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C15 - Katrina K #33

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43 changes: 37 additions & 6 deletions hash_practice/exercises.py
Original file line number Diff line number Diff line change
Expand Up @@ -2,18 +2,49 @@
def grouped_anagrams(strings):
""" 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 log m)
Space Complexity: O(n)
"""
Comment on lines 2 to 7

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👍 I would say either O(n) for time complexity if words are of limited length or O(n * m log m) if the words can be of arbitrary length because you're processing n words and sort each word of potentially m letters.

pass
if len(strings) < 1:
return []

anagrams = dict()

for string in strings:
key_string = "".join(sorted(string))

if key_string in anagrams:
anagrams[key_string].append(string)
else:
anagrams[key_string] = [string]

return [anagrams[group] for group in anagrams]


def top_k_frequent_elements(nums, k):
""" 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: At least O(n) assuming that max() is an O(1) look up but I strongly suspect it isn't
Space Complexity: At least O(n + k)
(Not sure about the big O here becuase I'm not certain how max works in this case)
"""
Comment on lines 24 to 30

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👍 Because you have a loop going k times and each time you find a maximum of the n elements. I would say this is O(nk)

pass
if len(nums) < 1:
return []

freq_dict = dict()

for num in nums:
if num in freq_dict:
freq_dict[num] += 1
else:
freq_dict[num] = 1

top_k = []
for i in range(k):
top_k.append(max(freq_dict, key=freq_dict.get))
freq_dict.pop(top_k[i])

return top_k


def valid_sudoku(table):
Expand Down