Pattern searching is an algorithm that involves searching for patterns such as strings, words, images, etc.
We use certain algorithms to do the search process. The complexity of pattern searching varies from algorithm to algorithm. They are very useful when performing a search in a database. The Pattern Searching algorithm is useful for finding patterns in substrings of larger strings. This process can be accomplished using a variety of algorithms that we are going to discuss in this blog.
Introduction to Pattern Searching – Data Structure and Algorithm Tutorial
Features of Pattern Searching Algorithm:
- Pattern searching algorithms should recognize familiar patterns quickly and accurately.
- Recognize and classify unfamiliar patterns.
- Identify patterns even when partly hidden.
- Recognize patterns quickly with ease, and with automaticity.
Naive pattern searching is the simplest method among other pattern-searching algorithms. It checks for all characters of the main string to the pattern. This algorithm is helpful for smaller texts. It does not need any pre-processing phases. We can find the substring by checking once for the string. It also does not occupy extra space to perform the operation.
Compare text characters with pattern characters
The time complexity of Naive Pattern Search method is O(m*n). The m is the size of pattern and n is the size of the main string.
C++
#include <bits/stdc++.h>
using namespace std;
void search( char * pat, char * txt)
{
int M = strlen (pat);
int N = strlen (txt);
for ( int i = 0; i <= N - M; i++) {
int j;
for (j = 0; j < M; j++)
if (txt[i + j] != pat[j])
break ;
if (j
== M)
cout << "Pattern found at index " << i << endl;
}
}
int main()
{
char txt[] = "AABAACAADAABAAABAA" ;
char pat[] = "AABA" ;
search(pat, txt);
return 0;
}
|
Java
class GFG {
static void search( char [] pat, char [] txt)
{
int M = pat.length;
int N = txt.length;
for ( int i = 0 ; i <= N - M; i++) {
int j;
for (j = 0 ; j < M; j++)
if (txt[i + j] != pat[j])
break ;
if (j == M)
System.out.println( "Pattern found at index "
+ i);
}
}
public static void main(String[] args)
{
char txt[] = "AABAACAADAABAAABAA" .toCharArray();
char pat[] = "AABA" .toCharArray();
search(pat, txt);
}
}
|
Python3
def search(pat, txt):
M = len (pat)
N = len (txt)
for i in range (N - M):
for j in range (M):
k = j + 1
if (txt[i + j] ! = pat[j]):
break
if (k = = M):
print ( "Pattern found at index " , i)
txt = "AABAACAADAABAAABAA"
pat = "AABA"
search(pat, txt)
|
C#
using System;
public class GFG {
public static void search( char [] pat, char [] txt)
{
int M = pat.Length;
int N = txt.Length;
for ( int i = 0; i <= N - M; i++) {
int j;
for (j = 0; j < M; j++)
if (txt[i + j] != pat[j])
break ;
if (j == M)
Console.WriteLine( "Pattern found at index "
+ i);
}
}
static public void Main()
{
char [] txt = "AABAACAADAABAAABAA" .ToCharArray();
char [] pat = "AABA" .ToCharArray();
search(pat, txt);
}
}
|
Javascript
function search(pat, txt)
{
let M = pat.length;
let N = txt.length;
for (let i = 0; i <= N - M; i++) {
let j = 0;
for (j = 0; j < M; j++)
if (txt[i + j] != pat[j])
break ;
if (j == M)
console.log( "Pattern found at index" ,i);
}
}
let txt = "AABAACAADAABAAABAA" ;
let pat = "AABA" ;
search(pat, txt);
|
Output
Pattern found at index 0
Pattern found at index 9
Pattern found at index 13
Time Complexity: O(N*M)
Auxiliary Space: O(1)
KMP algorithm is used to find a “Pattern” in a “Text”. This algorithm compares character by character from left to right. But whenever a mismatch occurs, it uses a preprocessed table called “Prefix Table” to skip characters comparison while matching. Sometimes prefix table is also known as LPS Table. Here LPS stands for “Longest proper Prefix which is also Suffix”.
How to use LPS Table
We use the LPS table to decide how many characters are to be skipped for comparison when a mismatch has occurred.
When a mismatch occurs, check the LPS value of the previous character of the mismatched character in the pattern. If it is ‘0’ then start comparing the first character of the pattern with the next character to the mismatched character in the text. If it is not ‘0’ then start comparing the character which is at an index value equal to the LPS value of the previous character to the mismatched character in pattern with the mismatched character in the Text.
Example of KMP algorithm
Compare first character of pattern with first character of text from left to right
Compare first character of pattern with next character of text
Compare pattern[0] and pattern[1] values
Compare pattern[0] with next characters in text.
