\chapter{String algorithms}

\index{string}
\index{alphabet}

A string $s$ of length $n$
is a sequence of characters
$s[1],s[2],\ldots,s[n]$.

An \key{alphabet} is a set of characters
that may appear in strings.
For example, the alphabet
$\{\texttt{A},\texttt{B},\ldots,\texttt{Z}\}$
consists of the capital letters of English.

\index{substring}

A \key{substring} consists of consecutive
characters in a string.
The number of substrings in a string is $n(n+1)/2$.
For example, \texttt{ORITH} is a substring
in \texttt{ALGORITHM}, and it corresponds
to \texttt{ALG\underline{ORITH}M}.

\index{subsequence}

A \key{subsequence} is a subset of characters
in a string in their original order.
The number of subsequences in a string is $2^n-1$.
For example, \texttt{LGRHM} is a subsequece
in \texttt{ALGORITHM}, and it corresponds
to \texttt{A\underline{LG}O\underline{R}IT\underline{HM}}.

\index{prefix}
\index{suffix}

A \key{prefix} is a subtring that contains the first
character of a string,
and a \key{suffix} is a substring that contains the last character.
For example, the prefixes of
\texttt{STORY} are \texttt{S}, \texttt{ST},
\texttt{STO}, \texttt{STOR} and \texttt{STORY},
and the suffixes are \texttt{Y}, \texttt{RY},
\texttt{ORY}, \texttt{TORY} and \texttt{STORY}.
A prefix or a suffix is \key{proper}
if it is not the whole string.

\index{rotation}

A \key{rotation} can be generated by moving
characters one by one from the beginning to the end
in a string (or vice versa).
For example, the rotations of \texttt{STORY} are
\texttt{STORY},
\texttt{TORYS},
\texttt{ORYST},
\texttt{RYSTO} and
\texttt{YSTOR}.

\index{period}

A \key{period} is a prefix of a string such that
we can construct the string by repeating the period.
The last repetition may be partial and contain
only a prefix of the period.
Often it is interesting to find the \key{shortest period}
of a string.
For example, the shortest period of
\texttt{ABCABCA} is \texttt{ABC}.
In this case, we first repeat the period twice
and then partially.

\index{border}

A \key{border} is a string that is both
a prefix and a suffix of a string.
For example, the borders for \texttt{ABADABA}
are \texttt{A}, \texttt{ABA} and \texttt{ABADABA}.
Often we want to find the \key{longest border}
that is not the whole string.

\index{lexicographical order}

Usually we compare string using the \key{lexicographical order}
that corresponds to the alphabetical order.
It means that $x<y$ if either $x$ is a proper prefix of $y$,
or there is an index $k$ such that
$x[i]=y[i]$ when $i<k$ and $x[k]<y[k]$.

\section{Trie structure}

\index{trie}

A \key{trie} is a tree structure that
maintains a set of strings.
Strings are stored in a trie as chains
of characters that start at the root
of the tree.
If two strings have a common prefix,
they also share a chain in the tree.

For example, the following trie corresponds
to the set
$\{\texttt{CANAL},\texttt{CANDY},\texttt{THE},\texttt{THERE}\}$:

\begin{center}
\begin{tikzpicture}[scale=0.9]
\node[draw, circle] (1) at (0,20) {$\phantom{1}$};
\node[draw, circle] (2) at (-1.5,19) {$\phantom{1}$};
\node[draw, circle] (3) at (1.5,19) {$\phantom{1}$};
\node[draw, circle] (4) at (-1.5,17.5) {$\phantom{1}$};
\node[draw, circle] (5) at (-1.5,16) {$\phantom{1}$};
\node[draw, circle] (6) at (-2.5,14.5) {$\phantom{1}$};
\node[draw, circle] (7) at (-0.5,14.5) {$\phantom{1}$};
\node[draw, circle] (8) at (-2.5,13) {*};
\node[draw, circle] (9) at (-0.5,13) {*};
\node[draw, circle] (10) at (1.5,17.5) {$\phantom{1}$};
\node[draw, circle] (11) at (1.5,16) {*};
\node[draw, circle] (12) at (1.5,14.5) {$\phantom{1}$};
\node[draw, circle] (13) at (1.5,13) {*};

