cphb/luku04.tex

\chapter{Data structures}

\index{data structure}

A \key{data structure} is a way to store
data in the memory of the computer.
It is important to choose a suitable
data structure for a problem,
because each data structure has its own
advantages and disadvantages.
The crucial question is: which operations
are efficient in the chosen data structure?

This chapter introduces the most important
data structures in the C++ standard library.
It is a good idea to use the standard library
whenever possible,
because it will save a lot of time.
Later in the book we will learn more sophisticated
data structures that are not available
in the standard library.

\section{Dynamic array}

\index{dynamic array}
\index{vector}
\index{vector@\texttt{vector}}

A \key{dynamic array} is an array whose
size can be changed during the execution
of the code.
The most popular dynamic array in C++ is
the \key{vector} structure (\texttt{vector}),
that can be used almost like a regular array.

The following code creates an empty vector and
adds three elements to it:

\begin{lstlisting}
vector<int> v;
v.push_back(3); // [3]
v.push_back(2); // [3,2]
v.push_back(5); // [3,2,5]
\end{lstlisting}

After this, the elements can be accessed like in a regular array:

\begin{lstlisting}
cout << v[0] << "\n"; // 3
cout << v[1] << "\n"; // 2
cout << v[2] << "\n"; // 5
\end{lstlisting}

The function \texttt{size} returns the number of elements in the vector.
The following code iterates through
the vector and prints all elements in it:

\begin{lstlisting}
for (int i = 0; i < v.size(); i++) {
    cout << v[i] << "\n";
}
\end{lstlisting}

\begin{samepage}
A shorter way to iterate trough a vector is as follows:

\begin{lstlisting}
for (auto x : v) {
    cout << x << "\n";
}
\end{lstlisting}
\end{samepage}

The function \texttt{back} returns the last element
in the vector, and
the function \texttt{pop\_back} removes the last element:

\begin{lstlisting}
vector<int> v;
v.push_back(5);
v.push_back(2);
cout << v.back() << "\n"; // 2
v.pop_back();
cout << v.back() << "\n"; // 5
\end{lstlisting}

The following code creates a vector with five elements:

\begin{lstlisting}
vector<int> v = {2,4,2,5,1};
\end{lstlisting}

Another way to create a vector is to give the number
of elements and the initial value for each element:

\begin{lstlisting}
// size 10, initial value 0
vector<int> v(10);
\end{lstlisting}
\begin{lstlisting}
// size 10, initial value 5
vector<int> v(10, 5);
\end{lstlisting}

The internal implementation of the vector
uses a regular array.
If the size of the vector increases and
the array becomes too small,
a new array is allocated and all the
elements are copied to the new array.
However, this doesn't happen often and the
time complexity of
\texttt{push\_back} is $O(1)$ on average.

\index{string}
\index{string@\texttt{string}}

Also the \key{string} structure (\texttt{string})
is a dynamic array that can be used almost like a vector.
In addition, there is special syntax for strings
that is not available in other data structures.
Strings can be combined using the \texttt{+} symbol.
The function $\texttt{substr}(k,x)$ returns the substring
that begins at index $k$ and has length $x$.
The function $\texttt{find}(\texttt{t})$ finds the position
where a substring \texttt{t} appears in the string.

The following code presents some string operations:

\begin{lstlisting}
string a = "hatti";
string b = a+a;
cout << b << "\n"; // hattihatti
b[5] = 'v';
cout << b << "\n"; // hattivatti
string c = b.substr(3,4);
cout << c << "\n"; // tiva
\end{lstlisting}

\section{Set structure}

\index{set}
\index{set@\texttt{set}}
\index{unordered\_set@\texttt{unordered\_set}}

A \key{set} is a data structure that
contains a collection of elements.
The basic operations in a set are element
insertion, search and removal.

C++ contains two set implementations:
\texttt{set} and \texttt{unordered\_set}.
The structure \texttt{set} is based on a balanced
binary tree and the time complexity of its
operations is $O(\log n)$.
The structure \texttt{unordered\_set} uses a hash table,
and the time complexity of its operations is $O(1)$ on average.

The choice which set implementation to use
is often a matter of taste.
The benefit in the \texttt{set} structure
is that it maintains the order of the elements
and provides functions that are not available
in \texttt{unordered\_set}.
On the other hand, \texttt{unordered\_set} is
often more efficient.