Compare pattern[2] with mismatched characters in text.
How the KMP Algorithm Works
Let’s take a look on working example of KMP Algorithm to find a Pattern in a Text.
LPS table
Define variables
Compare A with B
Compare A with C
Compare A with D
Compare A with A
Compare B with B
Compare C with D
Compare A with D
Implementation of the KMP algorithm:
C++
#include <bits/stdc++.h>
void computeLPSArray( char * pat, int M, int * lps);
void KMPSearch( char * pat, char * txt)
{
int M = strlen (pat);
int N = strlen (txt);
int lps[M];
computeLPSArray(pat, M, lps);
int i = 0;
int j = 0;
while ((N - i) >= (M - j)) {
if (pat[j] == txt[i]) {
j++;
i++;
}
if (j == M) {
printf ( "Found pattern at index %d " , i - j);
j = lps[j - 1];
}
else if (i < N && pat[j] != txt[i]) {
if (j != 0)
j = lps[j - 1];
else
i = i + 1;
}
}
}
void computeLPSArray( char * pat, int M, int * lps)
{
int len = 0;
lps[0] = 0;
int i = 1;
while (i < M) {
if (pat[i] == pat[len]) {
len++;
lps[i] = len;
i++;
}
else
{
if (len != 0) {
len = lps[len - 1];
}
else
{
lps[i] = 0;
i++;
}
}
}
}
int main()
{
char txt[] = "ABABDABACDABABCABAB" ;
char pat[] = "ABABCABAB" ;
KMPSearch(pat, txt);
return 0;
}
|
Java
public class KMP_String_Matching {
void KMPSearch(String pat, String txt)
{
int M = pat.length();
int N = txt.length();
int lps[] = new int [M];
int j = 0 ;
computeLPSArray(pat, M, lps);
int i = 0 ;
while (i < N) {
if (pat.charAt(j) == txt.charAt(i)) {
j++;
i++;
}
if (j == M) {
System.out.println( "Found pattern " + "at index " + (i - j));
j = lps[j - 1 ];
}
else if (i < N && pat.charAt(j) != txt.charAt(i)) {
if (j != 0 )
j = lps[j - 1 ];
else
i = i + 1 ;
}
}
}
void computeLPSArray(String pat, int M, int lps[])
{
int len = 0 ;
int i = 1 ;
lps[ 0 ] = 0 ;
while (i < M) {
if (pat.charAt(i) == pat.charAt(len)) {
len++;
lps[i] = len;
i++;
}
else
{
if (len != 0 ) {
len = lps[len - 1 ];
}
else
{
lps[i] = len;
i++;
}
}
}
}
public static void main(String[] args)
{
String txt = "ABABDABACDABABCABAB" ;
String pat = "ABABCABAB" ;
new KMP_String_Matching().KMPSearch(pat, txt);
}
}
|
Python3
def computeLPSArray(pat, M, lps):
len = 0
lps[ 0 ]
i = 1
while i < M:
if pat[i] = = pat[ len ]:
len + = 1
lps[i] = len
i + = 1
else :
if len ! = 0 :
len = lps[ len - 1 ]
else :
lps[i] = 0
i + = 1
def KMPSearch(pat, txt):
M = len (pat)
N = len (txt)
lps = [ 0 ] * M
j = 0
computeLPSArray(pat, M, lps)
i = 0
while (N - i) > = (M - j):
if pat[j] = = txt[i]:
j + = 1
i + = 1
if j = = M:
print ( "Found pattern at index:" , i - j)
j = lps[j - 1 ]
elif i < N and pat[j] ! = txt[i]:
if j ! = 0 :
j = lps[j - 1 ]
else :
i + = 1
txt = "ABABDABACDABABCABAB"
pat = "ABABCABAB"
KMPSearch(pat, txt)
|
C#
using System;
using System.Collections.Generic;
public class GFG {
public static void KMPSearch( char [] pat, char [] txt)
{
int M = pat.Length;
int N = txt.Length;
int [] lps = new int [M];
computeLPSArray(pat, M, lps);
int i = 0;
int j = 0;
while ((N - i) >= (M - j)) {
if (pat[j] == txt[i]) {
j++;
i++;
}
if (j == M) {
int temp = i - j;
Console.WriteLine( "Found pattern at index "
+ temp);
j = lps[j - 1];
}
else if (i < N && pat[j] != txt[i]) {
if (j != 0)
j = lps[j - 1];
else
i = i + 1;
}
}
}
public static void computeLPSArray( char [] pat, int M,
int [] lps)
{
int len = 0;
lps[0] = 0;
int i = 1;
while (i < M) {
if (pat[i] == pat[len]) {
len++;
lps[i] = len;
i++;
}
else
{
if (len != 0) {
len = lps[len - 1];
}
else
{
lps[i] = 0;
i++;
}
}
}
}
static public void Main()
{
char [] txt = "ABABDABACDABABCABAB" .ToCharArray();
char [] pat = "ABABCABAB" .ToCharArray();
KMPSearch(pat, txt);
}
}
|
Javascript
function computeLPSArray(pat, M, lps)
{
let len = 0;
lps[0] = 0;
let i = 1;
