Introduction to Pattern Searching – Data Structure and Algorithm Tutorial

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 algorithm

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++

// C++ program for Naive Pattern
// Searching algorithm
#include <bits/stdc++.h>
using namespace std;

void search(char* pat, char* txt)
{
    int M = strlen(pat);
    int N = strlen(txt);

    /* A loop to slide pat[] one by one */
    for (int i = 0; i <= N - M; i++) {
        int j;

        /* For current index i, check for pattern match */
        for (j = 0; j < M; j++)
            if (txt[i + j] != pat[j])
                break;

        if (j
            == M) // if pat[0...M-1] = txt[i, i+1, ...i+M-1]
            cout << "Pattern found at index " << i << endl;
    }
}

// Driver's Code
int main()
{
    char txt[] = "AABAACAADAABAAABAA";
    char pat[] = "AABA";

    // Function call
    search(pat, txt);
    return 0;
}

Java

// Java program for Naive Pattern
// Searching algorithm
class GFG {

  static void search(char[] pat, char[] txt)
  {
    int M = pat.length;
    int N = txt.length;

    /* A loop to slide pat[] one by one */
    for (int i = 0; i <= N - M; i++) {
      int j;

      /* For current index i, check for pattern match
             */
      for (j = 0; j < M; j++)
        if (txt[i + j] != pat[j])
          break;

      // if pat[0...M-1] = txt[i, i+1, ...i+M-1]
      if (j == M)
        System.out.println("Pattern found at index "
                           + i);
    }
  }

  // Driver's Code

  public static void main(String[] args)
  {
    char txt[] = "AABAACAADAABAAABAA".toCharArray();

    char pat[] = "AABA".toCharArray();

    // Function call
    search(pat, txt);
  }
}

// This code is contributed by karandeep1234

Python3

# Python program for above approach
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)

# This code is contributed by ishankhandelwals.

C#

using System;

public class GFG {

  public static void search(char[] pat, char[] txt)
  {
    int M = pat.Length;
    int N = txt.Length;

    /* A loop to slide pat[] one by one */
    for (int i = 0; i <= N - M; i++) {
      int j;

      /* For current index i, check for pattern match
             */
      for (j = 0; j < M; j++)
        if (txt[i + j] != pat[j])
          break;

      if (j == M) // if pat[0...M-1] = txt[i, i+1,
        // ...i+M-1]
        Console.WriteLine("Pattern found at index "
                          + i);
    }
  }

  static public void Main()
  {

    char[] txt = "AABAACAADAABAAABAA".ToCharArray();
    char[] pat = "AABA".ToCharArray();

    // Function call
    search(pat, txt);
  }
}
// This code is contributed by akashish__

Javascript

// JS program for Naive Pattern
// Searching algorithm
function search(pat, txt)
{
    let M = pat.length;
    let N = txt.length;

    /* A loop to slide pat[] one by one */
    for (let i = 0; i <= N - M; i++) {
        let j = 0;

        /* For current index i, check for pattern match */
        for (j = 0; j < M; j++)
            if (txt[i + j] != pat[j])
                break;
        if (j == M) // if pat[0...M-1] = txt[i, i+1, ...i+M-1]
            console.log("Pattern found at index",i);
    }
}

// Driver's Code
    let txt = "AABAACAADAABAAABAA";
    let pat = "AABA";

    // Function call
    search(pat, txt);

    // This code is contributed by ishankhandelwals.

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

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++

// C++ program for implementation of KMP pattern searching
// algorithm
#include <bits/stdc++.h>

void computeLPSArray(char* pat, int M, int* lps);

// Prints occurrences of txt[] in pat[]
void KMPSearch(char* pat, char* txt)
{
    int M = strlen(pat);
    int N = strlen(txt);

    // create lps[] that will hold the longest prefix suffix
    // values for pattern
    int lps[M];

    // Preprocess the pattern (calculate lps[] array)
    computeLPSArray(pat, M, lps);

    int i = 0; // index for txt[]
    int j = 0; // index for pat[]
    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];
        }

        // mismatch after j matches
        else if (i < N && pat[j] != txt[i]) {
            // Do not match lps[0..lps[j-1]] characters,
            // they will match anyway
            if (j != 0)
                j = lps[j - 1];
            else
                i = i + 1;
        }
    }
}

// Fills lps[] for given pattern pat[0..M-1]
void computeLPSArray(char* pat, int M, int* lps)
{
    // length of the previous longest prefix suffix
    int len = 0;

    lps[0] = 0; // lps[0] is always 0

    // the loop calculates lps[i] for i = 1 to M-1
    int i = 1;
    while (i < M) {
        if (pat[i] == pat[len]) {
            len++;
            lps[i] = len;
            i++;
        }
        else // (pat[i] != pat[len])
        {
            // This is tricky. Consider the example.
            // AAACAAAA and i = 7. The idea is similar
            // to search step.
            if (len != 0) {
                len = lps[len - 1];

                // Also, note that we do not increment
                // i here
            }
            else // if (len == 0)
            {
                lps[i] = 0;
                i++;
            }
        }
    }
}

// Driver program to test above function
int main()
{
    char txt[] = "ABABDABACDABABCABAB";
    char pat[] = "ABABCABAB";
    KMPSearch(pat, txt);
    return 0;
}

Java

// Java program for implementation of KMP pattern searching
// algorithm
public class KMP_String_Matching {
    void KMPSearch(String pat, String txt)
    {
        int M = pat.length();
        int N = txt.length();

        // create lps[] that will hold the longest prefix suffix
        // values for pattern
        int lps[] = new int[M];
        int j = 0; // index for pat[]

        // Preprocess the pattern (calculate lps[] array)
        computeLPSArray(pat, M, lps);

        int i = 0; // index for txt[]
        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];
            }

