Ari #7
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CheezItMan
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Sep 18, 2019
CheezItMan
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CheezItMan
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Really nice, I like how you thought to break down the Sudoku problem into smaller methods. You got all the problems here. Nice work!
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| # Checks whether there is any | ||
| # duplicate in current column or not. | ||
| def in_column?(table) |
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These in_row and in_column methods don't look very dry here.
| end | ||
| end | ||
| end | ||
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Minor thing: I suggest a return true here.
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| # Checks whether there is any duplicate | ||
| # in current 3×3 box or not. | ||
| def in_box?(table) |
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This is a really slick method! Nice
| # in the case of a tie it will select the first occurring element. | ||
| # Time Complexity: O(n) | ||
| # Space Complexity: O(n) | ||
| def top_k_frequent_elements(list, k) |
| # Time Complexity: ? | ||
| # Space Complexity: ? | ||
| # in the case of a tie it will select the first occurring element. | ||
| # Time Complexity: O(n) |
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Looking at your later loop, I would say this is O(k * n)
| # Each subarray will have strings which are anagrams of each other | ||
| # Time Complexity: ? | ||
| # Space Complexity: ? | ||
| # Time Complexity: O(n) |
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Assuming the words are small, this is correct.
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