\path[draw,thick,->] (1) -- node[font=\small,label=\texttt{C}] {} (2);
\path[draw,thick,->] (1) -- node[font=\small,label=\texttt{T}] {} (3);
\path[draw,thick,->] (2) -- node[font=\small,label=left:\texttt{A}] {} (4);
\path[draw,thick,->] (4) -- node[font=\small,label=left:\texttt{N}] {} (5);
\path[draw,thick,->] (5) -- node[font=\small,label=left:\texttt{A}] {} (6);
\path[draw,thick,->] (5) -- node[font=\small,label=right:\texttt{D}] {} (7);
\path[draw,thick,->] (6) -- node[font=\small,label=left:\texttt{L}] {}(8);
\path[draw,thick,->] (7) -- node[font=\small,label=right:\texttt{Y}] {} (9);
\path[draw,thick,->] (3) -- node[font=\small,label=right:\texttt{H}] {} (10);
\path[draw,thick,->] (10) -- node[font=\small,label=right:\texttt{E}] {} (11);
\path[draw,thick,->] (11) -- node[font=\small,label=right:\texttt{R}] {} (12);
\path[draw,thick,->] (12) -- node[font=\small,label=right:\texttt{E}] {} (13);
\end{tikzpicture}
\end{center}
The character * in a node means that
a string ends at the node.
This character is needed because a string
may be a prefix of another string.
For example, in this trie, \texttt{THE}
is a suffix of \texttt{THERE}.

Inserting and searching a string in a trie take $O(n)$ time
where $n$ is the length of the string.
Both operations can be implemented by
starting at the root node and following the
chain of characters that appear in the string.
If needed, new nodes will be added to the trie.

Tries can be used for searching both strings
and prefixes of strings.
In addition, it is possible to calculate numbers
of strings that correspond to each prefix,
which can be useful in some applications.

A trie can be stored as an array
\begin{lstlisting}
int t[N][A];
\end{lstlisting}
where $N$ is the maximum number of nodes
(the total length of the string to be stored)
and $A$ is the size of the alphabet.
The nodes of a trie are numbered
$1,2,3,\ldots$ so that the number of the root is 1,
and $\texttt{t}[s][c]$ is the next node in chain
from node $s$ using character $c$.

\section{String hashing}

\index{hashing}
\index{string hashing}

\key{String hashing} is a technique that
allows us to efficiently check whether two
substrings in a string are equal.
The idea is to compare hash values of the
substrings instead of their individual characters.

\subsubsection*{Calculating hash values}

\index{hash value}
\index{polynomial hashing}

A \key{hash value} of a string is
a number that is calculated from the characters
of the string.
If two strings are the same,
their hash values are also the same,
which makes it possible to compare strings
based on their hash values.

A usual way to implement string hashing
is to use polynomial hashing, which means
that the hash value is calculated using the formula
\[(c[1] A^{n-1} + c[2] A^{n-2} + \cdots + c[n] A^0) \bmod B  ,\]
where $c[1],c[2],\ldots,c[n]$
are the codes of the characters in the string,
and $A$ and $B$ are pre-chosen constants.

For example, the codes of the characters
in the string \texttt{ALLEY} are:
\begin{center}
\begin{tikzpicture}[scale=0.7]
\draw (0,0) grid (5,2);

\node at (0.5, 1.5) {\texttt{A}};
\node at (1.5, 1.5) {\texttt{L}};
\node at (2.5, 1.5) {\texttt{L}};
\node at (3.5, 1.5) {\texttt{E}};
\node at (4.5, 1.5) {\texttt{Y}};

\node at (0.5, 0.5) {65};
\node at (1.5, 0.5) {76};
\node at (2.5, 0.5) {76};
\node at (3.5, 0.5) {69};
\node at (4.5, 0.5) {89};

\end{tikzpicture}
\end{center}

If $A=3$ and $B=97$, the hash value
for the string \texttt{ALLEY} is

\[(65 \cdot 3^4 + 76 \cdot 3^3 + 76 \cdot 3^2 + 69 \cdot 3^1 + 89 \cdot 3^0) \bmod 97 = 52.\]

\subsubsection*{Preprocessing}

To efficiently calculate hash values of substrings,
we need to preprocess the string.
It turns out that using polynomial hashing,
we can calculate the hash value of any substring
in $O(1)$ time after an $O(n)$ time preprocessing.

The idea is to construct an array $h$ such that
$h[k]$ contains the hash value for the prefix
of the string that ends at index $k$.
The array values can be recursively calculated as follows:
\[
\begin{array}{lcl}
h[0] & = & 0 \\
h[k] & = & (h[k-1] A + c[k]) \bmod B \\
\end{array}
\]
In addition, we construct an array $p$
where $p[k]=A^k \bmod B$:
\[
\begin{array}{lcl}
p[0] & = & 1 \\
p[k] & = & (p[k-1] A) \bmod B. \\
\end{array}
\]
Constructing these arrays takes $O(n)$ time.
After this, the hash value for a substring
of the string
that begins at index $a$ and ends at index $b$
can be calculated in $O(1)$ time using the formula
\[(h[b]-h[a-1] p[b-a+1]) \bmod B.\]

\subsubsection*{Using hash values}

We can efficiently compare strings using hash values.
Instead of comparing the real contents of the strings,
the idea is to compare their hash values.
If the hash values are equal,
the strings are \emph{probably} equal,
and if the hash values are different,
the strings are \emph{certainly} different.

