\begin{minipage}[t]{8in}
\begin{tabular}{ll}
void  & insertAfter\index{Sequence!insertAfter}(Item e, Item after\_item);
\end{tabular}\\
 
		
 \parbox[b]{2in} \ \ 
\begin{minipage}[t]{4in}If the reference count of \verb+store+ is greater
			   than 1, then other collection objects are also 
			   using
			   it as their implementation.  Rather than changing
			   its contents, a copy will be made ($O(n)$), 
			   and the reference
			   counts will be updated.  Finally, 
			   insert $e$ after the first occurrence of 
			   \verb+after\_item+.  If there is no such 
			   occurrence, a NULL pointer is passed to
			   the {\Container}'s 
			   {\em insertAfter(Item,ContainerNode*)} method. 
			   Complexity depends on the storage medium. \end{minipage}
\end{minipage}
\vspace{0.2in}


\begin{minipage}[t]{8in}
\begin{tabular}{ll}
void  & insertBefore\index{Sequence!insertBefore}(Item e, Item before\_item);
\end{tabular}\\
 
		
 \parbox[b]{2in} \ \ 
\begin{minipage}[t]{4in} Similar to {\em insertAfter()} except that $e$ is 
		  inserted before the first occurrence of 
		  \verb+before\_item+. \end{minipage}
\end{minipage}
\vspace{0.2in}


\begin{minipage}[t]{8in}
\begin{tabular}{ll}
void  & prepend\index{Sequence!prepend}(Item e);
\end{tabular}\\
 
 \parbox[b]{2in} \ \ 
\begin{minipage}[t]{4in}After considering reference counts as above, 
			   call \rar{store}{prepend(e)}, which should
			   insert $e$ at the beginning.  Note that this
			   usually has the same result as calling 
			   {\em insert}.\end{minipage}
\end{minipage}
\vspace{0.2in}


\begin{minipage}[t]{8in}
\begin{tabular}{ll}
Item  & element\index{Sequence!element}(int i);
\end{tabular}\\
 
			
 \parbox[b]{2in} \ \ 
\begin{minipage}[t]{4in} Return the $i$th element, or the last element if
			  $i $>$ \mbox{this-$>$size()}$  $O(n)$.  \end{minipage}
\end{minipage}
\vspace{0.2in}
 

\begin{minipage}[t]{8in}
\begin{tabular}{ll}
SequenceBase$<$Item,Impl$>$  & subsequence\index{Sequence!subsequence}(int i, int j);
\end{tabular}\\
 
			
 \parbox[b]{2in} \ \ 
\begin{minipage}[t]{4in} Construct a new sequence and initialize
			  it with $s[i] \ldots s[j]$, where $s[p]$ is the 
			  $p$th element of the current sequence. $O(n)$\end{minipage}
\end{minipage}
\vspace{0.2in}