while (i < M) {
if (pat[i] == pat[len]) {
len++;
lps[i] = len;
i++;
}
else
{
if (len != 0) {
len = lps[len - 1];
}
else
{
lps[i] = 0;
i++;
}
}
}
}
function KMPSearch(pat, txt) {
let M = pat.length;
let N = txt.length
let lps = [];
computeLPSArray(pat, M, lps);
let i = 0;
let j = 0;
while ((N - i) >= (M - j)) {
if (pat[j] == txt[i]) {
j++;
i++;
}
if (j == M) {
console.log( "Found pattern at index:" , i - j);
j = lps[j - 1];
}
else if (i < N && pat[j] != txt[i])
{
if (j != 0)
j = lps[j - 1];
else
i = i + 1;
}
}
}
let txt = "ABABDABACDABABCABAB" ;
let pat = "ABABCABAB" ;
KMPSearch(pat, txt);
|
Output
Found pattern at index 10
Time complexity: O(n + m)
Auxiliary Space: O(M)
Rabin-Karp algorithm is an algorithm used for searching/matching patterns in the text using a hash function. Unlike Naive string-matching algorithm, it does not travel through every character in the initial phase rather it filters the characters that do not match and then perform the comparison.
Rabin-Karp compares a string’s hash values, rather than the strings themselves. For efficiency, the hash value of the next position in the text is easily computed from the hash value of the current position.
Working of Rabin-Karp algorithm
- Initially calculate the hash value of the pattern P.
- Start iterating from the start of the string:
- Calculate the hash value of the current substring having length m.
- If the hash value of the current substring and the pattern are same check if the substring is same as the pattern.
- If they are same, store the starting index as a valid answer. Otherwise, continue for the next substrings.
- Return the starting indices as the required answer.
Example of Rabin Karp
Below is the implementation of the Rabin-Karp algorithm.
C++
#include <bits/stdc++.h>
using namespace std;
#define d 256
void search( char pat[], char txt[], int q)
{
int M = strlen (pat);
int N = strlen (txt);
int i, j;
int p = 0;
int t = 0;
int h = 1;
for (i = 0; i < M - 1; i++)
h = (h * d) % q;
for (i = 0; i < M; i++) {
p = (d * p + pat[i]) % q;
t = (d * t + txt[i]) % q;
}
for (i = 0; i <= N - M; i++) {
if (p == t) {
for (j = 0; j < M; j++) {
if (txt[i + j] != pat[j]) {
break ;
}
}
if (j == M)
cout << "Pattern found at index " << i
<< endl;
}
if (i < N - M) {
t = (d * (t - txt[i] * h) + txt[i + M]) % q;
if (t < 0)
t = (t + q);
}
}
}
int main()
{
char txt[] = "GEEKS FOR GEEKS" ;
char pat[] = "GEEK" ;
int q = INT_MAX;
search(pat, txt, q);
return 0;
}
|
Java
import java.io.*;
import java.lang.*;
import java.util.*;
public class GFG {
public final static int d = 256 ;
public static void search(String pat, String txt, int q)
{
int M = pat.length();
int N = txt.length();
int i, j;
int p = 0 ;
int t = 0 ;
int h = 1 ;
for (i = 0 ; i < M - 1 ; i++)
h = (h * d) % q;
for (i = 0 ; i < M; i++) {
p = (d * p + pat.charAt(i)) % q;
t = (d * t + txt.charAt(i)) % q;
}
for (i = 0 ; i <= N - M; i++) {
if (p == t) {
for (j = 0 ; j < M; j++) {
if (txt.charAt(i + j)
!= pat.charAt(j)) {
break ;
}
}
if (j == M) {
System.out.println(
"Pattern found at index " + i);
}
}
if (i < N - M) {
t = (d * (t - txt.charAt(i) * h)
+ txt.charAt(i + M))
% q;
if (t < 0 )
t = (t + q);
}
}
}
public static void main(String[] args)
{
String txt = "GEEKS FOR GEEKS" ;
String pat = "GEEK" ;
int q = 101 ;
search(pat, txt, q);
}
}
|
Python3
d = 256
def search(pat, txt, q):
M = len (pat)
N = len (txt)
p = 0
t = 0
h = 1
for i in range (M - 1 ):
h = (h * d) % q
for i in range (M):
p = (d * p + ord (pat[i])) % q
t = (d * t + ord (txt[i])) % q
for i in range (N - M + 1 ):
if p = = t:
for j in range (M):
if txt[i + j] ! = pat[j]:
break
if j = = M - 1 :
