# Tag Info

11

The teacher is obviously using a better way. Check/read more about time complexity for lists in python here (python wiki). In your case, you are going for a time complexity: $O(n)$ for iteration $O(n)$ for .count() $O(n)$ for intermediate pop $O(n)$ for .index for a final (worst case) time: $O(n^2)$ (see explanation in comments from ...

9

Your teacher's approach is better than yours, the main reason is the one noted in the answer by @hjpotter92. However, their approach can be optimized even more. Note that there is the $O(k)$ check from not in. This is because when you check if an element is in a list, it goes through the whole list and tries to find it. A set on the other hand stores ...

5

numbers.pop(numbers.index(i)) is equivalent to numbers.remove(i). Given that your example lists are sorted and that your solution would fail for example [1, 3, 1, 2, 3, 2] (which it turns into [3, 1, 3, 2], not removing the duplicate 3), I'm going to assume that the input being sorted is part of the task specification. In which case that should be taken ...

1

The following are a few suggestions on how I'd write the SQL. Source Control If you don't already have a database project, create one in Visual Studio. Then check it in to source control. Microsoft Azure DevOps Services is free & private for teams of 5 or less (this is per project, so 5 developers per project). Then you'll be able to track changes you ...

1

It is difficult to tell "which one is better". As Jan mentioned in a comment: "tell which one is better" - what does "better" mean? performance? code readability? ... ? The notion of better is subjective. I am drawn to the first solution as it uses the fewest lines, though readability suffers because the lines are quite long ...

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