The following code creates a set
that consists of integers,
and shows how to use it.
The function \texttt{insert} adds an element to the set,
the function \texttt{count} returns how many times an
element appears in the set,
and the function \texttt{erase} removes an element from the set.

\begin{lstlisting}
set<int> s;
s.insert(3);
s.insert(2);
s.insert(5);
cout << s.count(3) << "\n"; // 1
cout << s.count(4) << "\n"; // 0
s.erase(3);
s.insert(4);
cout << s.count(3) << "\n"; // 0
cout << s.count(4) << "\n"; // 1
\end{lstlisting}

A set can be used mostly like a vector,
but it is not possible to access
the elements using the \texttt{[]} notation.
The following code creates a set,
prints the number of elements in it, and then
iterates through all the elements:
\begin{lstlisting}
set<int> s = {2,5,6,8};
cout << s.size() << "\n"; // 4
for (auto x : s) {
    cout << x << "\n";
}
\end{lstlisting}

An important property of a set is
that all the elements are distinct.
Thus, the function \texttt{count} always returns
either 0 (the element is not in the set)
or 1 (the element is in the set),
and the function \texttt{insert} never adds
an element to the set if it is
already in the set.
The following code illustrates this:

\begin{lstlisting}
set<int> s;
s.insert(5);
s.insert(5);
s.insert(5);
cout << s.count(5) << "\n"; // 1
\end{lstlisting}

\index{multiset@\texttt{multiset}}
\index{unordered\_multiset@\texttt{unordered\_multiset}}

C++ also contains the structures
\texttt{multiset} and \texttt{unordered\_multiset}
that work otherwise like \texttt{set}
and \texttt{unordered\_set}
but they can contain multiple copies of an element.
For example, in the following code all copies
of the number 5 are added to the set:

\begin{lstlisting}
multiset<int> s;
s.insert(5);
s.insert(5);
s.insert(5);
cout << s.count(5) << "\n"; // 3
\end{lstlisting}
The function \texttt{erase} removes
all instances of an element
from a \texttt{multiset}:
\begin{lstlisting}
s.erase(5);
cout << s.count(5) << "\n"; // 0
\end{lstlisting}
Often, only one instance should be removed,
which can be done as follows:
\begin{lstlisting}
s.erase(s.find(5));
cout << s.count(5) << "\n"; // 2
\end{lstlisting}

\section{Map structure}

\index{hakemisto@hakemisto}
\index{map@\texttt{map}}
\index{unordered\_map@\texttt{unordered\_map}}

A \key{map} is a generalized array
that consists of key-value-pairs.
While the keys in a regular array are always
the successive integers $0,1,\ldots,n-1$,
where $n$ is the size of the array,
the keys in a map can be of any data type and
they don't have to be successive values.

C++ contains two map implementations that
correspond to the set implementations:
the structure
\texttt{map} is based on a balanced
binary tree and accessing an element
takes $O(\log n)$ time,
while the structure
\texttt{unordered\_map} uses a hash map
and accessing an element takes $O(1)$ time on average.

The following code creates a map
where the keys are strings and the values are integers:

\begin{lstlisting}
map<string,int> m;
m["monkey"] = 4;
m["banana"] = 3;
m["harpsichord"] = 9;
cout << m["banana"] << "\n"; // 3
\end{lstlisting}

If a value of a key is requested
but the map doesn't contain it,
the key is automatically added to the map with
a default value.
For example, in the following code,
the key ''aybabtu'' with value 0
is added to the map.

\begin{lstlisting}
map<string,int> m;
cout << m["aybabtu"] << "\n"; // 0
\end{lstlisting}
The function \texttt{count} determines
if a key exists in the map:
\begin{lstlisting}
if (m.count("aybabtu")) {
    cout << "key exists in the map";
}
\end{lstlisting}
The following code prints all keys and values
in the map:
\begin{lstlisting}
for (auto x : m) {
    cout << x.first << " " << x.second << "\n";
}
\end{lstlisting}

\section{Iterators and ranges}

\index{iterator}

Many functions in the C++ standard library
are given iterators to data structures,
and iterators often correspond to ranges.
An \key{iterator} is a variable that points
to an element in a data structure.