Using hashing, we can often make a brute force
algorithm efficient.
As an example, let's consider a brute force
algorithm that calculates how many times
a string $p$ occurs as a substring in
a string $s$.
The algorithm goes through all locations
where $p$ can occur, and compares the strings
character by character.
The time complexity of such an algorithm is $O(n^2)$.

However, we can make the algorithm more efficient
using hashing, because the algorithm compares
substrings of strings.
Using hashing, each comparison only takes $O(1)$ time,
because only hash values of the strings are compared.
This results in an algorithm with time complexity $O(n)$,
which is the best possible time complexity for this problem.

By combining hashing and \emph{binary search},
it is also possible to check the lexicographic order of
two strings in logarithmic time.
This can be done by finding out the length
of the common prefix of the strings using binary search.
Once we know the common prefix,
the next character after the prefix
indicates the order of the strings.

\subsubsection*{Collisions and parameters}

\index{collision}

An evident risk in comparing hash values is
\key{collision}, which means that two strings have
different contents but equal hash values.
In this case, based on the hash values it seems that
the strings are equal, but in reality they aren't,
and the algorithm may give incorrect results.

Collisions are always possible,
because the number of different strings is larger
than the number of different hash values.
However, the probability of a collision is small
if the constants $A$ and $B$ are carefully chosen.
There are two goals: the hash values should be
evenly distributed for the strings,
and the number of different hash values should
be large enough.

A good solution is to use large random numbers
as constants.
A usual way is to choose constants that are
near $10^9$, for example
\[
\begin{array}{lcl}
A & = & 911382323 \\
B & = & 972663749 \\
\end{array}
\]
This choice ensures that the hash values
are distributed evenly enough in the range $0 \ldots B-1$.
The benefit in $10^9$ is that
the \texttt{long long} type can be used
for calculating the hash values,
because the products $AB$ and $BB$ fit in \texttt{long long}.
But is it enough to have $10^9$ different hash values?

Let's consider three scenarios where hashing can be used:

\textit{Scenario 1:} Strings $x$ and $y$ are compared with
each other.
The probability of a collision is $1/B$ assuming that
all hash values are equally probable.

\textit{Tapaus 2:} A string $x$ is compared with strings
$y_1,y_2,\ldots,y_n$.
The probability for one or more collisions is

\[1-(1-1/B)^n.\]

\textit{Tapaus 3:} Strings $x_1,x_2,\ldots,x_n$
are compared with each other.
The probability for one or more collisions is
\[ 1 - \frac{B \cdot (B-1) \cdot (B-2) \cdots (B-n+1)}{B^n}.\]

The following table shows the collision probabilities
when the value of $B$ varies and $n=10^6$:

\begin{center}
\begin{tabular}{rrrr}
constant $B$ & scenario 1 & scenario 2 & scenario 3 \\
\hline
$10^3$ & $0.001000$ & $1.000000$ & $1.000000$ \\
$10^6$ & $0.000001$ & $0.632121$ & $1.000000$ \\
$10^9$ & $0.000000$ & $0.001000$ & $1.000000$ \\
$10^{12}$ & $0.000000$ & $0.000000$ & $0.393469$ \\
$10^{15}$ & $0.000000$ & $0.000000$ & $0.000500$ \\
$10^{18}$ & $0.000000$ & $0.000000$ & $0.000001$ \\
\end{tabular}
\end{center}

The table shows that in scenario 1,
the probability of a collision is negligible
when $B \approx 10^9$.
In scenario 2, a collision is possible but the
probability is still quite small.
However, in scenario 3 the situation is very different:
a collision will almost always happen when
$B \approx 10^9$.

\index{birthday paradox}

The phenomenon in scenario 3 is known as the
\key{birthday paradox}: if there are $n$ people
in a room, the probability that some two people
have the same birthday is large even if $n$ is quite small.
In hashing, correspondingly, when all hash values are compared
with each other, the probability that some two
hash values are the same is large.

A good way to make the probability of a collision
smaller is to calculate \emph{multiple} hash values
using different parameters.
It is very unlikely that a collision would occur
in all hash values at the same time.
For example, two hash values with parameter
$B \approx 10^9$ correspond to one hash
value with parameter $B \approx 10^{18}$,
which makes the probability of a collision very small.

Some people use constants $B=2^{32}$ and $B=2^{64}$,
which is convenient, because operations with 32 and 64
bit integers are calculated modulo $2^{32}$ and $2^{64}$.
However, this is not a good choice, because it is possible
to construct inputs that always generate collisions when
constants of the form $2^x$ are used\footnote{
J. Pachocki and Jakub Radoszweski:
''Where to use and how not to use polynomial string hashing''.
\textit{Olympiads in Informatics}, 2013.
}.

