Gloria Scissors#42
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Jan 21, 2022
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Nice work Gloria, you hit the main learning goals here. Take a look at my comments concerning time/space complexity.
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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: ? | ||
| """ | ||
| pass | ||
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| Time Complexity: O(n) checking each word | ||
| Space Complexity: O(n) at worse case might have to create a node | ||
| """ |
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👍 , The time complexity is O(n) assuming the words are of limited length like real English words (no word is longer than so many characters) Otherwise it would be O(nm) where m is the word length.
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| # Time and Space O(n) might need to append all the values | ||
| my_sorted_list = sorted(frequent_elements.values(),reverse=True) |
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Since you're sorting here that would be O(n log n)
| if value in numbers: | ||
| k_list.append(key) | ||
| #after you matchremove from numbers so it doesn't match twice | ||
| numbers.remove(value) |
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Note that removing a value from an array is an O(n) operation and this is in a loop...
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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(nlogn) | ||
| Space Complexity: O(logn) |
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👍 This works, but see my notes on time/space complexity.
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