            // mismatch after j matches
            else if (i < N && pat.charAt(j) != txt.charAt(i)) {
                // Do not match lps[0..lps[j-1]] characters,
                // they will match anyway
                if (j != 0)
                    j = lps[j - 1];
                else
                    i = i + 1;
            }
        }
    }

    void computeLPSArray(String pat, int M, int lps[])
    {
        // length of the previous longest prefix suffix
        int len = 0;
        int i = 1;
        lps[0] = 0; // lps[0] is always 0

        // the loop calculates lps[i] for i = 1 to M-1
        while (i < M) {
            if (pat.charAt(i) == pat.charAt(len)) {
                len++;
                lps[i] = len;
                i++;
            }
            else // (pat[i] != pat[len])
            {
                // This is tricky. Consider the example.
                // AAACAAAA and i = 7. The idea is similar
                // to search step.
                if (len != 0) {
                    len = lps[len - 1];

                    // Also, note that we do not increment
                    // i here
                }
                else // if (len == 0)
                {
                    lps[i] = len;
                    i++;
                }
            }
        }
    }

    // Driver program to test above function
    public static void main(String[] args)
    {
        String txt = "ABABDABACDABABCABAB";
        String pat = "ABABCABAB";
        new KMP_String_Matching().KMPSearch(pat, txt);
    }
}

Python3

# Python program for implementation of KMP pattern searching
# algorithm
def computeLPSArray(pat, M, lps):
    len = 0  # length of the previous longest prefix suffix

    lps[0]  # lps[0] is always 0
    i = 1

    # the loop calculates lps[i] for i = 1 to M-1
    while i < M:
        if pat[i] == pat[len]:
            len += 1
            lps[i] = len
            i += 1
        else:
            # This is tricky. Consider the example.
            # AAACAAAA and i = 7. The idea is similar
            # to search step.
            if len != 0:
                len = lps[len-1]

            # Also, note that we do not increment i here
            else:
                lps[i] = 0
                i += 1

def KMPSearch(pat, txt):
    M = len(pat)
    N = len(txt)

    # create lps[] that will hold the longest prefix suffix
    # values for pattern
    lps = [0]*M
    j = 0  # index for pat[]

    # Preprocess the pattern (calculate lps[] array)
    computeLPSArray(pat, M, lps)

    i = 0  # index for txt[]
    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]

        # mismatch after j matches
        elif i < N and pat[j] != txt[i]:
            # Do not match lps[0..lps[j-1]] characters,
            # they will match anyway
            if j != 0:
                j = lps[j-1]
            else:
                i += 1

txt = "ABABDABACDABABCABAB"
pat = "ABABCABAB"
KMPSearch(pat, txt)

# This code is contributed by ishankhandelwals.

C#

using System;
using System.Collections.Generic;

public class GFG {

  // Prints occurrences of txt[] in pat[]
  public static void KMPSearch(char[] pat, char[] txt)
  {
    int M = pat.Length;
    int N = txt.Length;

    // create lps[] that will hold the longest prefix
    // suffix values for pattern
    int[] lps = new int[M];

    // Preprocess the pattern (calculate lps[] array)
    computeLPSArray(pat, M, lps);

    int i = 0; // index for txt[]
    int j = 0; // index for pat[]
    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];
      }

      // mismatch after j matches
      else if (i < N && pat[j] != txt[i]) {
        // Do not match lps[0..lps[j-1]] characters,
        // they will match anyway
        if (j != 0)
          j = lps[j - 1];
        else
          i = i + 1;
      }
    }
  }

  // Fills lps[] for given pattern pat[0..M-1]
  public static void computeLPSArray(char[] pat, int M,
                                     int[] lps)
  {
    // length of the previous longest prefix suffix
    int len = 0;

    lps[0] = 0; // lps[0] is always 0

    // the loop calculates lps[i] for i = 1 to M-1
    int i = 1;
    while (i < M) {
      if (pat[i] == pat[len]) {
        len++;
        lps[i] = len;
        i++;
      }
      else // (pat[i] != pat[len])
      {
        // This is tricky. Consider the example.
        // AAACAAAA and i = 7. The idea is similar
        // to search step.
        if (len != 0) {
          len = lps[len - 1];

          // Also, note that we do not increment
          // i here
        }
        else // if (len == 0)
        {
          lps[i] = 0;
          i++;
        }
      }
    }
  }

  static public void Main()
  {

    char[] txt = "ABABDABACDABABCABAB".ToCharArray();
    char[] pat = "ABABCABAB".ToCharArray();
    KMPSearch(pat, txt);
  }
}

// This code is contributed by akashish__

Javascript

// JS program for implementation of KMP pattern searching
// algorithm
// Prlets occurrences of txt[] in pat[]
function computeLPSArray(pat, M, lps)
{

    // length of the previous longest prefix suffix
    let len = 0;
    lps[0] = 0; // lps[0] is always 0
    // the loop calculates lps[i] for i = 1 to M-1
    let i = 1;
    while (i < M) {
        if (pat[i] == pat[len]) {
            len++;
            lps[i] = len;
            i++;
        }
        else // (pat[i] != pat[len])
        {

            // This is tricky. Consider the example.
            // AAACAAAA and i = 7. The idea is similar
            // to search step.
            if (len != 0) {
                len = lps[len - 1];

                // Also, note that we do not increment
                // i here
            }
            else // if (len == 0)
            {
                lps[i] = 0;
                i++;
            }
        }
    }
}
function KMPSearch(pat, txt) {
    let M = pat.length;
    let N = txt.length

    // create lps[] that will hold the longest prefix suffix
    // values for pattern
    let lps = [];

    // Preprocess the pattern (calculate lps[] array)
    computeLPSArray(pat, M, lps);
    let i = 0; // index for txt[]
    let j = 0; // index for pat[]
    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];
        }

        // mismatch after j matches
        else if (i < N && pat[j] != txt[i])
        {

            // Do not match lps[0..lps[j-1]] characters,
            // they will match anyway
            if (j != 0)
                j = lps[j - 1];
            else
                i = i + 1;
        }
    }
}

// Fills lps[] for given pattern pat[0..M-1]
// Driver program to test above function
let txt = "ABABDABACDABABCABAB";
let pat = "ABABCABAB";
KMPSearch(pat, txt);

// This code is contributed by ishankhandelwals.