\section{Z-algorithm}

\index{Z-algorithm}
\index{Z-array}

The \key{Z-algorithm} generates a \key{Z-array}
for the string, that contains for each index $k$
in the string the length of the longest substring
that begins at index $k$ and is a prefix of the string.
Many string problems can be efficiently solved
using the Z-algorithm.

It is often a matter of taste whether to use
the Z-algorithm or string hashing.
Unlike hashing, the Z-algorithm always works
and there is no risk for collisions.
On the other hand, the Z-algorithm is more difficult
to implement and some problems can only be solved
using hashing.

\subsubsection*{Description}

The Z-algorithm constructs a Z-array that
indicates for each position the length of the
longest substring that is also a prefix of the string.
For example, the Z-array for the string
\texttt{ACBACDACBACBACDA} is as follows:

\begin{center}
\begin{tikzpicture}[scale=0.7]
\draw (0,0) grid (16,2);

\node at (0.5, 1.5) {\texttt{A}};
\node at (1.5, 1.5) {\texttt{C}};
\node at (2.5, 1.5) {\texttt{B}};
\node at (3.5, 1.5) {\texttt{A}};
\node at (4.5, 1.5) {\texttt{C}};
\node at (5.5, 1.5) {\texttt{D}};
\node at (6.5, 1.5) {\texttt{A}};
\node at (7.5, 1.5) {\texttt{C}};
\node at (8.5, 1.5) {\texttt{B}};
\node at (9.5, 1.5) {\texttt{A}};
\node at (10.5, 1.5) {\texttt{C}};
\node at (11.5, 1.5) {\texttt{B}};
\node at (12.5, 1.5) {\texttt{A}};
\node at (13.5, 1.5) {\texttt{C}};
\node at (14.5, 1.5) {\texttt{D}};
\node at (15.5, 1.5) {\texttt{A}};

\node at (0.5, 0.5) {--};
\node at (1.5, 0.5) {0};
\node at (2.5, 0.5) {0};
\node at (3.5, 0.5) {2};
\node at (4.5, 0.5) {0};
\node at (5.5, 0.5) {0};
\node at (6.5, 0.5) {5};
\node at (7.5, 0.5) {0};
\node at (8.5, 0.5) {0};
\node at (9.5, 0.5) {7};
\node at (10.5, 0.5) {0};
\node at (11.5, 0.5) {0};
\node at (12.5, 0.5) {2};
\node at (13.5, 0.5) {0};
\node at (14.5, 0.5) {0};
\node at (15.5, 0.5) {1};

\footnotesize
\node at (0.5, 2.5) {1};
\node at (1.5, 2.5) {2};
\node at (2.5, 2.5) {3};
\node at (3.5, 2.5) {4};
\node at (4.5, 2.5) {5};
\node at (5.5, 2.5) {6};
\node at (6.5, 2.5) {7};
\node at (7.5, 2.5) {8};
\node at (8.5, 2.5) {9};
\node at (9.5, 2.5) {10};
\node at (10.5, 2.5) {11};
\node at (11.5, 2.5) {12};
\node at (12.5, 2.5) {13};
\node at (13.5, 2.5) {14};
\node at (14.5, 2.5) {15};
\node at (15.5, 2.5) {16};

\end{tikzpicture}
\end{center}

For example, the position 7 contains the value 5,
because the substring \texttt{ACBAC} of length 5
is a prefix of the string,
but the substring \texttt{ACBACB} of length 6
is not a prefix of the string.

The Z-algorithm scans the string from the left
to the right, and calculates for each position
the length of the longest substring that
is a prefix of the string.
The algorithm compares the first characters
of the string
and the active substring with each other to
find the length of the common prefix.

A straightforward implementation would yield
an algorithm with time complexity $O(n^2)$
because the common prefixes may be long.
However, the Z-algorithm has one important
optimization which ensures that the time complexity
is only $O(n)$.
The idea is to maintain a range $[x,y]$ such that
the substring from $x$ to $y$ is a prefix of
the string and $y$ is as large as possible.
Since the Z-array already contains information
about the characters in the range $[x,y]$,
it is not needed to process them again later in the algorithm.

The time complexity of the Z-algorithm is $O(n)$,
because the algorithm always compares substrings
character by character only from index $y+1$.
If the characters match, the value of $y$ increases,
and it is not needed to inspect the character again,
but the information in the Z-array can be used.