Often used iterators are \texttt{begin}
and \texttt{end} that define a range that contains
all elements in a data structure.
The iterator \texttt{begin} points to
the first element in the data structure,
and the iterator \texttt{end} points to
the position \emph{after} the last element.
The situation looks as follows:

\begin{center}
\begin{tabular}{llllllllll}
\{ & 3, & 4, & 6, & 8, & 12, & 13, & 14, & 17 & \} \\
& $\uparrow$ & & & & & & & & $\uparrow$ \\
& \multicolumn{3}{l}{\texttt{s.begin()}} & & & & & & \texttt{s.end()} \\
\end{tabular}
\end{center}

Note the asymmetry in the iterators:
\texttt{s.begin()} points to an element in the data structure,
while \texttt{s.end()} points outside the data structure.
Thus, the range defined by the iterators is \emph{half-open}.

\subsubsection{Handling ranges}

Iterators are used in C++ standard library functions
that work with ranges of data structures.
Usually, we want to process all elements in a
data structure, so the iterators
\texttt{begin} and \texttt{end} are given for the function.

For example, the following code sorts a vector
using the function \texttt{sort},
then reverses the order of the elements using the function
\texttt{reverse}, and finally shuffles the order of
the elements using the function \texttt{random\_shuffle}.

\index{sort@\texttt{sort}}
\index{reverse@\texttt{reverse}}
\index{random\_shuffle@\texttt{random\_shuffle}}

\begin{lstlisting}
sort(v.begin(), v.end());
reverse(v.begin(), v.end());
random_shuffle(v.begin(), v.end());
\end{lstlisting}

These functions can also be used with a regular array.
In this case, the functions are given pointers to the array
instead of iterators:

\newpage
\begin{lstlisting}
sort(t, t+n);
reverse(t, t+n);
random_shuffle(t, t+n);
\end{lstlisting}

\subsubsection{Set iterators}

Iterators are often used when accessing
elements in a set.
The following code creates an iterator
\texttt{it} that points to the first element in the set:
\begin{lstlisting}
set<int>::iterator it = s.begin();
\end{lstlisting}
A shorter way to write the code is as follows:
\begin{lstlisting}
auto it = s.begin();
\end{lstlisting}
The element to which an iterator points
can be accessed through the \texttt{*} symbol.
For example, the following code prints
the first element in the set:

\begin{lstlisting}
auto it = s.begin();
cout << *it << "\n";
\end{lstlisting}

Iterators can be moved using operators
\texttt{++} (forward) and \texttt{---} (backward),
meaning that the iterator moves to the next
or previous element in the set.

The following code prints all elements in the set:
\begin{lstlisting}
for (auto it = s.begin(); it != s.end(); it++) {
    cout << *it << "\n";
}
\end{lstlisting}
The following code prints the last element in the set:
\begin{lstlisting}
auto it = s.end();
it--;
cout << *it << "\n";
\end{lstlisting}

The function $\texttt{find}(x)$ returns an iterator
that points to an element whose value is $x$.
However, if the set doesn't contain $x$,
the iterator will be \texttt{end}.

\begin{lstlisting}
auto it = s.find(x);
if (it == s.end()) cout << "x is missing";
\end{lstlisting}

The function $\texttt{lower\_bound}(x)$ returns
an iterator to the smallest element in the set
whose value is at least $x$.
Correspondingly, 
the function $\texttt{upper\_bound}(x)$
returns an iterator to the smallest element
in the set whose value is larger than $x$.
If such elements do not exist,
the return value of the functions will be \texttt{end}.
These functions are not supported by the
\texttt{unordered\_set} structure that
doesn't maintain the order of the elements.

\begin{samepage}
For example, the following code finds the element
nearest to $x$:

\begin{lstlisting}
auto a = s.lower_bound(x);
if (a == s.begin() && a == s.end()) {
    cout << "joukko on tyhjä\n";
} else if (a == s.begin()) {
    cout << *a << "\n";
} else if (a == s.end()) {
    a--;
    cout << *a << "\n";
} else {
    auto b = a; b--;
    if (x-*b < *a-x) cout << *b << "\n";
    else cout << *a << "\n";
}
\end{lstlisting}

