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/* 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 on 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"; | |
int q = 101; // A prime number | |
search(pat, txt, q); | |
return 0; | |
} | |
// This is code is contributed by rathbhupendra |
Sorry, I couldn't make it to GFG
#include <bits/stdc++.h>
using namespace std;
int main()
{
string txt = "AABAACBAA", pat = "BAA";
int hpat=0,htxt=0;
int d = 26;
int p = 5381;
for (int i = 0; i < pat.size(); i++)
{
hpat *= d;
hpat = hpat + (((pat[i] - 'A' + 1)) % p);
}
int l=0, r = 0;
while (r < txt.size()){
htxt *= d;
htxt = htxt + ((txt[r] - 'A' + 1) % p);
if(r-l+1==pat.size()){
if(htxt==hpat)
cout << "Match at " << l;
htxt = htxt - (((txt[l] - 'A' + 1) * pow(d, r - l)));
l++;
}
r++;
}
return 0;
}
Can you explain this : for (int i = 0; i < pat.size(); i++)
{
hpat *= d;
hpat = hpat + (((pat[i] - 'A' + 1)) % p);
}
GFG solution
🤣🤣🤣🤣
import java.nio.file.Path;
import java.util.*;
public class Solution
{
private static final int d = 26; // Base value of alphabets
private static final int p = 5381; // Large prime number
public static void search(String pat, String txt)
{
int patHash = 0; // Hash value of pattern
int txtHash = 0; // Hash value of text
for (int i = 0; i < pat.length(); i++) // Generating Hash values for pattern and first window text
{
patHash = patHash * d;
txtHash = txtHash * d;
patHash = patHash + ((pat.charAt(i) - 'A' + 1) % p);
txtHash = txtHash + ((txt.charAt(i) - 'A' + 1) % p);
}
for (int i = 0; i < txt.length() - pat.length(); i++) // Loop of text size minus window
{
if (patHash == txtHash)
{
System.out.println("Pattern found at index " + i);
}
if (i < txt.length() - pat.length())
{
txtHash = txtHash - ((txt.charAt(i) - 'A' + 1) * (int)Math.pow(d, pat.length() - 1)); // Subtracting first element from current hash of d^window-1
txtHash = txtHash * d + (txt.charAt(i + pat.length()) - 'A' + 1); // multiplying obtained hash with d to left shift the number and then adding the next new element
}
}
}
public static void main(String []args)
{
String txt = "GEEKS FOR GEEKS";
String pat = "GEEK";
search(pat, txt);
}
}
import java.nio.file.Path; import java.util.*; public class Solution { private static final int d = 26; // Base value of alphabets private static final int p = 5381; // Large prime number public static void search(String pat, String txt) { int patHash = 0; // Hash value of pattern int txtHash = 0; // Hash value of text for (int i = 0; i < pat.length(); i++) // Generating Hash values for pattern and first window text { patHash = patHash * d; txtHash = txtHash * d; patHash = patHash + ((pat.charAt(i) - 'A' + 1) % p); txtHash = txtHash + ((txt.charAt(i) - 'A' + 1) % p); } for (int i = 0; i < txt.length() - pat.length(); i++) // Loop of text size minus window { if (patHash == txtHash) { System.out.println("Pattern found at index " + i); } if (i < txt.length() - pat.length()) { txtHash = txtHash - ((txt.charAt(i) - 'A' + 1) * (int)Math.pow(d, pat.length() - 1)); // Subtracting first element from current hash of d^window-1 txtHash = txtHash * d + (txt.charAt(i + pat.length()) - 'A' + 1); // multiplying obtained hash with d to left shift the number and then adding the next new element } } } public static void main(String []args) { String txt = "GEEKS FOR GEEKS"; String pat = "GEEK"; search(pat, txt); } }
for (int i = 0; i < txt.length() - pat.length(); i++)
I think this loop should run till ( i<= txt.length() - pat.length() ) as your loop is not able to match the pattern present at the end of the string
correct me if I'm wrong
https://leetcode.com/problems/longest-duplicate-substring/
apply this and solve for practice
#include <bits/stdc++.h>
using namespace std;
int main() { string txt = "AABAACBAA", pat = "BAA"; int hpat=0,htxt=0; int d = 26; int p = 5381; for (int i = 0; i < pat.size(); i++) { hpat *= d; hpat = hpat + (((pat[i] - 'A' + 1)) % p); } int l=0, r = 0; while (r < txt.size()){ htxt *= d; htxt = htxt + ((txt[r] - 'A' + 1) % p); if(r-l+1==pat.size()){ if(htxt==hpat) cout << "Match at " << l; htxt = htxt - (((txt[l] - 'A' + 1) * pow(d, r - l))); l++; } r++; } return 0; }
this will give runtime error: signed integer overflow: 408244051 * 26 cannot be represented in type 'int' .