\subsubsection*{Example}

Let's construct the following Z-array using
the Z-algorithm:

\begin{center}
\begin{tikzpicture}[scale=0.7]
\draw (0,0) grid (16,2);

\node at (0.5, 1.5) {A};
\node at (1.5, 1.5) {C};
\node at (2.5, 1.5) {B};
\node at (3.5, 1.5) {A};
\node at (4.5, 1.5) {C};
\node at (5.5, 1.5) {D};
\node at (6.5, 1.5) {A};
\node at (7.5, 1.5) {C};
\node at (8.5, 1.5) {B};
\node at (9.5, 1.5) {A};
\node at (10.5, 1.5) {C};
\node at (11.5, 1.5) {B};
\node at (12.5, 1.5) {A};
\node at (13.5, 1.5) {C};
\node at (14.5, 1.5) {D};
\node at (15.5, 1.5) {A};

\node at (0.5, 0.5) {--};
\node at (1.5, 0.5) {?};
\node at (2.5, 0.5) {?};
\node at (3.5, 0.5) {?};
\node at (4.5, 0.5) {?};
\node at (5.5, 0.5) {?};
\node at (6.5, 0.5) {?};
\node at (7.5, 0.5) {?};
\node at (8.5, 0.5) {?};
\node at (9.5, 0.5) {?};
\node at (10.5, 0.5) {?};
\node at (11.5, 0.5) {?};
\node at (12.5, 0.5) {?};
\node at (13.5, 0.5) {?};
\node at (14.5, 0.5) {?};
\node at (15.5, 0.5) {?};

\footnotesize
\node at (0.5, 2.5) {1};
\node at (1.5, 2.5) {2};
\node at (2.5, 2.5) {3};
\node at (3.5, 2.5) {4};
\node at (4.5, 2.5) {5};
\node at (5.5, 2.5) {6};
\node at (6.5, 2.5) {7};
\node at (7.5, 2.5) {8};
\node at (8.5, 2.5) {9};
\node at (9.5, 2.5) {10};
\node at (10.5, 2.5) {11};
\node at (11.5, 2.5) {12};
\node at (12.5, 2.5) {13};
\node at (13.5, 2.5) {14};
\node at (14.5, 2.5) {15};
\node at (15.5, 2.5) {16};

\end{tikzpicture}
\end{center}

The first interesting position is 7 where the
length of the common prefix is 5.
The corresponding range in the string is $[7,11]$:

\begin{center}
\begin{tikzpicture}[scale=0.7]
\fill[color=lightgray] (6,0) rectangle (7,1);
\draw (0,0) grid (16,2);

\node at (0.5, 1.5) {A};
\node at (1.5, 1.5) {C};
\node at (2.5, 1.5) {B};
\node at (3.5, 1.5) {A};
\node at (4.5, 1.5) {C};
\node at (5.5, 1.5) {D};
\node at (6.5, 1.5) {A};
\node at (7.5, 1.5) {C};
\node at (8.5, 1.5) {B};
\node at (9.5, 1.5) {A};
\node at (10.5, 1.5) {C};
\node at (11.5, 1.5) {B};
\node at (12.5, 1.5) {A};
\node at (13.5, 1.5) {C};
\node at (14.5, 1.5) {D};
\node at (15.5, 1.5) {A};

\node at (0.5, 0.5) {--};
\node at (1.5, 0.5) {0};
\node at (2.5, 0.5) {0};
\node at (3.5, 0.5) {2};
\node at (4.5, 0.5) {0};
\node at (5.5, 0.5) {0};
\node at (6.5, 0.5) {5};
\node at (7.5, 0.5) {?};
\node at (8.5, 0.5) {?};
\node at (9.5, 0.5) {?};
\node at (10.5, 0.5) {?};
\node at (11.5, 0.5) {?};
\node at (12.5, 0.5) {?};
\node at (13.5, 0.5) {?};
\node at (14.5, 0.5) {?};
\node at (15.5, 0.5) {?};

\draw [decoration={brace}, decorate, line width=0.5mm] (6,3.00) -- (11,3.00);

\node at (6.5,3.50) {$x$};
\node at (10.5,3.50) {$y$};


\footnotesize
\node at (0.5, 2.5) {1};
\node at (1.5, 2.5) {2};
\node at (2.5, 2.5) {3};
\node at (3.5, 2.5) {4};
\node at (4.5, 2.5) {5};
\node at (5.5, 2.5) {6};
\node at (6.5, 2.5) {7};
\node at (7.5, 2.5) {8};
\node at (8.5, 2.5) {9};
\node at (9.5, 2.5) {10};
\node at (10.5, 2.5) {11};
\node at (11.5, 2.5) {12};
\node at (12.5, 2.5) {13};
\node at (13.5, 2.5) {14};
\node at (14.5, 2.5) {15};
\node at (15.5, 2.5) {16};

