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%\usepackage{opensans}          % Can't make this work so far. Shame. Would be lovely.
\usepackage[normalem]{ulem}     % For underlining links
% Most of the following are not required for the majority
% of cheat sheets but are needed for some symbol support.
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%\usepackage[english,german,french,spanish,italian]{babel}              % Languages

% Document Info
\author{jasondias}
\pdfinfo{
  /Title (comp250.pdf)
  /Creator (Cheatography)
  /Author (jasondias)
  /Subject (COMP250 Cheat Sheet)
}

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% Header and Footer
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\fancyhead{} % Set header to blank
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\fancyhead[L]{
\noindent
\begin{multicols}{3}
\begin{tabulary}{5.8cm}{C}
    \SetRowColor{DarkBackground}
    \vspace{-7pt}
    {\parbox{\dimexpr\textwidth-2\fboxsep\relax}{\noindent
        \hspace*{-6pt}\includegraphics[width=5.8cm]{/web/www.cheatography.com/public/images/cheatography_logo.pdf}}
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\columnbreak
\begin{tabulary}{11cm}{L}
    \vspace{-2pt}\large{\bf{\textcolor{DarkBackground}{\textrm{COMP250 Cheat Sheet}}}} \\
    \normalsize{by \textcolor{DarkBackground}{jasondias} via \textcolor{DarkBackground}{\uline{cheatography.com/21209/cs/5468/}}}
\end{tabulary}
\end{multicols}}

\fancyfoot[L]{ \footnotesize
\noindent
\begin{multicols}{3}
\begin{tabulary}{5.8cm}{LL}
  \SetRowColor{FootBackground}
  \mymulticolumn{2}{p{5.377cm}}{\bf\textcolor{white}{Cheatographer}}  \\
  \vspace{-2pt}jasondias \\
  \uline{cheatography.com/jasondias} \\
  \end{tabulary}
\vfill
\columnbreak
\begin{tabulary}{5.8cm}{L}
  \SetRowColor{FootBackground}
  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Cheat Sheet}}  \\
   \vspace{-2pt}Published 20th October, 2015.\\
   Updated 9th May, 2016.\\
   Page {\thepage} of \pageref{LastPage}.
\end{tabulary}
\vfill
\columnbreak
\begin{tabulary}{5.8cm}{L}
  \SetRowColor{FootBackground}
  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Sponsor}}  \\
  \SetRowColor{white}
  \vspace{-5pt}
  %\includegraphics[width=48px,height=48px]{dave.jpeg}
  Measure your website readability!\\
  www.readability-score.com
\end{tabulary}
\end{multicols}}




\begin{document}
\raggedright
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\begin{multicols*}{4}

\begin{tabularx}{3.833cm}{x{1.40753 cm} x{2.02547 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{3.833cm}}{\bf\textcolor{white}{Algorithms}}  \tn
% Row 0
\SetRowColor{LightBackground}
Definition & unambiguous procedure executed in a finite number of steps \tn 
% Row Count 3 (+ 3)
% Row 1
\SetRowColor{white}
What makes a good algorithm? & Correctness, Speed, Space, Simplicity \tn 
% Row Count 5 (+ 2)
% Row 2
\SetRowColor{LightBackground}
Speed: & time it takes to solve problem \tn 
% Row Count 7 (+ 2)
% Row 3
\SetRowColor{white}
Space: & amount of memory required \tn 
% Row Count 9 (+ 2)
% Row 4
\SetRowColor{LightBackground}
Simplicity: & easy to understand, easy to implement, easy to debug, modify, update \tn 
% Row Count 12 (+ 3)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{3.833cm}{p{0.6066 cm} x{1.2132 cm} x{1.2132 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{3}{x{3.833cm}}{\bf\textcolor{white}{Running Time}}  \tn
% Row 0
\SetRowColor{LightBackground}
\seqsplit{Definition} & measurement of the speed of an algorithm &  \tn 
% Row Count 3 (+ 3)
% Row 1
\SetRowColor{white}
\seqsplit{Dependent} \seqsplit{variables:} & size of input \& content of input &  \tn 
% Row Count 6 (+ 3)
% Row 2
\SetRowColor{LightBackground}
Best Case: & time on the easiest input of fixed size & meaningless \tn 
% Row Count 9 (+ 3)
% Row 3
\SetRowColor{white}
Average Case: & time on average input & good measure, hard to calculate \tn 
% Row Count 11 (+ 2)
% Row 4
\SetRowColor{LightBackground}
Worst Case: & time on most difficult input & good for safety critical systems, easy to estimate \tn 