print ( "Pattern found at index " + str (i))
if i < N - M:
t = (d * (t - ord (txt[i]) * h) + ord (txt[i + M])) % q
if t < 0 :
t = (t + q)
txt = "GEEKS FOR GEEKS"
pat = "GEEK"
q = float ( 'inf' )
search(pat, txt, q)
|
C#
using System;
class GFG {
public static int d = 256;
public static void search( string pat, string txt, int q)
{
int M = pat.Length;
int N = txt.Length;
int i, j;
int p = 0;
int t = 0;
int h = 1;
for (i = 0; i < M - 1; i++)
h = (h * d) % q;
for (i = 0; i < M; i++) {
p = (d * p + pat[i]) % q;
t = (d * t + txt[i]) % q;
}
for (i = 0; i <= N - M; i++) {
if (p == t) {
for (j = 0; j < M; j++) {
if (txt[i + j] != pat[j]) {
break ;
}
}
if (j == M) {
Console.WriteLine(
"Pattern found at index " + i);
}
}
if (i < N - M) {
t = (d * (t - txt[i] * h) + txt[i + M]) % q;
if (t < 0)
t = (t + q);
}
}
}
public static void Main( string [] args)
{
string txt = "GEEKS FOR GEEKS" ;
string pat = "GEEK" ;
int q = 101;
search(pat, txt, q);
}
}
|
Javascript
const d = 256;
function search(pat, txt, q) {
const M = pat.length;
const N = txt.length;
let p = 0;
let t = 0;
let h = 1;
for (let i = 0; i < M - 1; i++) {
h = (h * d) % q;
}
for (let i = 0; i < M; i++) {
p = (d * p + pat.charCodeAt(i)) % q;
t = (d * t + txt.charCodeAt(i)) % q;
}
for (let i = 0; i <= N - M; i++) {
if (p === t) {
for (j = 0; j < M; j++) {
if (txt.charAt(i + j) !== pat.charAt(j)) {
break ;
}
}
if (j === M)
console.log( "Pattern found at index " + i);
}
if (i < N - M) {
t = (d * (t - txt.charCodeAt(i) * h) + txt.charCodeAt(i + M)) % q;
if (t < 0)
t = (t + q);
}
}
}
const txt = "GEEKS FOR GEEKS" ;
const pat = "GEEK" ;
const q = Number.MAX_SAFE_INTEGER;
search(pat, txt, q);
|
Output
Pattern found at index 0
Pattern found at index 10
Time Complexity:
- The average and best-case running time of the Rabin-Karp algorithm is O(n+m), but its worst-case time is O(nm).
- The worst case of the Rabin-Karp algorithm occurs when all characters of pattern and text are the same as the hash values of all the substrings of txt[] match with the hash value of pat[].
Space Complexity :
The space complexity of the Rabin-Karp algorithm is O(1), which means that it is a constant amount of memory that is required, regardless of the size of the input text and pattern. This is because the algorithm only needs to store a few variables that are updated as the algorithm progresses through the text and pattern. Specifically, the algorithm needs to store the hash value of the pattern, the hash value of the current window in the text, and a few loop counters and temporary variables. Since the size of these variables is fixed, the space complexity is constant.
This algorithm finds all occurrences of a pattern in a text in linear time. Let length of text be n and of pattern be m, then total time taken is O(m + n) with linear space complexity. Z algorithm works by maintaining an auxiliary array called the Z array. This Z array stores the length of the longest substring, starting from the current index, that also it’s prefix.
What is Z Array?
For a string str[0..n-1], Z array is of same length as string. An element Z[i] of Z array stores length of the longest substring starting from str[i] which is also a prefix of str[0..n-1]. The first entry of Z array is meaning less as complete string is always prefix of itself.
Example:
Index 0 1 2 3 4 5 6 7 8 9 10 11
Text a a b c a a b x a a a z
Z values X 1 0 0 3 1 0 0 2 2 1 0
How to construct Z array?
A Simple Solution is to run two nested loops, the outer loop goes to every index and the inner loop finds length of the longest prefix that matches the substring starting at current index. The time complexity of this solution is O(n2).