\end{tikzpicture}
\end{center}

The benefit in the range $[7,11]$ is that the
algorithm can calculate the subsequent values
for the Z-array more efficiently.
Since the ranges $[1,5]$ and $[7,11]$ contain
the same characters, also the Z-array will
contain similar values.
First, the values at indices 8 and 9
correspond to the values at indices 2 and 3:

\begin{center}
\begin{tikzpicture}[scale=0.7]
\fill[color=lightgray] (7,0) rectangle (9,1);
\draw (0,0) grid (16,2);

\node at (0.5, 1.5) {A};
\node at (1.5, 1.5) {C};
\node at (2.5, 1.5) {B};
\node at (3.5, 1.5) {A};
\node at (4.5, 1.5) {C};
\node at (5.5, 1.5) {D};
\node at (6.5, 1.5) {A};
\node at (7.5, 1.5) {C};
\node at (8.5, 1.5) {B};
\node at (9.5, 1.5) {A};
\node at (10.5, 1.5) {C};
\node at (11.5, 1.5) {B};
\node at (12.5, 1.5) {A};
\node at (13.5, 1.5) {C};
\node at (14.5, 1.5) {D};
\node at (15.5, 1.5) {A};

\node at (0.5, 0.5) {--};
\node at (1.5, 0.5) {0};
\node at (2.5, 0.5) {0};
\node at (3.5, 0.5) {2};
\node at (4.5, 0.5) {0};
\node at (5.5, 0.5) {0};
\node at (6.5, 0.5) {5};
\node at (7.5, 0.5) {0};
\node at (8.5, 0.5) {0};
\node at (9.5, 0.5) {?};
\node at (10.5, 0.5) {?};
\node at (11.5, 0.5) {?};
\node at (12.5, 0.5) {?};
\node at (13.5, 0.5) {?};
\node at (14.5, 0.5) {?};
\node at (15.5, 0.5) {?};


\draw [decoration={brace}, decorate, line width=0.5mm] (6,3.00) -- (11,3.00);

\node at (6.5,3.50) {$x$};
\node at (10.5,3.50) {$y$};


\footnotesize
\node at (0.5, 2.5) {1};
\node at (1.5, 2.5) {2};
\node at (2.5, 2.5) {3};
\node at (3.5, 2.5) {4};
\node at (4.5, 2.5) {5};
\node at (5.5, 2.5) {6};
\node at (6.5, 2.5) {7};
\node at (7.5, 2.5) {8};
\node at (8.5, 2.5) {9};
\node at (9.5, 2.5) {10};
\node at (10.5, 2.5) {11};
\node at (11.5, 2.5) {12};
\node at (12.5, 2.5) {13};
\node at (13.5, 2.5) {14};
\node at (14.5, 2.5) {15};
\node at (15.5, 2.5) {16};


\draw[thick,<->] (7.5,-0.25) .. controls (7,-1.25) and (2,-1.25) .. (1.5,-0.25);
\draw[thick,<->] (8.5,-0.25) .. controls (8,-1.25) and (3,-1.25) .. (2.5,-0.25);
\end{tikzpicture}
\end{center}

After this, the value for index 10 can be
calculated using the value at index 4.
The value at index 4 is 2,
so the first two characters
in the substring match the beginning of the string.
However, the characters after index $y=11$ have
not been inspected yet.

\begin{center}
\begin{tikzpicture}[scale=0.7]
\fill[color=lightgray] (9,0) rectangle (10,1);
\draw (0,0) grid (16,2);

\node at (0.5, 1.5) {A};
\node at (1.5, 1.5) {C};
\node at (2.5, 1.5) {B};
\node at (3.5, 1.5) {A};
\node at (4.5, 1.5) {C};
\node at (5.5, 1.5) {D};
\node at (6.5, 1.5) {A};
\node at (7.5, 1.5) {C};
\node at (8.5, 1.5) {B};
\node at (9.5, 1.5) {A};
\node at (10.5, 1.5) {C};
\node at (11.5, 1.5) {B};
\node at (12.5, 1.5) {A};
\node at (13.5, 1.5) {C};
\node at (14.5, 1.5) {D};
\node at (15.5, 1.5) {A};

\node at (0.5, 0.5) {--};
\node at (1.5, 0.5) {0};
\node at (2.5, 0.5) {0};
\node at (3.5, 0.5) {2};
\node at (4.5, 0.5) {0};
\node at (5.5, 0.5) {0};
\node at (6.5, 0.5) {5};
\node at (7.5, 0.5) {0};
\node at (8.5, 0.5) {0};
\node at (9.5, 0.5) {?};
\node at (10.5, 0.5) {?};
\node at (11.5, 0.5) {?};
\node at (12.5, 0.5) {?};
\node at (13.5, 0.5) {?};
\node at (14.5, 0.5) {?};
\node at (15.5, 0.5) {?};

\draw [decoration={brace}, decorate, line width=0.5mm] (6,3.00) -- (11,3.00);