% Row Count 15 (+ 4)
\hhline{>{\arrayrulecolor{DarkBackground}}---}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{3.833cm}{x{1.0299 cm} x{2.4031 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{3.833cm}}{\bf\textcolor{white}{Loop Invariants}}  \tn
% Row 0
\SetRowColor{LightBackground}
Definition & loop property that holds before and after every iteration of a loop. \tn 
% Row Count 3 (+ 3)
% Row 1
\SetRowColor{white}
\mymulticolumn{2}{x{3.833cm}}{Steps:} \tn 
% Row Count 4 (+ 1)
% Row 2
\SetRowColor{LightBackground}
1. \seqsplit{Initialization} & If it is true prior to the iteration of the loop \tn 
% Row Count 6 (+ 2)
% Row 3
\SetRowColor{white}
2. Maintenance & If it is true before an iteration of the loop, it remains true before the next iteration \tn 
% Row Count 10 (+ 4)
% Row 4
\SetRowColor{LightBackground}
3. Termination & When the loop terminates, the invariant gives us a useful property that helps show the algorithm is correct \tn 
% Row Count 14 (+ 4)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{3.833cm}{p{0.75526 cm} x{2.67774 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{3.833cm}}{\bf\textcolor{white}{In Place Sorting}}  \tn
% Row 0
\SetRowColor{LightBackground}
\seqsplit{Definition:} & Uses only a constant amount of memory in addition of that used to store the input \tn 
% Row Count 3 (+ 3)
% Row 1
\SetRowColor{white}
\seqsplit{Importance:} & Great for large data sets that take up large amounts of memory \tn 
% Row Count 5 (+ 2)
% Row 2
\SetRowColor{LightBackground}
\seqsplit{Examples:} & Selection Sort, Insertion Sort (Only moving elements around the array) \tn 
% Row Count 8 (+ 3)
% Row 3
\SetRowColor{white}
\seqsplit{MergeSort:} & Not in place: new array required \tn 
% Row Count 10 (+ 2)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{3.833cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{QuickSort}}  \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{3.833cm}}{Algorithm partition(A, start, stop) \newline Input: An array A, indices start and stop. \newline Output: Returns an index j and rearranges the elements of A \newline such that for all i\textless{}j, A{[}i{]} ! A{[}j{]} and \newline for all k\textgreater{}i, A{[}k{]} " A{[}j{]}. \newline pivot \# A{[}stop{]} \newline left \# start \newline right \# stop - 1 \newline while left ! right do \newline while left ! right and A{[}left{]} ! pivot) do left \# left + 1 \newline  while (left ! right and A{[}right{]} " pivot) do right \# right -1 \newline if (left \textless{} right ) then exchange A{[}left{]} \$ A{[}right{]} \newline exchange A{[}stop{]} \$ A{[}left{]} \newline return left} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{3.833cm}}{Time Complexities: \newline • Worse case: \newline – Already sorted array (either increasing or decreasing) \newline – T(n) = T(n-1) + c n + d \newline – T(n) is O(n2) \newline • Average case: If the array is in random order, the \newline pivot splits the array in roughly equal parts, so the \newline average running time is O(n log n) \newline • Advantage over mergeSort: \newline – constant hidden in O(n log n) are smaller for quickSort. \newline Thus it is faster by a constant factor \newline – QuickSort is easy to do "in-place"}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{3.833cm}{p{0.51495 cm} x{2.91805 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{3.833cm}}{\bf\textcolor{white}{QuickSort}}  \tn
% Row 0
\SetRowColor{LightBackground}
\seqsplit{Divide:} & choose an element of the array for pivot \tn 
% Row Count 2 (+ 2)
% Row 1
\SetRowColor{white}
 & divide into 3 sub-groups; those smaller, those larger and those equal to pivot \tn 
% Row Count 5 (+ 3)
% Row 2
\SetRowColor{LightBackground}
\seqsplit{Conquer} & recursively sort each group \tn 
% Row Count 7 (+ 2)
% Row 3
\SetRowColor{white}
\seqsplit{Combine} & concatenate the 3 lists \tn 