We can construct Z array in linear time. The idea is to maintain an interval [L, R] which is the interval with max R
such that [L, R] is prefix substring (substring which is also a prefix.
Steps for maintaining this interval are as follows –
- If i > R then there is no prefix substring that starts before i and ends after i, so we reset L and R and compute new [L, R] by comparing str[0..] to str[i..] and get Z[i] (= R-L+1).
- If i <= R then let K = i-L, now Z[i] >= min(Z[K], R-i+1) because str[i..] matches with str[K..] for atleast R-i+1 characters (they are in[L, R] interval which we know is a prefix substring).
Now two sub cases arise:
- If Z[K] < R-i+1 then there is no prefix substring starting at str[i] (otherwise Z[K] would be larger) so Z[i] = Z[K]and interval [L, R] remains same.
- If Z[K] >= R-i+1 then it is possible to extend the [L, R] interval thus we will set L as i and start matching from str[R] onwards and get new R then we will update interval [L, R] and calculate Z[i] (=R-L+1).
Construction of Z array
Below is the implementation of the Z algorithm:
C++
#include <iostream>
using namespace std;
void getZarr(string str, int Z[]);
void search(string text, string pattern)
{
string concat = pattern + "$" + text;
int l = concat.length();
int Z[l];
getZarr(concat, Z);
for ( int i = 0; i < l; ++i) {
if (Z[i] == pattern.length())
cout << "Pattern found at index "
<< i - pattern.length() - 1 << endl;
}
}
void getZarr(string str, int Z[])
{
int n = str.length();
int L, R, k;
L = R = 0;
for ( int i = 1; i < n; ++i) {
if (i > R) {
L = R = i;
while (R < n && str[R - L] == str[R])
R++;
Z[i] = R - L;
R--;
}
else {
k = i - L;
if (Z[k] < R - i + 1)
Z[i] = Z[k];
else {
L = i;
while (R < n && str[R - L] == str[R])
R++;
Z[i] = R - L;
R--;
}
}
}
}
int main()
{
string text = "GEEKS FOR GEEKS" ;
string pattern = "GEEK" ;
search(text, pattern);
return 0;
}
|
Java
import java.io.*;
class GFG
{
static void search(String text, String pattern)
{
String concat = pattern + "$" + text;
int l = concat.length();
int [] Z = new int [l];
getZarr(concat, Z);
for ( int i = 0 ; i < l; i++) {
if (Z[i] == pattern.length()) {
System.out.println(
"Pattern found at index "
+ (i - pattern.length() - 1 ));
}
}
}
static void getZarr(String str, int [] Z)
{
int n = str.length();
int L = 0 , R = 0 , k;
for ( int i = 1 ; i < n; ++i) {
if (i > R) {
L = R = i;
while (R < n
&& str.charAt(R - L)
== str.charAt(R)) {
R++;
}
Z[i] = R - L;
R--;
}
else {
k = i - L;
if (Z[k] < R - i + 1 )
Z[i] = Z[k];
else {
L = i;
while (R < n
&& str.charAt(R - L)
== str.charAt(R)) {
R++;
}
Z[i] = R - L;
R--;
}
}
}
}
public static void main(String[] args)
{
String text = "GEEKS FOR GEEKS" ;
String pattern = "GEEK" ;
search(text, pattern);
}
}
|
Python3
def getZarr(string, Z):
n = len (string)
L, R, k = 0 , 0 , 0
Z[ 0 ] = n
for i in range ( 1 , n):
if i > R:
L, R = i, i
while R < n and string[R - L] = = string[R]:
R + = 1
Z[i] = R - L
R - = 1
else :
k = i - L
if Z[k] < R - i + 1 :
Z[i] = Z[k]
else :
L = i
while R < n and string[R - L] = = string[R]:
R + = 1
Z[i] = R - L
R - = 1
def search(text, pattern):
concat = pattern + "$" + text
l = len (concat)
Z = [ 0 ] * l
getZarr(concat, Z)
for i in range (l):
if Z[i] = = len (pattern):
print ( "Pattern found at index" , i - len (pattern) - 1 )
if __name__ = = "__main__" :
text = "GEEKS FOR GEEKS"
pattern = "GEEK"
search(text, pattern)
|
C#
using System;
using System.Linq;
