

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36 changes: 30 additions & 6 deletions lib/exercises.rb
Original file line number Diff line number Diff line change
@@ -1,19 +1,43 @@

# 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(nm(log m)) + O(n) (sort_by for each string + length of strings array)
# Space Complexity: O(n) (hash size increases with input array length, no other data structures made)

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

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👍 Nice!

raise NotImplementedError, "Method hasn't been implemented yet!"
anagrams = {}
strings.each do |string|
abc_string = string.chars.sort_by(&:downcase).join
if !anagrams[abc_string]
anagrams[abc_string] = []
end
anagrams[abc_string] << string
end
puts anagrams
return anagrams.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(n) + O(n logn) ( sort is nlogn, list.each is n, map is n)
# Space Complexity: O(n) (multiple same sized arrays)
def top_k_frequent_elements(list, k)
Comment on lines +22 to 24

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👍 So the final time complexity is O(n log n) because that's the bigger term.

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

counted = {}
list.each do |num|
if counted[num]
counted[num] += 1
else
counted[num] = 1
end
end

sorted = counted.sort {|a,b| b[1]<=>a[1]}
keys = sorted.map { |n| n[0] }
Comment on lines +37 to +38

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Very compact!


return keys.slice(0, k)
end


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