import java.nio.file.Path; import java.util.*; public class Solution { private static final int d = 26; // Base value of alphabets private static final int p = 5381; // Large prime number public static void search(String pat, String txt) { int patHash = 0; // Hash value of pattern int txtHash = 0; // Hash value of text for (int i = 0; i < pat.length(); i++) // Generating Hash values for pattern and first window text { patHash = patHash * d; txtHash = txtHash * d; patHash = patHash + ((pat.charAt(i) - 'A' + 1) % p); txtHash = txtHash + ((txt.charAt(i) - 'A' + 1) % p); } for (int i = 0; i < txt.length() - pat.length(); i++) // Loop of text size minus window { if (patHash == txtHash) { System.out.println("Pattern found at index " + i); } if (i < txt.length() - pat.length()) { txtHash = txtHash - ((txt.charAt(i) - 'A' + 1) * (int)Math.pow(d, pat.length() - 1)); // Subtracting first element from current hash of d^window-1 txtHash = txtHash * d + (txt.charAt(i + pat.length()) - 'A' + 1); // multiplying obtained hash with d to left shift the number and then adding the next new element } } } public static void main(String []args) { String txt = "GEEKS FOR GEEKS"; String pat = "GEEK"; search(pat, txt); } }
You should change the loop condition to i <= txt.length() - pat.length() instead of i < txt.length() - pat.length(); the( <= )
package gfg;
import java.util.ArrayList;
import java.util.List;
public class RabinKarpRollHash
{
private static final int d = 26; // Base value of alphabets
private static final int p = 5381; // Large prime number
private List<Integer> rollHash(String pattern, String text)
{
List<Integer> indices = new ArrayList<Integer>();
int patHash = 0;
int txtHash = 0;
for (int i = 0,pow = pattern.length() - 1; i < pattern.length(); i++) // Generating Hash values for pattern and first window text
{
patHash += (pattern.charAt(i) - 'A' + 1) * (int)Math.pow(d, pow);
patHash %= p;
txtHash += (text.charAt(i) - 'A' + 1) * (int)Math.pow(d, pow--);
txtHash %= p;
}
for (int i = 0; i <= text.length() - pattern.length(); i++)
{
//declaration
String textToCompare = text.substring(i, i + pattern.length());
int pow = pattern.length() - 1;
//condition for matched pattern
if (patHash == txtHash && match(textToCompare, pattern))
{
indices.add(i + 1);
}
//calculate hash till last pattern string
if(i < text.length() - pattern.length())
{
//subtract
txtHash = txtHash - (((text.charAt(i) - 'A' + 1) * (int)Math.pow(d, pow)) % p);
//add
txtHash = ((txtHash * d) + (text.charAt(i + pattern.length()) - 'A' + 1)) % p;
}
//negative hash value
if (txtHash < 0)
{
txtHash += p;
}
}
return indices;
}
private boolean match(String text, String pattern)
{
boolean flag = true;
for (int i = 0; i < pattern.length(); i++)
{
if(text.charAt(i) != pattern.charAt(i))
{
flag = false;
break;
}
}
return flag;
}
public static void main(String[] args)
{
RabinKarpRollHash soln = new RabinKarpRollHash();
