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Rock - Priscille #35
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Rock - Priscille #35
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@@ -2,18 +2,40 @@ | |
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) | ||
""" | ||
pass | ||
anagrams_map = {} | ||
for word in strings: | ||
sorted_word = "".join(sorted(word)) | ||
if sorted_word not in anagrams_map: | ||
anagrams_map[sorted_word] = [word] | ||
else: | ||
anagrams_map[sorted_word].append(word) | ||
return list(anagrams_map.values()) | ||
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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) | ||
""" | ||
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There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 👍 Good use of the Given that I would say that this is O(nk) given the loop and |
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pass | ||
frequency_hash = {} | ||
most_frequent_items = [] | ||
if len(nums) == 0: | ||
return [] | ||
for item in nums: | ||
if item in frequency_hash: | ||
frequency_hash[item] += 1 | ||
else: | ||
frequency_hash[item] = 1 | ||
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for i in range(k): | ||
highest_frequency = max(frequency_hash, key = lambda num: frequency_hash[num]) | ||
most_frequent_items.append(highest_frequency) | ||
frequency_hash.pop(highest_frequency) | ||
return most_frequent_items | ||
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def valid_sudoku(table): | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 👍 Nice work, just a note since the board is always 9x9 and never grows, the time/space complexity can be considered O(1). |
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@@ -22,8 +44,52 @@ def valid_sudoku(table): | |
Each element can either be a ".", or a digit 1-9 | ||
The same digit cannot appear twice or more in the same | ||
row, column or 3x3 subgrid | ||
Time Complexity: ? | ||
Space Complexity: ? | ||
Time Complexity: O(n^2) | ||
Space Complexity: O(n^2) | ||
""" | ||
pass | ||
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#row | ||
for row in range(9): | ||
row_map = {} | ||
for column in range(9): | ||
tile = table[row][column] | ||
if tile == ".": | ||
continue | ||
elif tile not in row_map: | ||
row_map[tile] = 1 | ||
else: | ||
return False | ||
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#column | ||
for column in range(9): | ||
column_map = {} | ||
for row in range(9): | ||
tile = table[row][column] | ||
if tile == ".": | ||
continue | ||
elif tile not in column_map: | ||
column_map[tile] = 1 | ||
else: | ||
return False | ||
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#sub-boxes | ||
def squares(R, C): | ||
squares_map = {} | ||
for row in range(R, R+3): | ||
for column in range(C, C+3): | ||
tile= table[row][column] | ||
if tile == ".": | ||
continue | ||
elif tile not in squares_map: | ||
squares_map[tile] = 1 | ||
else: | ||
return False | ||
return True | ||
for row in range(0,9,3): | ||
for column in range(0,9,3): | ||
if not squares(row, column): | ||
return False | ||
return True | ||
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Choose a reason for hiding this comment
The reason will be displayed to describe this comment to others. Learn more.
👍 Just note, if the words are limited in length you have the time/space complexity right. If not then the time complexity is O(n * m log m) where m is the length of a word.