Daniela Sanchez - Paper #1
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| // Time Complexity: O(n^2) | ||
| // Space Complexity: O(n) | ||
| function grouped_anagrams(strings) { |
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👍 However this is not an O(n^2) solution. Instead it's O(n), if the words are limited in length (like to English words). If the words are not limited it's O(n * m log m) due to sorting.
| // Time Complexity: O(n^2) | ||
| // Space Complexity: O(n) | ||
| function top_k_frequent_elements(list, k) { |
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👍 But the time complexity is O(n) time complexity is O(n * k * m) where n is the number of elements and k the number of most frequent elements and m the size of the most frequent element. It can be written more efficiently, but this is an ok solution.
| // Time Complexity: O(n^2) | ||
| // Space Complexity: O(n) | ||
| function valid_sudoku(table) { |
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One note. Because Sudoku boards never change in size, scale isn't an issue!
| if (highest_freq in counting_hash) { | ||
| for (let i = 0; i < counting_hash[highest_freq].length; i++) { | ||
| if (output.length === k) { | ||
| continue | ||
| } |
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Maybe instead of counting down highest_freq each time, you simply find the largest value in the object and then remove that key-value pair. That might speed things up.
Hash Table Practice
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