Output

Found pattern at index 10 

Time complexity: O(n + m)
Auxiliary Space: O(M)

Rabin Karp algorithm:

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++

/* Following program is a C++ implementation of Rabin Karp
Algorithm given in the CLRS book */
#include <bits/stdc++.h>
using namespace std;

// d is the number of characters in the input alphabet
#define d 256

/* pat -> pattern
    txt -> text
    q -> A prime number
*/
void search(char pat[], char txt[], int q)
{
    int M = strlen(pat);
    int N = strlen(txt);
    int i, j;
    int p = 0; // hash value for pattern
    int t = 0; // hash value for txt
    int h = 1;

    // The value of h would be "pow(d, M-1)%q"
    for (i = 0; i < M - 1; i++)
        h = (h * d) % q;

    // Calculate the hash value of pattern and first
    // window of text
    for (i = 0; i < M; i++) {
        p = (d * p + pat[i]) % q;
        t = (d * t + txt[i]) % q;
    }

    // Slide the pattern over text one by one
    for (i = 0; i <= N - M; i++) {

        // Check the hash values of current window of text
        // and pattern. If the hash values match then only
        // check for characters one by one
        if (p == t) {
            /* Check for characters one by one */
            for (j = 0; j < M; j++) {
                if (txt[i + j] != pat[j]) {
                    break;
                }
            }

            // if p == t and pat[0...M-1] = txt[i, i+1,
            // ...i+M-1]

            if (j == M)
                cout << "Pattern found at index " << i
                     << endl;
        }

        // Calculate hash value for next window of text:
        // Remove leading digit, add trailing digit
        if (i < N - M) {
            t = (d * (t - txt[i] * h) + txt[i + M]) % q;

            // We might get negative value of t, converting
            // it to positive
            if (t < 0)
                t = (t + q);
        }
    }
}

/* Driver code */
int main()
{
    char txt[] = "GEEKS FOR GEEKS";
    char pat[] = "GEEK";

    // we mod to avoid overflowing of value but we should
    // take as big q as possible to avoid the collison
    int q = INT_MAX;

    // Function Call
    search(pat, txt, q);
    return 0;
}

// This is code is contributed by rathbhupendra

Java

import java.io.*;
import java.lang.*;
import java.util.*;

/* pat -> pattern
    txt -> text
    q -> A prime number
*/
public class GFG {
  // d is the number of characters in the input alphabet
  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; // hash value for pattern
    int t = 0; // hash value for txt
    int h = 1;

    // The value of h would be "pow(d, M-1)%q"
    for (i = 0; i < M - 1; i++)
      h = (h * d) % q;
    // Calculate the hash value of pattern and first
    // window of text
    for (i = 0; i < M; i++) {
      p = (d * p + pat.charAt(i)) % q;
      t = (d * t + txt.charAt(i)) % q;
    }

    // Slide the pattern over text one by one
    for (i = 0; i <= N - M; i++) {

      // Check the hash values of current window of
      // text and pattern. If the hash values match
      // then only check for characters one by one
      if (p == t) {
        /* Check for characters one by one */
        for (j = 0; j < M; j++) {
          if (txt.charAt(i + j)
              != pat.charAt(j)) {
            break;
          }
        }

        // if p == t and pat[0...M-1] = txt[i, i+1,
        // ...i+M-1]

        if (j == M) {
          System.out.println(
            "Pattern found at index " + i);
        }
      }
      // Calculate hash value for next window of text:
      // Remove leading digit, add trailing digit
      if (i < N - M) {
        t = (d * (t - txt.charAt(i) * h)
             + txt.charAt(i + M))
          % q;

        // We might get negative value of t,
        // converting it to positive
        if (t < 0)
          t = (t + q);
      }
    }
  }

  /* Driver code */
  public static void main(String[] args)
  {
    String txt = "GEEKS FOR GEEKS";
    String pat = "GEEK";

    // A prime number
    int q = 101;

    // Function Call
    search(pat, txt, q);
  }
}

// This code is contributed by ishankhandelwals.

Python3

# d is the number of characters in the input alphabet
d = 256

''' pat -> pattern
txt -> text
q -> A prime number '''
def search(pat, txt, q):

    M = len(pat)
    N = len(txt)
    p = 0 # hash value for pattern
    t = 0 # hash value for txt
    h = 1

    # The value of h would be "pow(d, M-1)%q"
    for i in range(M - 1):
        h = (h * d) % q

    # Calculate the hash value of pattern and first
    # window of text
    for i in range(M):
        p = (d * p + ord(pat[i])) % q
        t = (d * t + ord(txt[i])) % q

    # Slide the pattern over text one by one
    for i in range(N - M + 1):
        # Check the hash values of current window of text
        # and pattern. If the hash values match then only
        # check for characters one by one
        if p == t:
            # Check for characters one by one
            for j in range(M):
                if txt[i + j] != pat[j]:
                    break
            # if p == t and pat[0...M-1] = txt[i, i+1,
            # ...i+M-1]
            if j == M - 1:
                print("Pattern found at index " + str(i))

        # Calculate hash value for next window of text:
        # Remove leading digit, add trailing digit
        if i < N - M:
            t = (d * (t - ord(txt[i]) * h) + ord(txt[i + M])) % q
            # We might get negative value of t, converting
            # it to positive
            if t < 0:
                t = (t + q)

# Driver code
txt = "GEEKS FOR GEEKS"
pat = "GEEK"

# we mod to avoid overflowing of value but we should
# take as big q as possible to avoid the collison
q = float('inf')

# Function Call
search(pat, txt, q)

# This code is contributed by akashish__

C#

// C# code
using System;