The code goes through all possible cases
using the iterator \texttt{a}.
First, the iterator points to the smallest
element whose value is at least $x$.
If \texttt{a} is both \texttt{begin}
and \texttt{end} at the same time, the set is empty.
If \texttt{a} equals \texttt{begin},
the corresponding element is nearest to $x$.
If \texttt{a} equals \texttt{end},
the last element in the set is nearest to $x$.
If none of the previous cases is true,
the element nearest to $x$ is either the
element that corresponds to $a$ or the previous element.
\end{samepage}

\section{Other structures}

\subsubsection{Bitset}

\index{bitset}
\index{bitset@\texttt{bitset}}

A \key{bitset} (\texttt{bitset}) is an array
where each element is either 0 or 1.
For example, the following code creates
a bitset that contains 10 elements:
\begin{lstlisting}
bitset<10> s;
s[2] = 1;
s[5] = 1;
s[6] = 1;
s[8] = 1;
cout << s[4] << "\n"; // 0
cout << s[5] << "\n"; // 1
\end{lstlisting}

The benefit in using a bitset is that
it requires less memory than a regular array,
because each element in the bitset only
uses one bit of memory.
For example, 
if $n$ bits are stored as an \texttt{int} array,
$32n$ bits of memory will be used,
but a corresponding bitset only requires $n$ bits of memory.
In addition, the values in a bitset
can be efficiently manipulated using
bit operators, which makes it possible to
optimize algorithms.

The following code shows another way to create a bitset:
\begin{lstlisting}
bitset<10> s(string("0010011010"));
cout << s[4] << "\n"; // 0
cout << s[5] << "\n"; // 1
\end{lstlisting}

The function \texttt{count} returns the number
of ones in the bitset:

\begin{lstlisting}
bitset<10> s(string("0010011010"));
cout << s.count() << "\n"; // 4
\end{lstlisting}

The following code shows examples of using bit operations:
\begin{lstlisting}
bitset<10> a(string("0010110110"));
bitset<10> b(string("1011011000"));
cout << (a&b) << "\n"; // 0010010000
cout << (a|b) << "\n"; // 1011111110
cout << (a^b) << "\n"; // 1001101110
\end{lstlisting}

\subsubsection{Pakka}

\index{deque}
\index{deque@\texttt{deque}}

A \key{deque} (\texttt{deque}) is a dynamic array
whose size can be changed at both ends of the array.
Like a vector, a deque contains functions
\texttt{push\_back} and \texttt{pop\_back}, but
it also contains additional functions
\texttt{push\_front} and \texttt{pop\_front}
that are not available in a vector.

A deque can be used as follows:
\begin{lstlisting}
deque<int> d;
d.push_back(5); // [5]
d.push_back(2); // [5,2]
d.push_front(3); // [3,5,2]
d.pop_back(); // [3,5]
d.pop_front(); // [5]
\end{lstlisting}

The internal implementation of a deque
is more complex than the implementation of a vector.
For this reason, a deque is slower than a vector.
Still, the time complexity of adding and removing
elements is $O(1)$ on average at both ends.

\subsubsection{Pino}

\index{stack}
\index{stack@\texttt{stack}}

A \key{stack} (\texttt{stack})
is a data structure that provides two
$O(1)$ time operations:
adding an element to the top,
and removing an element from the top.
It is only possible to access the top
element of a stack.

The following code shows how a stack can be used:
\begin{lstlisting}
stack<int> s;
s.push(3);
s.push(2);
s.push(5);
cout << s.top(); // 5
s.pop();
cout << s.top(); // 2
\end{lstlisting}
\subsubsection{Queue}

\index{queue}
\index{queue@\texttt{queue}}

A \key{queue} (\texttt{queue}) also
provides two $O(1)$ time operations:
adding a new element to the end,
and removing the first element.
It is only possible to access the first
and the last element of a queue.

The following code shows how a queue can be used:
\begin{lstlisting}
queue<int> s;
s.push(3);
s.push(2);
s.push(5);
cout << s.front(); // 3
s.pop();
cout << s.front(); // 2
\end{lstlisting}

\subsubsection{Priority queue}

\index{priority queue}
\index{heap}
\index{priority\_queue@\texttt{priority\_queue}}

A \key{priority queue} (\texttt{priority\_queue})
maintains a set of elements.
The supported operations are insertion and,
depending on the type of the queue,
retrieval and removal of
either the minimum element or the maximum element.
The time complexity is $O(\log n)$
for insertion and removal and $O(1)$ for retrieval.