% Row Count 9 (+ 2)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{3.833cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Proofs by Induction (Examples)}}  \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{p{3.833cm}}{\vspace{1px}\centerline{\includegraphics[width=5.1cm]{/web/www.cheatography.com/public/uploads/jasondias_1445380086_Screen Shot 2015-10-20 at 6.28.15 PM.png}}} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{3.833cm}{x{0.92691 cm} x{2.50609 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{3.833cm}}{\bf\textcolor{white}{Object Orientated Programming}}  \tn
% Row 0
\SetRowColor{LightBackground}
\seqsplit{Definition:} & User defined types to complement primitive types \tn 
% Row Count 2 (+ 2)
% Row 1
\SetRowColor{white}
 & Called a class \tn 
% Row Count 3 (+ 1)
% Row 2
\SetRowColor{LightBackground}
Contains: & Data \& methods \tn 
% Row Count 4 (+ 1)
% Row 3
\SetRowColor{white}
Static members: & members shared by all objects of the class \tn 
% Row Count 6 (+ 2)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{3.833cm}{x{0.89258 cm} x{2.54042 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{3.833cm}}{\bf\textcolor{white}{Recursion Programming}}  \tn
% Row 0
\SetRowColor{LightBackground}
\seqsplit{Definition} & using methods that call themselves \tn 
% Row Count 2 (+ 2)
% Row 1
\SetRowColor{white}
\mymulticolumn{2}{x{3.833cm}}{Structure:} \tn 
% Row Count 3 (+ 1)
% Row 2
\SetRowColor{LightBackground}
base case & a simple occurrence that can be answered directly \tn 
% Row Count 5 (+ 2)
% Row 3
\SetRowColor{white}
recursive case & A more complex occurrence of	the problem that cannot be directly answered,	but can instead	be described in	terms of smaller occurrences of the same problem. \tn 
% Row Count 11 (+ 6)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{3.833cm}{p{0.51495 cm} x{2.91805 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{3.833cm}}{\bf\textcolor{white}{Divide \& Conquer}}  \tn
% Row 0
\SetRowColor{LightBackground}
\seqsplit{Divide} & the problem into sub problems that are similar to the original but smaller in size \tn 
% Row Count 3 (+ 3)
% Row 1
\SetRowColor{white}
\seqsplit{Conquer} & the sub-problems by solving them recursively. If they are small enough, solve them in a straightforward manner \tn 
% Row Count 7 (+ 4)
% Row 2
\SetRowColor{LightBackground}
\seqsplit{Combine} & the solutions to create a solution to the original problem \tn 
% Row Count 9 (+ 2)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{3.833cm}{x{1.51052 cm} x{1.92248 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{3.833cm}}{\bf\textcolor{white}{BIG O Definition}}  \tn
% Row 0
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{3.833cm}}{f(n) \& g(n) are two non negative functions defined on the natural numbers N} \tn 
% Row Count 2 (+ 2)
% Row 1
\SetRowColor{white}
f(n) is O(g(n)) if and only if: & there exists an integer n0 and a real number c  such that Ɐn\textgreater{}= n0 f(n) \textless{}= c * g(n) \tn 
% Row Count 6 (+ 4)
% Row 2
\SetRowColor{LightBackground}
 & N.B. The constant c must not depend on n \tn 
% Row Count 8 (+ 2)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{3.833cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Big O Visualization}}  \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{p{3.833cm}}{\vspace{1px}\centerline{\includegraphics[width=5.1cm]{/web/www.cheatography.com/public/uploads/jasondias_1445523925_Screen Shot 2015-10-22 at 10.25.11 AM.png}}} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{3.833cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Big O Recurrence}}  \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{p{3.833cm}}{\vspace{1px}\centerline{\includegraphics[width=5.1cm]{/web/www.cheatography.com/public/uploads/jasondias_1445529071_camera_capture.jpg}}} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{3.833cm}}{Sum of a\textasciicircum{}i from 0 to n = (a\textasciicircum{}(n+1) -1 )/(a-1)}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{3.833cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Limitations of Arrays}}  \tn
% Row 0
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{3.833cm}}{Size has to be known in advance} \tn 
% Row Count 1 (+ 1)
% Row 1
\SetRowColor{white}