List<Integer> ans = soln.rollHash("ecda", "ecdadecdacde");
System.out.println(ans);
}
}
This is the code that i wrote to my understanding level, any corrections are welcomed
package gfg;
import java.util.ArrayList;
public class RabinKarpRollHash
{
private ArrayList<Integer> rollHash(String pattern, String text)
{
ArrayList<Integer> indices = new ArrayList<Integer>();
int patHash = 0;
int txtHash = 0;
for (char c : pattern.toCharArray())
{
patHash += c;
}
for (int i = 0; i < pattern.length(); i++)
{
txtHash += text.charAt(i);
}
for (int i = 0; i <= text.length() - pattern.length(); i++)
{
//declaration
String textToCompare = text.substring(i, i + pattern.length());
//condition for matched pattern
if (patHash == txtHash && textToCompare.equals(pattern))
{
indices.add(i + 1);
}
//calculate hash till last pattern string
if(i < text.length() - pattern.length())
{
txtHash = txtHash - text.charAt(i) + text.charAt(i + pattern.length());
}
}
return indices;
}
public static void main(String[] args)
{
RabinKarpRollHash soln = new RabinKarpRollHash();
ArrayList<Integer> ans = soln.rollHash("egibbbdbgdiedabcbidgdiicahagehfihbiedaebchicaaiebhibhcaihbahfedaadacegggcchihgbfgccficffdgaeiebd", "egibbbdbgdiedabcbidgdiicahagehfihbiedaebchicaaiebhibhcaihbahfedaadacegggcchihgbfgccficffdgaeiebdibegibbbdbgdiedabcbidgdiicahagehfihbiedaebchicaaiebhibhcaihbahfedaadacegggcchihgbfgccficffdgaeiebdcfegibbbdbgdiedabcbidgdiicahagehfihbiedaebchicaaiebhibhcaihbahfedaadacegggcchihgbfgccficffdgaeiebdiiedegibbbdbgdiedabcbidgdiicahagehfihbiedaebchicaaiebhibhcaihbahfedaadacegggcchihgbfgccficffdgaeiebdeghegibbbdbgdiedabcbidgdiicahagehfihbiedaebchicaaiebhibhcaihbahfedaadacegggcchihgbfgccficffdgaeiebddgddfegibbbdbgdiedabcbidgdiicahagehfihbiedaebchicaaiebhibhcaihbahfedaadacegggcchihgbfgccficffdgaeiebdaedggagegibbbdbgdiedabcbidgdiicahagehfihbiedaebchicaaiebhibhcaihbahfedaadacegggcchihgbfgccficffdgaeiebdadeaaheabidegibbbdbgdiedabcbidgdiicahagehfihbiedaebchicaaiebhibhcaihbahfedaadacegggcchihgbfgccficffdgaeiebdbbegibbbdbgdiedabcbidgdiicahagehfihbiedaebchicaaiebhibhcaihbahfedaadacegggcchihgbfgccficffdgaeiebdhaegibbbdbgdiedabcbidgdiicahagehfihbiedaebchicaaiebhibhcaihbahfedaadacegggcchihgbfgccficffdgaeiebddegibbbdbgdiedabcbidgdiicahagehfihbiedaebchicaaiebhibhcaihbahfedaadacegggcchihgbfgccficffdgaeiebdeaaeaegibbbdbgdiedabcbidgdiicahagehfihbiedaebchicaaiebhibhcaihbahfedaadacegggcchihgbfgccficffdgaeiebdhegibbbdbgdiedabcbidgdiicahagehfihbiedaebchicaaiebhibhcaihbahfedaadacegggcchihgbfgccficffdgaeiebdiacegibbbdbgdiedabcbidgdiicahagehfihbiedaebchicaaiebhibhcaihbahfedaadacegggcchihgbfgccficffdgaeiebdhegibbbdbgdiedabcbidgdiicahagehfihbiedaebchicaaiebhibhcaihbahfedaadacegggcchihgbfgccficffdgaeiebddegibbbdbgdiedabcbidgdiicahagehfihbiedaebchicaaiebhibhcaihbahfedaadacegggcchihgbfgccficffdgaeiebdcegibbbdbgdiedabcbidgdiicahagehfihbiedaebchicaaiebhibhcaihbahfedaadacegggcchihgbfgccficffdgaeiebdaihhccegibbbdbgdiedabcbidgdiicahagehfihbiedaebchicaaiebhibhcaihbahfedaadacegggcchihgbfgccficffdgaeiebdfegibbbdbgdiedabcbidgdiicahagehfihbiedaebchicaaiebhibhcaihbahfedaadacegggcchihgbfgccficffdgaeiebdiegibbbdbgdiedabcbidgdiicahagehfihbiedaebchicaaiebhibhcaihbahfedaadacegggcchihgbfgccficffdgaeiebdaegibbbdbgdiedabcbidgdiicahagehfihbiedaebchicaaiebhibhcaihbahfedaadacegggcchihgbfgccficffdgaeiebdccaegibbbdbgdiedabcbidgdiicahagehfihbiedaebchicaaiebhibhcaihbahfedaadacegggcchihgbfgccficffdgaeiebdbegibbbdbgdiedabcbidgdiicahagehfihbiedaebchicaaiebhibhcaihbahfedaadacegggcchihgbfgccficffdgaeiebdhicgggiegibbbdbgdiedabcbidgdiicahagehfihbiedaebchicaaiebhibhcaihbahfedaadacegggcchihgbfgccficffdgaeiebdiagegibbbdbgdiedabcbidgdiicahagehfihbiedaebchicaaiebhibhcaihbahfedaadacegggcchihgbfgccficffdgaeiebdabfgdfhbhegibbbdbgdiedabcbidgdiicahagehfihbiedaebchicaaiebhibhcaihbahfedaadacegggcchihgbfgccficffdgaeiebdedegibbbdbgdiedabcbidgdiicahagehfihbiedaebchicaaiebhibhcaihbahfedaadacegggcchihgbfgccficffdgaeiebdheaegibbbdbgdiedabcbidgdiicahagehfihbiedaebchicaaiebhibhcaihbahfedaadacegggcchihgbfgccficffdgaeiebdgegibbbdbgdiedabcbidgdiicahagehfihbiedaebchicaaiebhibhcaihbahfedaadacegggcchihgbfgccficffdgaeiebdgdcegibbbdbgdiedabcbidgdiicahagehfihbiedaebchicaaiebhibhcaihbahfedaadacegggcchihgbfgccficffdgaeiebdaadb");
System.out.print(ans);
}
}
I tried this without hash
import java.util.ArrayList;
import java.util.Scanner;
public class rabinKarp {
public static void main(String args[]) {
Scanner sc = new Scanner(System.in);
System.out.print("Enter the string1:- ");
String str1 = sc.nextLine();
System.out.print("Enter the string2:- ");
String str2 = sc.nextLine();
sc.close();
if (str2.length() > str1.length()) {
return;
}
rabin(str1, str2);
}
private static final int BASE = 256;
private static final long PRIME = 10000007;
private static void rabin(String txt, String pat) {
// Precomputing BASE^(M-1) When Deleting the First Character It is useful
long h = 1;
for (int i = 0; i < pat.length() - 1; i++) {
h = (h * BASE) % PRIME;
}
ArrayList<Integer> indexes = new ArrayList<>();
long patHash = calculateHash(pat);
long windowHash = calculateHash(txt.substring(0, pat.length()));
for (int i = 0; i <= txt.length() - pat.length(); i++) {
if (windowHash == patHash) {
indexes.add(i);
}
System.out.println(patHash + " " + windowHash);
int ascciOld = txt.charAt(i);
// Remove first character from hash
// ( windowHash - (1 * h) % prime + prime ) prime
windowHash = (windowHash - (ascciOld * h) % PRIME + PRIME) % PRIME;
if (i + pat.length() < txt.length()) {
int ascciNew = txt.charAt(i + pat.length());
// (windowHash * Base + assciNew) % prime
windowHash = (windowHash * BASE + ascciNew) % PRIME;
}
}
System.out.println(indexes);
}
private static long calculateHash(String str) {
long hash = 0;
for (int i = 0; i < str.length(); i++) {
hash = (hash * BASE + str.charAt(i)) % PRIME;
}
return hash;
}
}
GFG solution