While a set structure efficiently supports 
all the operations of a priority queue,
the benefit in using a priority queue is
that it has smaller constant factors.
A priority queue is usually implemented using
a heap structure that is much simpler than a
balanced binary tree needed for an ordered set.

\begin{samepage}
As default, the elements in the C++
priority queue are sorted in decreasing order,
and it is possible to find and remove the
largest element in the queue.
The following code shows an example:

\begin{lstlisting}
priority_queue<int> q;
q.push(3);
q.push(5);
q.push(7);
q.push(2);
cout << q.top() << "\n"; // 7
q.pop();
cout << q.top() << "\n"; // 5
q.pop();
q.push(6);
cout << q.top() << "\n"; // 6
q.pop();
\end{lstlisting}
\end{samepage}

The following definition creates a priority queue
that supports finding and removing the minimum element:

\begin{lstlisting}
priority_queue<int,vector<int>,greater<int>> q;
\end{lstlisting}

\section{Comparison to sorting}

Often it's possible to solve a problem
using either data structures or sorting.
Sometimes there are remarkable differences
in the actual efficiency of these approaches,
which may be hidden in their time complexities.

Let us consider a problem where
we are given two lists $A$ and $B$
that both contain $n$ integers.
Our task is to calculate the number of integers
that belong to both of the lists.
For example, for the lists
\[A = [5,2,8,9,4] \hspace{10px} \textrm{and} \hspace{10px} B = [3,2,9,5],\]
the answer is 3 because the numbers 2, 5
and 9 belong to both of the lists.

A straightforward solution for the problem is
to go through all pairs of numbers in $O(n^2)$ time,
but next we will concentrate on
more efficient solutions.

\subsubsection{Solution 1}

We construct a set of the numbers in $A$,
and after this, iterate through the numbers
in $B$ and check for each number if it
also belongs to $A$.
This is efficient because the numbers in $A$
are in a set.
Using the \texttt{set} structure,
the time complexity of the algorithm is $O(n \log n)$.

\subsubsection{Solution 2}

It is not needed to maintain an ordered set,
so instead of the \texttt{set} structure
we can also use the \texttt{unordered\_set} structure.
This is an easy way to make the algorithm
more efficient because we only have to change
the data structure that the algorithm uses.
The time complexity of the new algorithm is $O(n)$.

\subsubsection{Solution 3}

Instead of data structures, we can use sorting.
First, we sort both lists $A$ and $B$.
After this, we iterate through both the lists
at the same time and find the common elements.
The time complexity of sorting is $O(n \log n)$,
and the rest of the algorithm works in $O(n)$ time,
so the total time complexity is $O(n \log n)$.

\subsubsection{Efficiency comparison}

The following table shows how efficient
the above algorithms are when $n$ varies and
the elements in the lists are random
integers between $1 \ldots 10^9$:

\begin{center}
\begin{tabular}{rrrr}
$n$ & solution 1 & solution 2 & solution 3 \\
\hline
$10^6$ & $1{,}5$ s & $0{,}3$ s & $0{,}2$ s \\
$2 \cdot 10^6$ & $3{,}7$ s & $0{,}8$ s & $0{,}3$ s \\
$3 \cdot 10^6$ & $5{,}7$ s & $1{,}3$ s & $0{,}5$ s \\
$4 \cdot 10^6$ & $7{,}7$ s & $1{,}7$ s & $0{,}7$ s \\
$5 \cdot 10^6$ & $10{,}0$ s & $2{,}3$ s & $0{,}9$ s \\
\end{tabular}
\end{center}

Solutions 1 and 2 are equal except that
solution 1 uses the \texttt{set} structure
and solution 2 uses the \texttt{unordered\_set} structure.
In this case, this choice has a big effect on
the running time becase solution 2
is 4–5 times faster than solution 1.

However, the most efficient solution is solution 3
that uses sorting.
It only uses half of the time compared to solution 2.
Interestingly, the time complexity of both
solution 1 and solution 3 is $O(n \log n)$,
but despite this, solution 3 is ten times faster.
The explanation for this is that
sorting is a simple procedure and it is done
only once at the beginning of solution 3,
and the rest of the algorithm works in linear time.
On the other hand,
solution 3 maintains a complex balanced binary tree
during the whole algorithm.