Fix minor typos

latex/arrays.tex

@@ -73,7 +73,7 @@ \section{#ArrayStack#: Fast Stack Operations Using an Array}
 \seclabel{arraystack}
 
 \index{ArrayStack@#ArrayStack#}%
-An #ArrayStack# implements the list interface using an array #a#, called
+An #ArrayStack# implements the #List# interface using an array #a#, called
 the \emph{backing array}.  The list element with index #i# is stored
 in #a[i]#.  At most times, #a# is larger than strictly necessary,
 so an integer #n# is used to keep track of the number of elements
@@ -305,7 +305,7 @@ \section{#ArrayQueue#: An Array-Based Queue}
 some integer $k$.  Less formally, the value $r$ is the remainder we get
 when we divide $a$ by $m$.
 \pcodeonly{In many programming languages, including C, C++, and Java,
-the mod operate is represented using the \% symbol.}
+the mod operator is represented using the \% symbol.}
 \notpcode{In many programming languages, including
 \javaonly{Java}\cpponly{C++}, the $\bmod$ operator is represented
 using the #%# symbol.\footnote{This is sometimes referred to as the
@@ -593,7 +593,7 @@ \subsection{Balancing}
   As in the proof of \lemref{arraystack-amortized}, this is sufficient
   to prove that the total time spent by #balance()# is $O(m)$.
 
-  We will perform our analysis using a technique knows as the
+  We will perform our analysis using a technique known as the
   \emph{potential method}.
   \index{potential}%
   \index{potential method}%

latex/intro.tex

@@ -272,7 +272,7 @@ \subsection{The #List# Interface: Linear Sequences}
     $#x#_{#i#},\ldots,#x#_{#n#-1}$; \\ 
     Set $#x#_{j+1}=#x#_j$, for all
     $j\in\{#n#-1,\ldots,#i#\}$, increment #n#, and set $#x#_i=#x#$
-  \item #remove(i)# remove the value $#x#_{#i#}$, displacing
+  \item #remove(i)#: remove the value $#x#_{#i#}$, displacing
     $#x#_{#i+1#},\ldots,#x#_{#n#-1}$; \\ 
     Set $#x#_{j}=#x#_{j+1}$, for all
     $j\in\{#i#,\ldots,#n#-2\}$ and decrement #n#
@@ -409,7 +409,7 @@ \subsection{Exponentials and Logarithms}
 \index{exponential}%
 The expression $b^x$ denotes the number $b$ raised to the power of $x$.
 If $x$ is a positive integer, then this is just the value of $b$
-multiplied by itself $x-1$ times:
+multiplied by itself $x$ times:
 \[
     b^x = \underbrace{b\times b\times \cdots \times b}_{x} \enspace .
 \]

latex/linkedlists.tex

@@ -692,13 +692,13 @@ \section{Discussion and Exercises}
   interface.}
   \begin{enumerate}
     \item Write a #DLList# method called #takeFirst(l2)#.
-       This method takes the first node from #l2# and appends it to the the
+       This method takes the first node from #l2# and appends it to the
        receiving list.  This is equivalent to #add(size(),l2.remove(0))#,
        except that it should not create a new node.
     \item Write a #DLList# static method, #merge(l1,l2)#, that takes two
        sorted lists #l1# and #l2#, merges them, and returns a new sorted
        list containing the result.  This causes #l1# and #l2# to be emptied
-       in the proces.  For example, if #l1# contains $a,c,d$ and #l2# contains
+       in the process.  For example, if #l1# contains $a,c,d$ and #l2# contains
        $b,e,f$, then this method returns a new list containing $a,b,c,d,e,f$.
     \item Write a #DLList# method #sort()# that sorts the elements
        contained in the list using the merge sort algorithm.
@@ -742,7 +742,7 @@ \section{Discussion and Exercises}
   \index{MinDeque@#MinDeque#}%
   Design and implement a #MinDeque# data structure that can store
   comparable elements and supports all the deque operations #addFirst(x)#,
-  #addLast(x)# #removeFirst()#, #removeLast()# and #size()#, and the
+  #addLast(x)#, #removeFirst()#, #removeLast()# and #size()#, and the
   #min()# operation, which returns the minimum value currently stored in
   the data structure.  All operations should run in constant amortized
   time.