\newsection
\section{Search \& Sort}
\subsection{Search}
\subsubsection{Linear search}
Linear search, as the name implies, searches through the entire array and has linear runtime, i.e. $\Theta(n)$. 

It works by simply iterating over an iterable object (usually array) and returns the first element (it can also be modified to return \textit{all} elements that match the search pattern) where the search pattern is matched.

\tc{n}

\subsubsection{Binary search}
If we want to search in a sorted array, however, we can use what is known as binary array, improving our runtime to logarithmic, i.e. $\Theta(\log(n))$.
It works using divide and conquer, hence it picks a pivot in the middle of the array (at $\floor{\frac{n}{2}}$) and there checks if the value is bigger than our search query $b$, i.e. if $A[m] < b$. This is repeated, until we have homed in on $b$. Pseudo-Code:

\begin{algorithm}
    \begin{spacing}{1.2}
        \caption{\textsc{binarySearch(b)}}
        \begin{algorithmic}[1]
            \State $l \gets 1$, $r \gets n$ \Comment{\textit{Left and right bound}}
            \While{$l \leq r$}
                \State $m \gets \floor{\frac{l + r}{2}}$
                \If{$A[m] = b$} \Return m \Comment{\textit{Element found}}
                \ElsIf{$A[m] > b$} $r \gets m - 1$ \Comment{\textit{Search to the left}}
                \Else \hspace{0.2em} $l \gets m + 1$ \Comment{\textit{Search to the right}}
                \EndIf
            \EndWhile
            \State \Return "Not found"
        \end{algorithmic}
    \end{spacing}
\end{algorithm}
\tc{\log(n)}

Proving runtime lower bounds (worst case runtime) for this kind of algorithm is done using a decision tree. It in fact is $\Omega(\log(n))$

% INFO: If =0, then there is an issue with math environment in the algorithm