Rock - Priscille #35
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Rock - Priscille #35
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CheezItMan
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Jan 10, 2022
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| 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) | ||
| """ |
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👍 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.
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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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👍 Good use of the max function!
Given that I would say that this is O(nk) given the loop and max function.
| 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): |
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👍 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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