Water - Christabel #16
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CheezItMan
reviewed
Apr 28, 2021
| hashes = [] | ||
| strings.each do |word| | ||
| word_hash = {} | ||
| word = word.chars.sort #sort is O(n)? |
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| # Time Complexity: O(2nm + p) where n is the length of the list passed to grouped_anagrams, and m represents the number of chars in each string and p is the number of unique char hashes | ||
| # == O(nm) | ||
| # Space Complexity: Um O(2n) == O(n) since I am creating an array equal to the length of the passed in list and the worst case size for the meta_hash is also the length of the passed in list? | ||
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| def grouped_anagrams(strings) |
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I'd say this is O(n * m log m) where m is the number of chars in each string and n is the length of the array.
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| # Time Complexity: O(3n) == O(n) where n is the length of the passed in list | ||
| # Space Complexity: O(n) since the worst case is that the elements hash is the full length of the passed in list | ||
| def top_k_frequent_elements(list, k) |
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This is O(n log n) because sorting is O(n log n)
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| # Time Complexity: oh goodness. this is a mean question. I'm going with O(1) since the grid is always 9x9, but let me show my work here... | ||
| # O(9) == O(1) for checking for uniqueness in all rows and columns | ||
| # O(9) == O(1) for building the 2darray of all subgrid values | ||
| # O(9^2) == O(1) for checking for uniqueness in all subgrid values | ||
| # Space Complexity: for lack of more confident alternate logic, I'm going with the same answer as above, O(1) since the grid is a fixed size so the worst case for the created row, column, and subgrid hash are all worst case space complexities of O(9) and the array of subgrid values is a fixed size of O(81) | ||
| def valid_sudoku(table) |
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👍 You are correct that the time/space complexity is O(1)
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