public class GFG {
static void search( string text, string pattern)
{
string concat = pattern + "$" + text;
int l = concat.Length;
int [] Z = new int [l];
GetZarr(concat, Z);
for ( int i = 0; i < l; i++) {
if (Z[i] == pattern.Length) {
Console.WriteLine(
"Pattern found at index "
+ (i - pattern.Length - 1));
}
}
}
static void GetZarr( string str, int [] Z)
{
int n = str.Length;
int L = 0, R = 0, k;
for ( int i = 1; i < n; ++i) {
if (i > R) {
L = R = i;
while (R < n && str[R - L] == str[R]) {
R++;
}
Z[i] = R - L;
R--;
}
else {
k = i - L;
if (Z[k] < R - i + 1)
Z[i] = Z[k];
else {
L = i;
while (R < n && str[R - L] == str[R]) {
R++;
}
Z[i] = R - L;
R--;
}
}
}
}
static public void Main()
{
string text = "GEEKS FOR GEEKS" ;
string pattern = "GEEK" ;
search(text, pattern);
}
}
|
Javascript
function search(text, pattern) {
let concat = pattern + "$" + text;
let l = concat.length;
let Z = [];
getZarr(concat, Z);
for (let i = 0; i < l; i++) {
if (Z[i] == pattern.length) {
console.log(`Pattern found at index ${i - pattern.length - 1}`);
}
}
}
function getZarr(str, Z) {
let n = str.length;
let L, R, k;
L = R = 0;
for (let i = 1; i < n; i++) {
if (i > R) {
L = R = i;
while (R < n && str[R - L] == str[R]) {
R++;
}
Z[i] = R - L;
R--;
} else {
k = i - L;
if (Z[k] < R - i + 1) {
Z[i] = Z[k];
}
else {
L = i;
while (R < n && str[R - L] == str[R]) {
R++;
}
Z[i] = R - L;
R--;
}
}
}
}
let text = "GEEKS FOR GEEKS" ;
let pattern = "GEEK" ;
search(text, pattern);
|
Output
Pattern found at index 0
Pattern found at index 10
Time Complexity: O(m+n), where m is length of pattern and n is length of text.
Auxiliary Space: O(m+n)
Aho-Corasick Algorithm finds all words in O(n + m + z) time where z is the total number of occurrences of words in text. The Aho–Corasick string matching algorithm formed the basis of the original Unix command “fgrep”.
Preprocessing: Build an automaton of all words in arr[] The automaton has mainly three functions:
Go To: This function simply follows edges of Trie of all words in arr[].
It is represented as 2D array g[][] where we store next state for current state and character.
Failure: This function stores all edges that are followed when current character doesn’t have edge in Trie.
It is represented as1D array f[] where we store next state for current state.
Output: Stores indexes of all words that end at current state.
It is represented as 1D array o[] where we store indices of all matching words as a bitmap for current state.
Matching: Traverse the given text over built automaton to find all matching words.
Preprocessing:
Illustration of Aho-Corasick algorithm
Preprocessing: We first Build a Trie (or Keyword Tree) of all words.
Build a Trie (or Keyword Tree) of all words.
- This part fills entries in goto g[][] and output o[].
- Next, we extend Trie into an automaton to support linear time matching.
Fills entries in goto g[][] and output o[]
- This part fills entries in failure f[] and output o[].
Go to: We build Trie. And for all characters which don’t have an edge at the root, we add an edge back to root.
Failure: For a state s, we find the longest proper suffix which is a proper prefix of some pattern. This is done using Breadth First Traversal of Trie.
Output: For a state s, indexes of all words ending at s are stored. These indexes are stored as bitwise map (by doing bitwise OR of values). This is also computing using Breadth First Traversal with Failure.