/* pat -> pattern
    txt -> text
    q -> A prime number
*/

class GFG {
  // d is the number of characters in the input alphabet
  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; // hash value for pattern
    int t = 0; // hash value for txt
    int h = 1;
    // The value of h would be "pow(d, M-1)%q"
    for (i = 0; i < M - 1; i++)
      h = (h * d) % q;
    // Calculate the hash value of pattern and first
    // window of text
    for (i = 0; i < M; i++) {
      p = (d * p + pat[i]) % q;
      t = (d * t + txt[i]) % q;
    }

    // Slide the pattern over text one by one
    for (i = 0; i <= N - M; i++) {

      // Check the hash values of current window of
      // text and pattern. If the hash values match
      // then only check for characters one by one
      if (p == t) {
        /* Check for characters one by one */
        for (j = 0; j < M; j++) {
          if (txt[i + j] != pat[j]) {
            break;
          }
        }

        // if p == t and pat[0...M-1] = txt[i, i+1,
        // ...i+M-1]

        if (j == M) {
          Console.WriteLine(
            "Pattern found at index " + i);
        }
      }
      // Calculate hash value for next window of text:
      // Remove leading digit, add trailing digit
      if (i < N - M) {
        t = (d * (t - txt[i] * h) + txt[i + M]) % q;

        // We might get negative value of t,
        // converting it to positive
        if (t < 0)
          t = (t + q);
      }
    }
  }

  /* Driver code */
  public static void Main(string[] args)
  {
    string txt = "GEEKS FOR GEEKS";
    string pat = "GEEK";

    // A prime number
    int q = 101;

    // Function Call
    search(pat, txt, q);
  }
}

// This code is contributed by akashish__

Javascript

// d is the number of characters in the input alphabet
const d = 256;

/* pat -> pattern
    txt -> text
    q -> A prime number
*/
function search(pat, txt, q) {
  const M = pat.length;
  const N = txt.length;
  let p = 0; // hash value for pattern
  let t = 0; // hash value for txt
  let h = 1;

  // The value of h would be "pow(d, M-1)%q"
  for (let i = 0; i < M - 1; i++) {
    h = (h * d) % q;
  }

  // Calculate the hash value of pattern and first
  // window of text
  for (let i = 0; i < M; i++) {
    p = (d * p + pat.charCodeAt(i)) % q;
    t = (d * t + txt.charCodeAt(i)) % q;
  }

  // Slide the pattern over text one by one
  for (let i = 0; i <= N - M; i++) {
    // Check the hash values of current window of text
    // and pattern. If the hash values match then only
    // check for characters one by one
    if (p === t) {
    /* Check for characters one by one */
    for (j = 0; j < M; j++) {
        if (txt.charAt(i + j) !== pat.charAt(j)) {
        break;
        }
    }

    // if p == t and pat[0...M-1] = txt[i, i+1,
    // ...i+M-1]

    if (j === M)
        console.log("Pattern found at index " + i);
    }

    // Calculate hash value for next window of text:
    // Remove leading digit, add trailing digit
    if (i < N - M) {
    t = (d * (t - txt.charCodeAt(i) * h) + txt.charCodeAt(i + M)) % q;

    // We might get negative value of t, converting
    // it to positive
    if (t < 0)
        t = (t + q);
    }
  }
}

/* Driver code */
const txt = "GEEKS FOR GEEKS";
const pat = "GEEK";

// we mod to avoid overflowing of value but we should
// take as big q as possible to avoid the collison
const q = Number.MAX_SAFE_INTEGER;

// Function Call
search(pat, txt, q);

// This code is contributed by ishankhandelwals.

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.

Z algorithm:

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 – 

1. 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).
2. 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++

// A C++ program that implements Z algorithm for pattern
// searching
#include <iostream>
using namespace std;

void getZarr(string str, int Z[]);

// prints all occurrences of pattern in text using Z algo
void search(string text, string pattern)
{
    // Create concatenated string "P$T"
    string concat = pattern + "$" + text;
    int l = concat.length();

    // Construct Z array
    int Z[l];
    getZarr(concat, Z);

    // now looping through Z array for matching condition
    for (int i = 0; i < l; ++i) {
        // if Z[i] (matched region) is equal to pattern
        // length we got the pattern
        if (Z[i] == pattern.length())
            cout << "Pattern found at index "
                 << i - pattern.length() - 1 << endl;
    }
}

// Fills Z array for given string str[]
void getZarr(string str, int Z[])
{
    int n = str.length();
    int L, R, k;

    // [L, R] make a window which matches with prefix of s
    L = R = 0;
    for (int i = 1; i < n; ++i) {
        // if i>R nothing matches so we will calculate.
        // Z[i] using naive way.
        if (i > R) {
            L = R = i;

            // R-L = 0 in starting, so it will start
            // checking from 0'th index. For example,
            // for "ababab" and i = 1, the value of R
            // remains 0 and Z[i] becomes 0. For string
            // "aaaaaa" and i = 1, Z[i] and R become 5
            while (R < n && str[R - L] == str[R])
                R++;
            Z[i] = R - L;
            R--;
        }
        else {
            // k = i-L so k corresponds to number which
            // matches in [L, R] interval.
            k = i - L;

            // if Z[k] is less than remaining interval
            // then Z[i] will be equal to Z[k].
            // For example, str = "ababab", i = 3, R = 5
            // and L = 2
            if (Z[k] < R - i + 1)
                Z[i] = Z[k];

            // For example str = "aaaaaa" and i = 2, R is 5,
            // L is 0
            else {
                // else start from R and check manually
                L = i;
                while (R < n && str[R - L] == str[R])
                    R++;
                Z[i] = R - L;
                R--;
            }
        }
    }
}

// Driver program
int main()
{
    string text = "GEEKS FOR GEEKS";
    string pattern = "GEEK";
    search(text, pattern);
    return 0;
}

Java

// A Java program that implements Z algorithm for pattern
// searching
import java.io.*;

class GFG
{

  // prints all occurrences of pattern in text using Z
  // algo
  static void search(String text, String pattern)
  {