\node at (6.5,3.50) {$x$};
\node at (10.5,3.50) {$y$};


\footnotesize
\node at (0.5, 2.5) {1};
\node at (1.5, 2.5) {2};
\node at (2.5, 2.5) {3};
\node at (3.5, 2.5) {4};
\node at (4.5, 2.5) {5};
\node at (5.5, 2.5) {6};
\node at (6.5, 2.5) {7};
\node at (7.5, 2.5) {8};
\node at (8.5, 2.5) {9};
\node at (9.5, 2.5) {10};
\node at (10.5, 2.5) {11};
\node at (11.5, 2.5) {12};
\node at (12.5, 2.5) {13};
\node at (13.5, 2.5) {14};
\node at (14.5, 2.5) {15};
\node at (15.5, 2.5) {16};

\draw[thick,<->] (9.5,-0.25) .. controls (9,-1.25) and (4,-1.25) .. (3.5,-0.25);
\end{tikzpicture}
\end{center}

The algorithm compares the substring
beginning at index $y+1=12$ character by character.
The previous values in the Z-array cannot be used,
because this is the first time the characters
after index 11 are inspected.
It turns out that the length of the common
prefix is 7, and the range $[x,y]$ will be updated:

\begin{center}
\begin{tikzpicture}[scale=0.7]
\fill[color=lightgray] (9,0) rectangle (10,1);
\draw (0,0) grid (16,2);

\node at (0.5, 1.5) {A};
\node at (1.5, 1.5) {C};
\node at (2.5, 1.5) {B};
\node at (3.5, 1.5) {A};
\node at (4.5, 1.5) {C};
\node at (5.5, 1.5) {D};
\node at (6.5, 1.5) {A};
\node at (7.5, 1.5) {C};
\node at (8.5, 1.5) {B};
\node at (9.5, 1.5) {A};
\node at (10.5, 1.5) {C};
\node at (11.5, 1.5) {B};
\node at (12.5, 1.5) {A};
\node at (13.5, 1.5) {C};
\node at (14.5, 1.5) {D};
\node at (15.5, 1.5) {A};

\node at (0.5, 0.5) {--};
\node at (1.5, 0.5) {0};
\node at (2.5, 0.5) {0};
\node at (3.5, 0.5) {2};
\node at (4.5, 0.5) {0};
\node at (5.5, 0.5) {0};
\node at (6.5, 0.5) {5};
\node at (7.5, 0.5) {0};
\node at (8.5, 0.5) {0};
\node at (9.5, 0.5) {7};
\node at (10.5, 0.5) {?};
\node at (11.5, 0.5) {?};
\node at (12.5, 0.5) {?};
\node at (13.5, 0.5) {?};
\node at (14.5, 0.5) {?};
\node at (15.5, 0.5) {?};

\draw [decoration={brace}, decorate, line width=0.5mm] (9,3.00) -- (16,3.00);

\node at (9.5,3.50) {$x$};
\node at (15.5,3.50) {$y$};


\footnotesize
\node at (0.5, 2.5) {1};
\node at (1.5, 2.5) {2};
\node at (2.5, 2.5) {3};
\node at (3.5, 2.5) {4};
\node at (4.5, 2.5) {5};
\node at (5.5, 2.5) {6};
\node at (6.5, 2.5) {7};
\node at (7.5, 2.5) {8};
\node at (8.5, 2.5) {9};
\node at (9.5, 2.5) {10};
\node at (10.5, 2.5) {11};
\node at (11.5, 2.5) {12};
\node at (12.5, 2.5) {13};
\node at (13.5, 2.5) {14};
\node at (14.5, 2.5) {15};
\node at (15.5, 2.5) {16};

% \draw[thick,<->] (9.5,-0.25) .. controls (9,-1.25) and (4,-1.25) .. (3.5,-0.25);
\end{tikzpicture}
\end{center}

After this, all subsequent values in the Z-array
can be calculated using the information in
the range $[x,y]$. All the remaining values can be
directly retrieved from the beginning of the Z-array:

\begin{center}
\begin{tikzpicture}[scale=0.7]
\draw (0,0) grid (16,2);

\node at (0.5, 1.5) {A};
\node at (1.5, 1.5) {C};
\node at (2.5, 1.5) {B};
\node at (3.5, 1.5) {A};
\node at (4.5, 1.5) {C};
\node at (5.5, 1.5) {D};
\node at (6.5, 1.5) {A};
\node at (7.5, 1.5) {C};
\node at (8.5, 1.5) {B};
\node at (9.5, 1.5) {A};
\node at (10.5, 1.5) {C};
\node at (11.5, 1.5) {B};
\node at (12.5, 1.5) {A};
\node at (13.5, 1.5) {C};
\node at (14.5, 1.5) {D};
\node at (15.5, 1.5) {A};

