deletions | additions       diff --git a/Complexity analysis.tex b/Complexity analysis.tex  index e1118cc..0fdf132 100644  --- a/Complexity analysis.tex  +++ b/Complexity analysis.tex 
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T(n) = R(n) + k_0 + k_1 n + k_2 n \log n  \]  This means that if, $R(n) \in O(n \log n)$, then $T(n) \in O(n \log n)$ as well. If not, then $T(n) \in O(R(n))$. Thus, if we can find $R$ in $O(n \log n)$, then our solution is $O(n \log n)$. For b), where $R$ is chosen arbitrarily, it can trivially be seen that $R(n) \in O(1)$. For a), the case is a bit more complex, but it should be possible to achieve $O(n \log n)$  by creating two lists containing all the intervals, sorting one by starting angle and one by ending angle, and then going through the lists in parallell, which would make $R(n) \approx 2n \log n + 2n$, meaning $R(n) \in n \log n$.