    // Create concatenated string "P$T"
    String concat = pattern + "$" + text;
    int l = concat.length();

    // Construct Z array
    int[] Z = new int[l];
    getZarr(concat, Z);

    // now looping through Z array for matching
    // condition
    for (int i = 0; i < l; i++) {
      // if Z[i] (matched region) is equal to pattern
      // length we got the pattern
      if (Z[i] == pattern.length()) {
        System.out.println(
          "Pattern found at index "
          + (i - pattern.length() - 1));
      }
    }
  }

  // Fills Z array for given string str[]
  static void getZarr(String str, int[] Z)
  {
    int n = str.length();
    // [L, R] make a window which matches with prefix of
    // s
    int L = 0, R = 0, k;

    for (int i = 1; i < n; ++i) {
      // if i>R nothing matches so we will calculate.
      // Z[i] using naive way.
      if (i > R) {
        L = R = i;
        // R-L = 0 in starting, so it will start
        // checking from 0'th index. For example,
        // for "ababab" and i = 1, the value of R
        // remains 0 and Z[i] becomes 0. For string
        // "aaaaaa" and i = 1, Z[i] and R become 5
        while (R < n
               && str.charAt(R - L)
               == str.charAt(R)) {
          R++;
        }
        Z[i] = R - L;
        R--;
      }
      else {
        // k = i-L so k corresponds to number which
        // matches in [L, R] interval.
        k = i - L;

        // if Z[k] is less than remaining interval
        // then Z[i] will be equal to Z[k].
        // For example, str = "ababab", i = 3, R = 5
        // and L = 2
        if (Z[k] < R - i + 1)
          Z[i] = Z[k];

        // For example str = "aaaaaa" and i = 2, R
        // is 5, L is 0
        else {
          // else start from R and check manually
          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);
  }
}

// This code is contributed by lokeshmvs21.

Python3

# A Python program that implements Z algorithm for pattern
# searching
# Fills Z array for given string str[]
def getZarr(string, Z):
    n = len(string)

    # [L, R] make a window which matches with prefix of s
    L, R, k = 0, 0, 0
    Z[0] = n

    for i in range(1, n):

      # if i>R nothing matches so we will calculate.
        # Z[i] using naive way.
        if i > R:
            L, R = i, i

            # R-L = 0 in starting, so it will start
            # checking from 0'th index. For example,
            # for "ababab" and i = 1, the value of R
            # remains 0 and Z[i] becomes 0. For string
            # "aaaaaa" and i = 1, Z[i] and R become 5
            while R < n and string[R - L] == string[R]:
                R += 1
            Z[i] = R - L
            R -= 1
        else:

          # k = i-L so k corresponds to number which
            # matches in [L, R] interval.
            k = i - L

            # if Z[k] is less than remaining interval
            # then Z[i] will be equal to Z[k].
            # For example, str = "ababab", i = 3, R = 5
            # and L = 2
            if Z[k] < R - i + 1:
                Z[i] = Z[k]

            # For example str = "aaaaaa" and i = 2, R is 5,
            # L is 0
            else:

              # else start from R and check manually
                L = i
                while R < n and string[R - L] == string[R]:
                    R += 1
                Z[i] = R - L
                R -= 1

# prints all occurrences of pattern in text using Z algo
def search(text, pattern):

  # Create concatenated string "P$T"
    concat = pattern + "$" + text
    l = len(concat)

    # Construct Z array
    Z = [0] * l
    getZarr(concat, Z)

    # now looping through Z array for matching condition
    for i in range(l):

      # if Z[i] (matched region) is equal to pattern
        # length we got the pattern
        if Z[i] == len(pattern):
            print("Pattern found at index", i - len(pattern) - 1)

# Driver program
if __name__ == "__main__":
    text = "GEEKS FOR GEEKS"
    pattern = "GEEK"
    search(text, pattern)

# This code is contributed by akashish__

C#

using System;
using System.Linq;

public class GFG {

  // prints all occurrences of pattern in text using Z
  // algo
  static void search(string text, string pattern)
  {
    // Create concatenated string "P$T"
    string concat = pattern + "$" + text;
    int l = concat.Length;

    // Construct Z array
    int[] Z = new int[l];
    GetZarr(concat, Z);

    // now looping through Z array for matching
    // condition
    for (int i = 0; i < l; i++) {
      // if Z[i] (matched region) is equal to
      // pattern length we got the pattern
      if (Z[i] == pattern.Length) {
        Console.WriteLine(
          "Pattern found at index "
          + (i - pattern.Length - 1));
      }
    }
  }

  // Fills Z array for given string str[]
  static void GetZarr(string str, int[] Z)
  {
    int n = str.Length;
    // [L, R] make a window which matches with
    // prefix of
    // s
    int L = 0, R = 0, k;

    for (int i = 1; i < n; ++i) {
      // if i>R nothing matches so we will
      // calculate. Z[i] using naive way.
      if (i > R) {
        L = R = i;
        // R-L = 0 in starting, so it will start
        // checking from 0'th index. For
        // example, for "ababab" and i = 1, the
        // value of R remains 0 and Z[i] becomes
        // 0. For string "aaaaaa" and i = 1,
        // Z[i] and R become 5
        while (R < n && str[R - L] == str[R]) {
          R++;
        }
        Z[i] = R - L;
        R--;
      }
      else {
        // k = i-L so k corresponds to number
        // which matches in [L, R] interval.
        k = i - L;

        // if Z[k] is less than remaining
        // interval then Z[i] will be equal to
        // Z[k]. For example, str = "ababab", i
        // = 3, R = 5 and L = 2
        if (Z[k] < R - i + 1)
          Z[i] = Z[k];

        // For example str = "aaaaaa" and i = 2,
        // R is 5, L is 0
        else {
          // else start from R and check
          // manually
          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);
  }
}
// This code is contributed by akashish__

Javascript

function search(text, pattern) {
  // Create concatenated string "P$T"
  let concat = pattern + "$" + text;
  let l = concat.length;

