Timeline for Given an array of integers, return the smallest positive integer not in it
Current License: CC BY-SA 3.0
2 events
| when toggle format | what | by | license | comment | |
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| Oct 29, 2017 at 21:58 | comment | added | Clearer | Sorting with a comparison based method is only $O(n \log n)$ in the worst case. It's trivial to make it $O(n)$ in the best case (even for multiple such cases at the same time), and gradually move from $O(n)$ to $O(n \log n)$ as the list is "less" sorted. | |
| Oct 29, 2017 at 7:51 | history | answered | Surt | CC BY-SA 3.0 |