\mymulticolumn{1}{x{3.833cm}}{memory required may be larger than number of elements} \tn 
% Row Count 3 (+ 2)
% Row 2
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{3.833cm}}{inserting or deleting an element takes up to O(n)} \tn 
% Row Count 4 (+ 1)
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{3.833cm}{p{0.6866 cm} x{2.7464 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{3.833cm}}{\bf\textcolor{white}{ADT: Queues}}  \tn
% Row 0
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{3.833cm}}{FIFO(First in first out)} \tn 
% Row Count 1 (+ 1)
% Row 1
\SetRowColor{white}
\mymulticolumn{2}{x{3.833cm}}{Any first come first serve service} \tn 
% Row Count 2 (+ 1)
% Row 2
\SetRowColor{LightBackground}
\seqsplit{Operations} & enqueue() - add to rear \tn 
% Row Count 4 (+ 2)
% Row 3
\SetRowColor{white}
 & dequeue() - removes object at front \tn 
% Row Count 6 (+ 2)
% Row 4
\SetRowColor{LightBackground}
 & front() - returns object at front \tn 
% Row Count 8 (+ 2)
% Row 5
\SetRowColor{white}
 & size() - returns number of objects O(n) \tn 
% Row Count 10 (+ 2)
% Row 6
\SetRowColor{LightBackground}
 & isEmpty() - returns true if empty \tn 
% Row Count 12 (+ 2)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{3.833cm}}{Double Ended Queues(deque): Allows for insertions and removals from front and back \newline - By adding reference to previous node - removals occur in O(1)}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{3.833cm}{p{0.78959 cm} x{2.64341 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{3.833cm}}{\bf\textcolor{white}{ATD: Stacks}}  \tn
% Row 0
\SetRowColor{LightBackground}
Def: & Operations allowed at only one end of the list (top) \tn 
% Row Count 2 (+ 2)
% Row 1
\SetRowColor{white}
 & LIFO: (Last in first out) \tn 
% Row Count 3 (+ 1)
% Row 2
\SetRowColor{LightBackground}
\seqsplit{Operations:} & push() - inserts element at top \tn 
% Row Count 5 (+ 2)
% Row 3
\SetRowColor{white}
 & pop() - removes object at top \tn 
% Row Count 6 (+ 1)
% Row 4
\SetRowColor{LightBackground}
 & top() - returns top element without removing it \tn 
% Row Count 8 (+ 2)
% Row 5
\SetRowColor{white}
 & size() - returns number of elements \tn 
% Row Count 10 (+ 2)
% Row 6
\SetRowColor{LightBackground}
 & isEmpty() - returns True if empty \tn 
% Row Count 12 (+ 2)
% Row 7
\SetRowColor{white}
\seqsplit{Applications} & page visited history in web browser \tn 
% Row Count 14 (+ 2)
% Row 8
\SetRowColor{LightBackground}
 & JVM - keeps track of chain of active elements (allows for rec) \tn 
% Row Count 17 (+ 3)
% Row 9
\SetRowColor{white}
\seqsplit{Performance:} & space used: O(n) \tn 
% Row Count 19 (+ 2)
% Row 10
\SetRowColor{LightBackground}
 & operations: O(1) \tn 
% Row Count 20 (+ 1)
% Row 11
\SetRowColor{white}
\seqsplit{Limitations} & max size must be defined prior \tn 
% Row Count 22 (+ 2)
% Row 12
\SetRowColor{LightBackground}
 & pushing to a full stack causes implementation specific error \tn 
% Row Count 24 (+ 2)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{3.833cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{BinarySearch}}  \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{3.833cm}}{BinarySearch(A{[}0..N-1{]}, value) \{ \newline       low = 0 \newline       high = N - 1 \newline       while (low \textless{}= high) \{ \newline         mid = (low + high) / 2 \newline           if (A{[}mid{]} \textgreater{} value) \newline               high = mid - 1 \newline           else if (A{[}mid{]} \textless{} value) \newline               low = mid + 1 \newline           else \newline               return mid \newline       \} \newline       return not\_found // value would be inserted at index "low" \newline   \}} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{3.833cm}}{Invariants:  \newline `value \textgreater{} A{[}i{]} for all i \textless{} low `  `value \textless{} A{[}i{]} for all i \textgreater{} high` \newline Worst case performance:	O(log n) \newline Best case performance:	O(1) \newline Average case performance:	O(log n)}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{3.833cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{BinarySearch (Recursive)}}  \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{3.833cm}}{int bsearch(int{[}{]} A, int i, int j, int x) \{  \newline if (i\textless{}j) \{ \newline int e = {[}(i+j)/2{]}; \newline if (A{[}e{]} \textgreater{} x) \{ \newline return bsearch(A,i,e-1); \newline \} else if (A{[}e{]} \textless{} x) \{ \newline return bsearch(A,e+1,j); \newline \} else \{ \newline return e; \newline \} \newline \} else \{ return -1; \}	 \newline \}} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{3.833cm}}{Time Complexity: log(base2)(n)}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{3.833cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Insertion Sort (Iterative)}}  \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{3.833cm}}{For i ← 1 to length(A) - 1 \newline  j ← i \newline  while j \textgreater{} 0 and A{[}j-1{]} \textgreater{} A{[}j{]} \newline  swap A{[}j{]} and A{[}j-1{]} \newline  j ← j - 1 \newline  end while \newline end for} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{3.833cm}}{Time complexity: O(n\textasciicircum{}2\textasciicircum{})}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{3.833cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Merge-then-sort}}  \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{3.833cm}}{Algorithm ListIntersection (A,m, B,n) \newline Input: Same as before \newline Output: Same as before \newline inter ← 0 \newline Array C{[}m+n{]}; \newline for i ← 0 to m-1 do C{[}i{]} ← A{[}i{]}; \newline for i ← 0 to n-1 do C{[} i+m {]} ← B{[}i{]}; \newline C ← sort( C, m+n ); \newline ptr ← 0 \newline while ( ptr \textless{} m+n-1 ) do \{ \newline  if ( C{[}ptr{]} = C{[}ptr+1{]} ) then \{ \newline  inter ← inter+1 \newline  ptr ← ptr+2 \newline  \} \newline  else ptr ← ptr+1 \newline \} \newline return inter} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{3.833cm}}{Time Complexity: (m+n) * (log(m+n)) + m + n -1}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{3.833cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{MergeSort (Recursive)}}  \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{3.833cm}}{MergeSort (A, p, r) // sort A{[}p..r{]} by divide \& conquer \newline if p \textless{} r \newline    then q ← ⎣(p+r)/2⎦ \newline       MergeSort (A, p, q) \newline       MergeSort (A, q+1, r) \newline       Merge (A, p, q, r)} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{3.833cm}}{Time Complexity:  2T(n/2) + k • n + c'}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{3.833cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Primitive Operations}}  \tn
% Row 0
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{3.833cm}}{assignment} \tn 
% Row Count 1 (+ 1)
% Row 1
\SetRowColor{white}
\mymulticolumn{1}{x{3.833cm}}{calling method} \tn 
% Row Count 2 (+ 1)
% Row 2
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{3.833cm}}{returning from method} \tn 
% Row Count 3 (+ 1)
% Row 3
\SetRowColor{white}
\mymulticolumn{1}{x{3.833cm}}{arithmetic} \tn 
% Row Count 4 (+ 1)
% Row 4
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{3.833cm}}{comparisons of primitive types} \tn 
% Row Count 5 (+ 1)
% Row 5
\SetRowColor{white}
\mymulticolumn{1}{x{3.833cm}}{conditional} \tn 
% Row Count 6 (+ 1)
% Row 6
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{3.833cm}}{indexing into array} \tn 
% Row Count 7 (+ 1)
% Row 7
\SetRowColor{white}
\mymulticolumn{1}{x{3.833cm}}{following object reference} \tn 
% Row Count 8 (+ 1)
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{3.833cm}}{Assume each primitive operation holds the same value = 1 primitive operation}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{3.833cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Prove Big - Oh}}  \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{p{3.833cm}}{\vspace{1px}\centerline{\includegraphics[width=5.1cm]{/web/www.cheatography.com/public/uploads/jasondias_1445524538_camera_capture.jpg}}} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}


% That's all folks
\end{multicols*}

\end{document}