  // Construct Z array
  let Z = [];
  getZarr(concat, Z);

  // now looping through Z array for matching condition
  for (let i = 0; i < l; i++) {
    // if Z[i] (matched region) is equal to pattern
    // length we got the pattern
    if (Z[i] == pattern.length) {
      console.log(`Pattern found at index ${i - pattern.length - 1}`);
    }
  }
}

// Fills Z array for given string str[]
function getZarr(str, Z) {
  let n = str.length;
  let L, R, k;

  // [L, R] make a window which matches with prefix of s
  L = R = 0;
  for (let i = 1; i < n; i++) {
    // if i>R nothing matches so we will calculate.
    // Z[i] using naive way.
    if (i > R) {
      L = R = i;

      // R-L = 0 in starting, so it will start
      // checking from 0'th index. For example,
      // for "ababab" and i = 1, the value of R
      // remains 0 and Z[i] becomes 0. For string
      // "aaaaaa" and i = 1, Z[i] and R become 5
      while (R < n && str[R - L] == str[R]) {
        R++;
      }
      Z[i] = R - L;
      R--;
    } else {
      // k = i-L so k corresponds to number which
      // matches in [L, R] interval.
      k = i - L;

      // if Z[k] is less than remaining interval
      // then Z[i] will be equal to Z[k].
      // For example, str = "ababab", i = 3, R = 5
      // and L = 2
      if (Z[k] < R - i + 1) {
        Z[i] = Z[k];
      }

      // For example str = "aaaaaa" and i = 2, R is 5,
      // L is 0
      else {
        // else start from R and check manually
        L = i;
        while (R < n && str[R - L] == str[R]) {
          R++;
        }
        Z[i] = R - L;
        R--;
      }
    }
  }
}

// Driver program
let text = "GEEKS FOR GEEKS";
let pattern = "GEEK";
search(text, pattern);

// This code is contributed by akashish__

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:

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++

// C++ program for implementation of
// Aho Corasick algorithm for String
// matching
#include <bits/stdc++.h>
using namespace std;

// Max number of states in the matching
// machine. Should be equal to the sum
// of the length of all keywords.

#define MAXS 500

// Maximum number of characters
// in input alphabet

#define MAXC 26

// OUTPUT FUNCTION IS IMPLEMENTED USING out[]
// Bit i in this mask is one if the word with
// index i appears when the machine enters
// this state.
int out[MAXS];

// FAILURE FUNCTION IS IMPLEMENTED USING f[]
int f[MAXS];

// GOTO FUNCTION (OR TRIE) IS
// IMPLEMENTED USING g[][]
int g[MAXS][MAXC];

// Builds the String matching machine.
// arr - array of words. The index of each keyword is
// important:
//"out[state] & (1 << i)" is > 0 if we just found
// word[i] in the text.
// Returns the number of states that the built machine
// has. States are numbered 0 up to the return value -
// 1, inclusive.

int buildMatchingMachine(string arr[], int k)
{
    // Initialize all values in output function as 0.
    memset(out, 0, sizeof out);

    // Initialize all values in goto function as -1.
    memset(g, -1, sizeof g);

    // Initially, we just have the 0 state
    int states = 1;

    // Convalues for goto function, i.e., fill g[][]
    // This is same as building a Trie for arr[]
    for (int i = 0; i < k; i++) {
        string word = arr[i];
        int currentState = 0;

        // Insert all characters of current
        // word in arr[]
        for (int j = 0; j < word.length(); j++) {
            int ch = word[j] - 'a';

            // Allocate a new node (create a new state)
            // if a node for ch doesn't exist.
            if (g[currentState][ch] == -1)
                g[currentState][ch] = states++;

            currentState = g[currentState][ch];
        }

        // Add current word in output function
        out[currentState] |= (1 << i);
    }

    // For all characters which don't have
    // an edge from root (or state 0) in Trie,
    // add a goto edge to state 0 itself
    for (int ch = 0; ch < MAXC; ch++)
        if (g[0][ch] == -1)
            g[0][ch] = 0;

    // Now, let's build the failure function
    // Initialize values in fail function
    memset(f, -1, sizeof f);

    // Failure function is computed in
    // breadth first order
    // using a queue
    queue<int> q;

    // Iterate over every possible input
    for (int ch = 0; ch < MAXC; ch++) {

        // All nodes of depth 1 have failure
        // function value as 0. For example,
        // in above diagram we move to 0
        // from states 1 and 3.
        if (g[0][ch] != 0) {
            f[g[0][ch]] = 0;
            q.push(g[0][ch]);
        }
    }

    // Now queue has states 1 and 3
    while (!q.empty()) {

        // Remove the front state from queue
        int state = q.front();
        q.pop();

        // For the removed state, find failure
        // function for all those characters
        // for which goto function is
        // not defined.
        for (int ch = 0; ch < MAXC; ch++) {

            // If goto function is defined for
            // character 'ch' and 'state'
            if (g[state][ch] != -1) {

                // Find failure state of removed state
                int failure = f[state];

                // Find the deepest node labeled by
                // proper suffix of String from root to
                // current state.
                while (g[failure][ch] == -1)
                    failure = f[failure];
                failure = g[failure][ch];
                f[g[state][ch]] = failure;

                // Merge output values
                out[g[state][ch]] |= out[failure];

                // Insert the next level node
                // (of Trie) in Queue
                q.push(g[state][ch]);
            }
        }
    }
    return states;
}

// Returns the next state the machine will transition to
// using goto and failure functions. currentState - The
// current state of the machine. Must be between
// 0 and the number of states - 1,
// inclusive.
// nextInput - The next character that enters into the
// machine.