Below is the implementation of the Aho-Corasick Algorithm:
C++
#include <bits/stdc++.h>
using namespace std;
#define MAXS 500
#define MAXC 26
int out[MAXS];
int f[MAXS];
int g[MAXS][MAXC];
int buildMatchingMachine(string arr[], int k)
{
memset (out, 0, sizeof out);
memset (g, -1, sizeof g);
int states = 1;
for ( int i = 0; i < k; i++) {
string word = arr[i];
int currentState = 0;
for ( int j = 0; j < word.length(); j++) {
int ch = word[j] - 'a' ;
if (g[currentState][ch] == -1)
g[currentState][ch] = states++;
currentState = g[currentState][ch];
}
out[currentState] |= (1 << i);
}
for ( int ch = 0; ch < MAXC; ch++)
if (g[0][ch] == -1)
g[0][ch] = 0;
memset (f, -1, sizeof f);
queue< int > q;
for ( int ch = 0; ch < MAXC; ch++) {
if (g[0][ch] != 0) {
f[g[0][ch]] = 0;
q.push(g[0][ch]);
}
}
while (!q.empty()) {
int state = q.front();
q.pop();
for ( int ch = 0; ch < MAXC; ch++) {
if (g[state][ch] != -1) {
int failure = f[state];
while (g[failure][ch] == -1)
failure = f[failure];
failure = g[failure][ch];
f[g[state][ch]] = failure;
out[g[state][ch]] |= out[failure];
q.push(g[state][ch]);
}
}
}
return states;
}
void searchWords(string arr[], int k, string text)
{
buildMatchingMachine(arr, k);
int currentState = 0;
for ( int i = 0; i < text.length(); i++) {
int ch = text[i] - 'a' ;
while (g[currentState][ch] == -1)
currentState = f[currentState];
currentState = g[currentState][ch];
if (out[currentState] == 0)
continue ;
for ( int j = 0; j < k; j++) {
if (out[currentState] & (1 << j))
cout << "Word " << arr[j]
<< " appears from "
<< i - arr[j].length() + 1 << " to "
<< i << endl;
}
}
}
int main()
{
string arr[] = { "he" , "she" , "hers" , "his" };
int k = sizeof (arr) / sizeof (arr[0]);
string text = "ahishers" ;
searchWords(arr, k, text);
return 0;
}
|
Java
import java.util.*;
class GFG {
static int MAXS = 500 ;
static int MAXC = 26 ;
static int [] out = new int [MAXS];
static int [] f = new int [MAXS];
static int [][] g = new int [MAXS][MAXC];
static int buildMatchingMachine(String arr[], int k)
{
Arrays.fill(out, 0 );
for ( int i = 0 ; i < MAXS; i++)
Arrays.fill(g[i], - 1 );
int states = 1 ;
for ( int i = 0 ; i < k; ++i) {
String word = arr[i];
int currentState = 0 ;
for ( int j = 0 ; j < word.length(); ++j) {
int ch = word.charAt(j) - 'a' ;
if (g[currentState][ch] == - 1 )
g[currentState][ch] = states++;
currentState = g[currentState][ch];
}
out[currentState] |= ( 1 << i);
}
for ( int ch = 0 ; ch < MAXC; ++ch)
if (g[ 0 ][ch] == - 1 )
g[ 0 ][ch] = 0 ;
Arrays.fill(f, - 1 );
Queue<Integer> q = new LinkedList<>();
for ( int ch = 0 ; ch < MAXC; ++ch) {
if (g[ 0 ][ch] != 0 ) {
f[g[ 0 ][ch]] = 0 ;
q.add(g[ 0 ][ch]);
}
}
while (!q.isEmpty()) {
int state = q.peek();
q.remove();
for ( int ch = 0 ; ch < MAXC; ++ch) {
if (g[state][ch] != - 1 ) {
int failure = f[state];
while (g[failure][ch] == - 1 )
failure = f[failure];
failure = g[failure][ch];
f[g[state][ch]] = failure;
out[g[state][ch]] |= out[failure];
q.add(g[state][ch]);
}
}
}
return states;
}
static int findNextState( int currentState,
char nextInput)
{
int answer = currentState;
int ch = nextInput - 'a' ;
while (g[answer][ch] == - 1 )
answer = f[answer];
return g[answer][ch];
}
static void searchWords(String arr[], int k,
String text)
{
buildMatchingMachine(arr, k);
int currentState = 0 ;
for ( int i = 0 ; i < text.length(); ++i) {
currentState = findNextState(currentState,
text.charAt(i));
if (out[currentState] == 0 )
continue ;
for ( int j = 0 ; j < k; ++j) {
if ((out[currentState] & ( 1 << j)) > 0 ) {
System.out.print(
"Word " + arr[j] + " appears from "
+ (i - arr[j].length() + 1 ) + " to "
+ i + "\n" );
}
}
}
}
public static void main(String[] args)
{
String arr[] = { "he" , "she" , "hers" , "his" };
String text = "ahishers" ;
int k = arr.length;
searchWords(arr, k, text);
}
}
|
Python
from collections import defaultdict
class AhoCorasick:
def __init__( self , words):
self .max_states = sum ([ len (word) for word in words])
self .max_characters = 26
self .out = [ 0 ] * ( self .max_states + 1 )
self .fail = [ - 1 ] * ( self .max_states + 1 )
self .goto = [
[ - 1 ] * self .max_characters for _ in range ( self .max_states + 1 )]
