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37 changes: 31 additions & 6 deletions hash_practice/exercises.py
Original file line number Diff line number Diff line change
Expand Up @@ -2,18 +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 log n)
Comment on lines 2 to +6

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👍 The time complexity is right if the words are all of limited length (like english words).

The space complexity is O(n) because you're building a dictionary of n words.

"""
pass
letters_hash = {}

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

return list(letters_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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👍 However your time complexity is O(nk) because you're looping through the words and k times finding the maximum word (most common).

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

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

sorted_k_values = []
for i in range(k):
highest_freq_value = max(highest_k_hash, key=highest_k_hash.get)
Comment on lines +36 to +37

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Note that you have a loop here which runs k times and inside it you're going through the dictionary (of max n elements) and finding the maximum entry. So O(nk)

sorted_k_values.append(highest_freq_value)
highest_k_hash.pop(highest_freq_value)

return sorted_k_values


def valid_sudoku(table):
Expand Down