// This function finds all occurrences of
// all array words in text.
void searchWords(string arr[], int k, string text)
{

    // Preprocess patterns.
    // Build machine with goto, failure
    // and output functions
    buildMatchingMachine(arr, k);

    // Initialize current state
    int currentState = 0;

    // Traverse the text through the
    // built machine to find all
    // occurrences of words in arr[]
    for (int i = 0; i < text.length(); i++) {
        int ch = text[i] - 'a';

        // If goto is not defined, use
        // failure function
        while (g[currentState][ch] == -1)
            currentState = f[currentState];
        currentState = g[currentState][ch];

        // If match not found, move to next state
        if (out[currentState] == 0)
            continue;

        // Match found, print all matching
        // words of arr[]
        // using output function.
        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;
        }
    }
}
// Driver code

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

// Java program for implementation of
// Aho Corasick algorithm for String
// matching
import java.util.*;

class GFG {

    // Max number of states in the matching
    // machine. Should be equal to the sum
    // of the length of all keywords.
    static int MAXS = 500;

    // Maximum number of characters
    // in input alphabet
    static int MAXC = 26;

    // OUTPUT FUNCTION IS IMPLEMENTED USING out[]
    // Bit i in this mask is one if the word with
    // index i appears when the machine enters
    // this state.
    static int[] out = new int[MAXS];

    // FAILURE FUNCTION IS IMPLEMENTED USING f[]
    static int[] f = new int[MAXS];

    // GOTO FUNCTION (OR TRIE) IS
    // IMPLEMENTED USING g[][]
    static int[][] g = new int[MAXS][MAXC];

    // Builds the String matching machine.
    // arr - array of words. The index of each keyword is
    // important:
    //         "out[state] & (1 << i)" is > 0 if we just
    //         found
    // word[i]          in the text.
    // Returns the number of states that the built machine
    // has. States are numbered 0 up to the return value -
    // 1, inclusive.
    static int buildMatchingMachine(String arr[], int k)
    {

        // Initialize all values in output function as 0.
        Arrays.fill(out, 0);

        // Initialize all values in goto function as -1.
        for (int i = 0; i < MAXS; i++)
            Arrays.fill(g[i], -1);

        // Initially, we just have the 0 state
        int states = 1;

        // Convalues for goto function, i.e., fill g[][]
        // This is same as building a Trie for arr[]
        for (int i = 0; i < k; ++i) {
            String word = arr[i];
            int currentState = 0;

            // Insert all characters of current
            // word in arr[]
            for (int j = 0; j < word.length(); ++j) {
                int ch = word.charAt(j) - 'a';

                // Allocate a new node (create a new state)
                // if a node for ch doesn't exist.
                if (g[currentState][ch] == -1)
                    g[currentState][ch] = states++;

                currentState = g[currentState][ch];
            }

            // Add current word in output function
            out[currentState] |= (1 << i);
        }

        // For all characters which don't have
        // an edge from root (or state 0) in Trie,
        // add a goto edge to state 0 itself
        for (int ch = 0; ch < MAXC; ++ch)
            if (g[0][ch] == -1)
                g[0][ch] = 0;

        // Now, let's build the failure function
        // Initialize values in fail function
        Arrays.fill(f, -1);

        // Failure function is computed in
        // breadth first order
        // using a queue
        Queue<Integer> q = new LinkedList<>();

        // Iterate over every possible input
        for (int ch = 0; ch < MAXC; ++ch) {

            // All nodes of depth 1 have failure
            // function value as 0. For example,
            // in above diagram we move to 0
            // from states 1 and 3.
            if (g[0][ch] != 0) {
                f[g[0][ch]] = 0;
                q.add(g[0][ch]);
            }
        }

        // Now queue has states 1 and 3
        while (!q.isEmpty()) {

            // Remove the front state from queue
            int state = q.peek();
            q.remove();

            // For the removed state, find failure
            // function for all those characters
            // for which goto function is
            // not defined.
            for (int ch = 0; ch < MAXC; ++ch) {

                // If goto function is defined for
                // character 'ch' and 'state'
                if (g[state][ch] != -1) {

                    // Find failure state of removed state
                    int failure = f[state];

                    // Find the deepest node labeled by
                    // proper suffix of String from root to
                    // current state.
                    while (g[failure][ch] == -1)
                        failure = f[failure];

                    failure = g[failure][ch];
                    f[g[state][ch]] = failure;

                    // Merge output values
                    out[g[state][ch]] |= out[failure];

                    // Insert the next level node
                    // (of Trie) in Queue
                    q.add(g[state][ch]);
                }
            }
        }
        return states;
    }

    // Returns the next state the machine will transition to
    // using goto and failure functions. currentState - The
    // current state of the machine. Must be between
    // 0 and the number of states - 1,
    // inclusive.
    // nextInput - The next character that enters into the
    // machine.
    static int findNextState(int currentState,
                             char nextInput)
    {
        int answer = currentState;
        int ch = nextInput - 'a';

        // If goto is not defined, use
        // failure function
        while (g[answer][ch] == -1)
            answer = f[answer];

        return g[answer][ch];
    }

    // This function finds all occurrences of
    // all array words in text.
    static void searchWords(String arr[], int k,
                            String text)
    {

        // Preprocess patterns.
        // Build machine with goto, failure
        // and output functions
        buildMatchingMachine(arr, k);

        // Initialize current state
        int currentState = 0;

        // Traverse the text through the
        // built machine to find all
        // occurrences of words in arr[]
        for (int i = 0; i < text.length(); ++i) {
            currentState = findNextState(currentState,
                                         text.charAt(i));

            // If match not found, move to next state
            if (out[currentState] == 0)
                continue;

            // Match found, print all matching
            // words of arr[]
            // using output function.
            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");
                }
            }
        }
    }

    // Driver code
    public static void main(String[] args)
    {
        String arr[] = { "he", "she", "hers", "his" };
        String text = "ahishers";
        int k = arr.length;

        searchWords(arr, k, text);
    }
}
// This code is wriiten by Sundaram.