for i in range ( len (words)):
words[i] = words[i].lower()
self .words = words
self .states_count = self .__build_matching_machine()
def __build_matching_machine( self ):
k = len ( self .words)
states = 1
for i in range (k):
word = self .words[i]
current_state = 0
for character in word:
ch = ord (character) - 97
if self .goto[current_state][ch] = = - 1 :
self .goto[current_state][ch] = states
states + = 1
current_state = self .goto[current_state][ch]
self .out[current_state] | = ( 1 << i)
for ch in range ( self .max_characters):
if self .goto[ 0 ][ch] = = - 1 :
self .goto[ 0 ][ch] = 0
queue = []
for ch in range ( self .max_characters):
if self .goto[ 0 ][ch] ! = 0 :
self .fail[ self .goto[ 0 ][ch]] = 0
queue.append( self .goto[ 0 ][ch])
while queue:
state = queue.pop( 0 )
for ch in range ( self .max_characters):
if self .goto[state][ch] ! = - 1 :
failure = self .fail[state]
while self .goto[failure][ch] = = - 1 :
failure = self .fail[failure]
failure = self .goto[failure][ch]
self .fail[ self .goto[state][ch]] = failure
self .out[ self .goto[state][ch]] | = self .out[failure]
queue.append( self .goto[state][ch])
return states
def __find_next_state( self , current_state, next_input):
answer = current_state
ch = ord (next_input) - 97
while self .goto[answer][ch] = = - 1 :
answer = self .fail[answer]
return self .goto[answer][ch]
def search_words( self , text):
text = text.lower()
current_state = 0
result = defaultdict( list )
for i in range ( len (text)):
current_state = self .__find_next_state(current_state, text[i])
if self .out[current_state] = = 0 :
continue
for j in range ( len ( self .words)):
if ( self .out[current_state] & ( 1 << j)) > 0 :
word = self .words[j]
result[word].append(i - len (word) + 1 )
return result
if __name__ = = "__main__" :
words = [ "he" , "she" , "hers" , "his" ]
text = "ahishers"
aho_chorasick = AhoCorasick(words)
result = aho_chorasick.search_words(text)
for word in result:
for i in result[word]:
print ( "Word" , word, "appears from" , i, "to" , i + len (word) - 1 )
|
Javascript
const MAXS = 500;
const MAXC = 26;
let out = new Array(MAXS).fill(0);
let f = new Array(MAXS).fill(-1);
let g = Array.from(Array(MAXS), () => new Array(MAXC).fill(-1));
function buildMatchingMachine(arr, k) {
out.fill(0);
g.forEach(row => row.fill(-1));
let states = 1;
for (let i = 0; i < k; i++) {
const word = arr[i];
let currentState = 0;
for (let j = 0; j < word.length; j++) {
const ch = word.charCodeAt(j) - 'a' .charCodeAt(0);
if (g[currentState][ch] === -1) g[currentState][ch] = states++;
currentState = g[currentState][ch];
}
out[currentState] |= 1 << i;
}
for (let ch = 0; ch < MAXC; ch++) {
if (g[0][ch] === -1) g[0][ch] = 0;
}
f.fill(-1);
const q = [];
for (let ch = 0; ch < MAXC; ch++) {
if (g[0][ch] !== 0) {
f[g[0][ch]] = 0;
q.push(g[0][ch]);
}
}
while (q.length) {
const state = q.shift();
for (let ch = 0; ch < MAXC; ch++) {
if (g[state][ch] !== -1) {
let failure = f[state];
while (g[failure][ch] === -1) failure = f[failure];
failure = g[failure][ch];
f[g[state][ch]] = failure;
out[g[state][ch]] |= out[failure];
q.push(g[state][ch]);
}
}
}
return states;
}
function searchWords(arr, k, text) {
buildMatchingMachine(arr, k);
let currentState = 0;
for (let i = 0; i < text.length; i++) {
const ch = text.charCodeAt(i) - 'a' .charCodeAt(0);
while (g[currentState][ch] === -1) currentState = f[currentState];
currentState = g[currentState][ch];
if (out[currentState] === 0) continue ;
for (let j = 0; j < k; j++) {
if (out[currentState] & (1 << j)) {
console.log(`Word ${arr[j]} appears from ${i - arr[j].length + 1} to ${i}`);
}
}
}
}
const arr = [ "he" , "she" , "hers" , "his" ];
const k = arr.length;
const text = "ahishers" ;
searchWords(arr, k, text);
|
Output
Word his appears from 1 to 3
Word he appears from 4 to 5
Word she appears from 3 to 5
Word hers appears from 4 to 7
Time Complexity: O(n + l + z), where ‘n’ is the length of the text, ‘l’ is the length of keywords, and ‘z’ is the number of matches.
Auxiliary Space: O(l * q), where ‘q’ is the length of the alphabet since that is the maximum number of children a node can have.
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