\node at (0.5, 0.5) {--};
\node at (1.5, 0.5) {0};
\node at (2.5, 0.5) {0};
\node at (3.5, 0.5) {2};
\node at (4.5, 0.5) {0};
\node at (5.5, 0.5) {0};
\node at (6.5, 0.5) {5};
\node at (7.5, 0.5) {0};
\node at (8.5, 0.5) {0};
\node at (9.5, 0.5) {7};
\node at (10.5, 0.5) {0};
\node at (11.5, 0.5) {0};
\node at (12.5, 0.5) {2};
\node at (13.5, 0.5) {0};
\node at (14.5, 0.5) {0};
\node at (15.5, 0.5) {1};

\draw [decoration={brace}, decorate, line width=0.5mm] (9,3.00) -- (16,3.00);

\node at (9.5,3.50) {$x$};
\node at (15.5,3.50) {$y$};


\footnotesize
\node at (0.5, 2.5) {1};
\node at (1.5, 2.5) {2};
\node at (2.5, 2.5) {3};
\node at (3.5, 2.5) {4};
\node at (4.5, 2.5) {5};
\node at (5.5, 2.5) {6};
\node at (6.5, 2.5) {7};
\node at (7.5, 2.5) {8};
\node at (8.5, 2.5) {9};
\node at (9.5, 2.5) {10};
\node at (10.5, 2.5) {11};
\node at (11.5, 2.5) {12};
\node at (12.5, 2.5) {13};
\node at (13.5, 2.5) {14};
\node at (14.5, 2.5) {15};
\node at (15.5, 2.5) {16};

\end{tikzpicture}
\end{center}

\subsubsection{Using the Z-array}

As an example, let's solve a problem
where our task is to calculate
the number of times a string $p$
occurs as a substring in a string $s$.
Previously, we solved this problem
using string hashing, but the Z-algorithm
provides another way to solve the problem.

A usual idea when using the Z-algorithm
is to construct a string that consists of
several strings separated by special characters.
In this problem, we can construct a string
$p$\texttt{\#}$s$,
where $p$ and $s$ are separated by a special
character \texttt{\#} that doesn't occur
in the strings.
After this, the Z-array for the string
$p$\texttt{\#}$s$ indicates the positions
where $p$ occurs in $s$.
Such positions are those positions in the Z-array
that contain the value $p$.

\begin{samepage}
For example, if $s=$\texttt{HATTIVATTI} and $p=$\texttt{ATT},
the Z-array is as follows:

\begin{center}
\begin{tikzpicture}[scale=0.7]
\draw (0,0) grid (14,2);

\node at (0.5, 1.5) {A};
\node at (1.5, 1.5) {T};
\node at (2.5, 1.5) {T};
\node at (3.5, 1.5) {\#};
\node at (4.5, 1.5) {H};
\node at (5.5, 1.5) {A};
\node at (6.5, 1.5) {T};
\node at (7.5, 1.5) {T};
\node at (8.5, 1.5) {I};
\node at (9.5, 1.5) {V};
\node at (10.5, 1.5) {A};
\node at (11.5, 1.5) {T};
\node at (12.5, 1.5) {T};
\node at (13.5, 1.5) {I};

\node at (0.5, 0.5) {--};
\node at (1.5, 0.5) {0};
\node at (2.5, 0.5) {0};
\node at (3.5, 0.5) {0};
\node at (4.5, 0.5) {0};
\node at (5.5, 0.5) {3};
\node at (6.5, 0.5) {0};
\node at (7.5, 0.5) {0};
\node at (8.5, 0.5) {0};
\node at (9.5, 0.5) {0};
\node at (10.5, 0.5) {3};
\node at (11.5, 0.5) {0};
\node at (12.5, 0.5) {0};
\node at (13.5, 0.5) {0};

\footnotesize
\node at (0.5, 2.5) {1};
\node at (1.5, 2.5) {2};
\node at (2.5, 2.5) {3};
\node at (3.5, 2.5) {4};
\node at (4.5, 2.5) {5};
\node at (5.5, 2.5) {6};
\node at (6.5, 2.5) {7};
\node at (7.5, 2.5) {8};
\node at (8.5, 2.5) {9};
\node at (9.5, 2.5) {10};
\node at (10.5, 2.5) {11};
\node at (11.5, 2.5) {12};
\node at (12.5, 2.5) {13};
\node at (13.5, 2.5) {14};
\end{tikzpicture}
\end{center}
\end{samepage}
The positions 6 and 11 contain the value 3,
which means that the substring \texttt{ATT}
occurs in the corresponding positions
in the string \texttt{HATTIVATTI}.

The time complexity of the resulting algorithm
is $O(n)$, because it suffices to construct and
go through the Z-array.