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Rock - Hannah L. #39

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37 changes: 30 additions & 7 deletions hash_practice/exercises.py
Original file line number Diff line number Diff line change
Expand Up @@ -2,19 +2,43 @@
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)
Space Complexity: O(n)
"""
Comment on lines 2 to 7

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👍 Your time complexity is correct, if the words are limited in length, like regular english words. If the words could be any length then this is O(n * m log m) because you're sorting each word.

pass
anagrams_hash = {}

for word in strings:
sorted_word = "".join(sorted(word))
if sorted_word in anagrams_hash:
anagrams_hash[sorted_word].append(word)
else:
anagrams_hash[sorted_word] = [word]

return list(anagrams_hash.values())

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: O(n)
Space Complexity: O(n)
Comment on lines 19 to +23

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Because you find the max word k times the time complexity here is O(nk)

"""
pass
frequency_hash = {}
if nums == []:
return nums

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

return_list = []
for i in range(k):
highest_value = max(frequency_hash, key=frequency_hash.get)
return_list.append(highest_value)
frequency_hash.pop(highest_value)

return return_list

def valid_sudoku(table):
""" This method will return the true if the table is still
Expand All @@ -26,4 +50,3 @@ def valid_sudoku(table):
Space Complexity: ?
"""
pass