Rabin-Karp Algorithm for Pattern Searching
Last Updated :
06 Sep, 2023
Given a text T[0. . .n-1] and a pattern P[0. . .m-1], write a function search(char P[], char T[]) that prints all occurrences of P[] present in T[] using Rabin Karp algorithm. You may assume that n > m.
Examples:
Input: T[] = “THIS IS A TEST TEXT”, P[] = “TEST”
Output: Pattern found at index 10
Input: T[] = “AABAACAADAABAABA”, P[] = “AABA”
Output: Pattern found at index 0
Pattern found at index 9
Pattern found at index 12
Rabin-Karp Algorithm:
In the Naive String Matching algorithm, we check whether every substring of the text of the pattern’s size is equal to the pattern or not one by one.
Like the Naive Algorithm, the Rabin-Karp algorithm also check every substring. But unlike the Naive algorithm, the Rabin Karp algorithm matches the hash value of the pattern with the hash value of the current substring of text, and if the hash values match then only it starts matching individual characters. So Rabin Karp algorithm needs to calculate hash values for the following strings.
- Pattern itself
- All the substrings of the text of length m which is the size of pattern.
How is Hash Value calculated in Rabin-Karp?
Hash value is used to efficiently check for potential matches between a pattern and substrings of a larger text. The hash value is calculated using a rolling hash function, which allows you to update the hash value for a new substring by efficiently removing the contribution of the old character and adding the contribution of the new character. This makes it possible to slide the pattern over the text and calculate the hash value for each substring without recalculating the entire hash from scratch.
Here’s how the hash value is typically calculated in Rabin-Karp:
Step 1: Choose a suitable base and a modulus:
- Select a prime number ‘p‘ as the modulus. This choice helps avoid overflow issues and ensures a good distribution of hash values.
- Choose a base ‘b‘ (usually a prime number as well), which is often the size of the character set (e.g., 256 for ASCII characters).
Step 2: Initialize the hash value:
- Set an initial hash value ‘hash‘ to 0.
Step 3: Calculate the initial hash value for the pattern:
- Iterate over each character in the pattern from left to right.
- For each character ‘c’ at position ‘i’, calculate its contribution to the hash value as ‘c * (bpattern_length – i – 1) % p’ and add it to ‘hash‘.
- This gives you the hash value for the entire pattern.
Step 4: Slide the pattern over the text:
- Start by calculating the hash value for the first substring of the text that is the same length as the pattern.
Step 5: Update the hash value for each subsequent substring:
- To slide the pattern one position to the right, you remove the contribution of the leftmost character and add the contribution of the new character on the right.
- The formula for updating the hash value when moving from position ‘i’ to ‘i+1’ is:
hash = (hash - (text[i - pattern_length] * (bpattern_length - 1)) % p) * b + text[i]
Step 6: Compare hash values:
- When the hash value of a substring in the text matches the hash value of the pattern, it’s a potential match.
- If the hash values match, we should perform a character-by-character comparison to confirm the match, as hash collisions can occur.
Below is the Illustration of above algorithm:
Step-by-step approach:
- Initially calculate the hash value of the pattern.
- Start iterating from the starting 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.
Below is the implementation of the above approach:
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;
}
|
C
#include <stdio.h>
#include <string.h>
#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)
printf ( "Pattern found at index %d \n" , i);
}
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 = 101;
search(pat, txt, q);
return 0;
}
|
Java
public class Main {
public final static int d = 256 ;
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)
i = 0
j = 0
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
else :
j + = 1
if j = = M:
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
if __name__ = = '__main__' :
txt = "GEEKS FOR GEEKS"
pat = "GEEK"
q = 101
search(pat, txt, q)
|
C#
using System;
public class GFG {
public readonly static int d = 256;
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 txt = "GEEKS FOR GEEKS" ;
String pat = "GEEK" ;
int q = 101;
search(pat, txt, q);
}
}
|
Javascript
<script>
let d = 256;
function search(pat, txt, q)
{
let M = pat.length;
let N = txt.length;
let i, j;
let p = 0;
let t = 0;
let h = 1;
for (i = 0; i < M - 1; i++)
h = (h * d) % q;
for (i = 0; i < M; i++)
{
p = (d * p + pat[i].charCodeAt()) % q;
t = (d * t + txt[i].charCodeAt()) % 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)
document.write( "Pattern found at index " +
i + "<br/>" );
}
if (i < N - M)
{
t = (d * (t - txt[i].charCodeAt() * h) +
txt[i + M].charCodeAt()) % q;
if (t < 0)
t = (t + q);
}
}
}
let txt = "GEEKS FOR GEEKS" ;
let pat = "GEEK" ;
let q = 101;
search(pat, txt, q);
</script>
|
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 T[] match with the hash value of P[].
Auxiliary Space: O(1)
Limitations of Rabin-Karp Algorithm
Spurious Hit: When the hash value of the pattern matches with the hash value of a window of the text but the window is not the actual pattern then it is called a spurious hit. Spurious hit increases the time complexity of the algorithm. In order to minimize spurious hit, we use good hash function. It greatly reduces the spurious hit.
Related Posts:
Searching for Patterns | Set 1 (Naive Pattern Searching)
Searching for Patterns | Set 2 (KMP Algorithm)
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