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HashPractice Paper Karla T. #49
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Original file line number | Diff line number | Diff line change |
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@@ -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(1) | ||
""" | ||
pass | ||
|
||
dict = {} | ||
for string in strings: | ||
sorted_string = ''.join(sorted(string)) | ||
if sorted_string in dict: | ||
dict[sorted_string].append(string) | ||
else: | ||
dict[sorted_string] = [string] | ||
return_list = [] | ||
for key, value in dict.items(): | ||
return_list.append(value) | ||
return return_list | ||
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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 log n) | ||
Space Complexity: O(n) | ||
""" | ||
Comment on lines
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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. 👍 I would also say that the time complexity is O(nk) since you have the nested loop in lines 37 - 40 (max has to loop through the list). |
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pass | ||
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frequency_dict = {} | ||
if nums == []: | ||
return nums | ||
for num in nums: | ||
if num in frequency_dict: | ||
frequency_dict[num] += 1 | ||
else: | ||
frequency_dict[num] = 1 | ||
return_list = [] | ||
for i in range(k): | ||
highest_value = max(frequency_dict, key = frequency_dict.get) | ||
return_list.append(highest_value) | ||
frequency_dict.pop(highest_value) | ||
return return_list | ||
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def valid_sudoku(table): | ||
""" This method will return the true if the table is still | ||
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@@ -25,5 +49,5 @@ def valid_sudoku(table): | |
Time Complexity: ? | ||
Space Complexity: ? | ||
""" | ||
pass | ||
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👍 However space complexity is O(n) since you're building a dictionary of all the words.