Hunter Smith - Sprint-Challenge--Data-Structures-Python #1
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Tulf
reviewed
Jan 25, 2019
| self.left.depth_first_for_each(cb) | ||
| if self.right != None: | ||
| self.right.depth_first_for_each(cb) | ||
| return |
| # if name_1 == name_2: | ||
| # duplicates.append(name_1) | ||
| a_set = set(names_1) | ||
| b_set = set(names_2) |
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nice you can also use intersection or a for loop with an if all of these would be O(n)
| O(n) | ||
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| 5. What is the runtime complexity of the provided code in `names.py`? | ||
| O(n^2) since it is looping through each element of each list to compare to each element of the other |
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@Tulf