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Nice work Mackenzie, you hit the learning goals here. Well done.
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| # Time Complexity: O(nm(log m)) + O(n) (sort_by for each string + length of strings array) | ||
| # Space Complexity: O(n) (hash size increases with input array length, no other data structures made) | ||
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| def grouped_anagrams(strings) |
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| # Time Complexity: O(n) + O(n logn) ( sort is nlogn, list.each is n, map is n) | ||
| # Space Complexity: O(n) (multiple same sized arrays) | ||
| def top_k_frequent_elements(list, k) |
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👍 So the final time complexity is O(n log n) because that's the bigger term.
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| sorted = counted.sort {|a,b| b[1]<=>a[1]} | ||
| keys = sorted.map { |n| n[0] } |
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Hash Table Practice
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Comprehension Questions