Python

# Python program for implementation of
# Aho-Corasick algorithm for string matching

# defaultdict is used only for storing the final output
# We will return a dictionary where key is the matched word
# and value is the list of indexes of matched word
from collections import defaultdict

# For simplicity, Arrays and Queues have been implemented using lists.
# If you want to improve performance try using them instead


class AhoCorasick:
    def __init__(self, words):

        # Max number of states in the matching machine.
        # Should be equal to the sum of the length of all keywords.
        self.max_states = sum([len(word) for word in words])

        # Maximum number of characters.
        # Currently supports only alphabets [a, z]
        self.max_characters = 26

        # OUTPUT FUNCTION IS IMPLEMENTED USING out []
        # Bit i in this mask is 1 if the word with
        # index i appears when the machine enters this state.
        # Lets say, a state outputs two words "he" and "she" and
        # in our provided words list, he has index 0 and she has index 3
        # so value of out[state] for this state will be 1001
        # It has been initialized to all 0.
        # We have taken one extra state for the root.
        self.out = [0]*(self.max_states + 1)

        # FAILURE FUNCTION IS IMPLEMENTED USING fail []
        # There is one value for each state + 1 for the root
        # It has been initialized to all -1
        # This will contain the fail state value for each state
        self.fail = [-1]*(self.max_states + 1)

        # GOTO FUNCTION (OR TRIE) IS IMPLEMENTED USING goto [[]]
        # Number of rows = max_states + 1
        # Number of columns = max_characters i.e 26 in our case
        # It has been initialized to all -1.
        self.goto = [
            [-1]*self.max_characters for _ in range(self.max_states + 1)]

        # Convert all words to lowercase
        # so that our search is case insensitive
        for i in range(len(words)):
            words[i] = words[i].lower()

        # All the words in dictionary which will be used to create Trie
        # The index of each keyword is important:
        # "out[state] & (1 << i)" is > 0 if we just found word[i]
        # in the text.
        self.words = words

        # Once the Trie has been built, it will contain the number
        # of nodes in Trie which is total number of states required <= max_states
        self.states_count = self.__build_matching_machine()

    # Builds the String matching machine.
    # Returns the number of states that the built machine has.
    # States are numbered 0 up to the return value - 1, inclusive.

    def __build_matching_machine(self):
        k = len(self.words)

        # Initially, we just have the 0 state
        states = 1

        # Convalues for goto function, i.e., fill goto
        # This is same as building a Trie for words[]
        for i in range(k):
            word = self.words[i]
            current_state = 0

            # Process all the characters of the current word
            for character in word:
                ch = ord(character) - 97  # Ascii value of 'a' = 97

                # Allocate a new node (create a new state)
                # if a node for ch doesn't exist.
                if self.goto[current_state][ch] == -1:
                    self.goto[current_state][ch] = states
                    states += 1

                current_state = self.goto[current_state][ch]

            # Add current word in output function
            self.out[current_state] |= (1 << i)

        # For all characters which don't have
        # an edge from root (or state 0) in Trie,
        # add a goto edge to state 0 itself
        for ch in range(self.max_characters):
            if self.goto[0][ch] == -1:
                self.goto[0][ch] = 0

        # Failure function is computed in
        # breadth first order using a queue
        queue = []

        # Iterate over every possible input
        for ch in range(self.max_characters):

            # All nodes of depth 1 have failure
            # function value as 0. For example,
            # in above diagram we move to 0
            # from states 1 and 3.
            if self.goto[0][ch] != 0:
                self.fail[self.goto[0][ch]] = 0
                queue.append(self.goto[0][ch])

        # Now queue has states 1 and 3
        while queue:

            # Remove the front state from queue
            state = queue.pop(0)

            # For the removed state, find failure
            # function for all those characters
            # for which goto function is not defined.
            for ch in range(self.max_characters):

                # If goto function is defined for
                # character 'ch' and 'state'
                if self.goto[state][ch] != -1:

                    # Find failure state of removed state
                    failure = self.fail[state]

                    # Find the deepest node labeled by proper
                    # suffix of String from root to current state.
                    while self.goto[failure][ch] == -1:
                        failure = self.fail[failure]

                    failure = self.goto[failure][ch]
                    self.fail[self.goto[state][ch]] = failure

                    # Merge output values
                    self.out[self.goto[state][ch]] |= self.out[failure]

                    # Insert the next level node (of Trie) in Queue
                    queue.append(self.goto[state][ch])

        return states

    # Returns the next state the machine will transition to using goto
    # and failure functions.
    # current_state - The current state of the machine. Must be between
    # 0 and the number of states - 1, inclusive.
    # next_input - The next character that enters into the machine.

    def __find_next_state(self, current_state, next_input):
        answer = current_state
        ch = ord(next_input) - 97  # Ascii value of 'a' is 97

        # If goto is not defined, use
        # failure function
        while self.goto[answer][ch] == -1:
            answer = self.fail[answer]

        return self.goto[answer][ch]

    # This function finds all occurrences of all words in text.

    def search_words(self, text):
        # Convert the text to lowercase to make search case insensitive
        text = text.lower()

        # Initialize current_state to 0
        current_state = 0

        # A dictionary to store the result.
        # Key here is the found word
        # Value is a list of all occurrences start index
        result = defaultdict(list)

        # Traverse the text through the built machine
        # to find all occurrences of words
        for i in range(len(text)):
            current_state = self.__find_next_state(current_state, text[i])

            # If match not found, move to next state
            if self.out[current_state] == 0:
                continue

            # Match found, store the word in result dictionary
            for j in range(len(self.words)):
                if (self.out[current_state] & (1 << j)) > 0:
                    word = self.words[j]

                    # Start index of word is (i-len(word)+1)
                    result[word].append(i-len(word)+1)

        # Return the final result dictionary
        return result


# Driver code
if __name__ == "__main__":
    words = ["he", "she", "hers", "his"]
    text = "ahishers"

    # Create an Object to initialize the Trie
    aho_chorasick = AhoCorasick(words)

    # Get the result
    result = aho_chorasick.search_words(text)

    # Print the result
    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}`);
      }
    }
  }
}

// Driver code
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.