Count distinct substrings of a string using Rabin Karp algorithm
Last Updated :
04 Oct, 2023
Given a string, return the number of distinct substrings using Rabin Karp Algorithm.
Examples:
Input : str = “aba”
Output : 5
Explanation :
Total number of distinct substring are 5 - "a", "ab", "aba", "b" ,"ba"
Input : str = “abcd”
Output : 10
Explanation :
Total number of distinct substring are 10 - "a", "ab", "abc", "abcd", "b", "bc", "bcd", "c", "cd", "d"
Approach:
Prerequisite: Rabin-Karp Algorithm for Pattern Searching
Calculate the current hash value of the current character and store
in a dictionary/map to avoid repetition.
To compute the hash (rolling hash) as done in Rabin-Karp algorithm follow:
The hash function suggested by Rabin and Karp calculates an integer value. The integer value for a string is the numeric value of a string. For example, if all possible characters are from 1 to 10, the numeric value of “122” will be 122. The number of possible characters is higher than 10 (256 in general) and the pattern length can be large. So the numeric values cannot be practically stored as an integer. Therefore, the numeric value is calculated using modular arithmetic to make sure that the hash values can be stored in an integer variable (can fit in memory words). To do rehashing, we need to take off the most significant digit and add the new least significant digit in the hash value. Rehashing is done using the following formula.
hash( txt[s+1 .. s+m] ) = ( d ( hash( txt[s .. s+m-1]) – txt[s]*h ) + txt[s + m] ) mod q
hash( txt[s .. s+m-1] ) : Hash value at shift s.
hash( txt[s+1 .. s+m] ): Hash value at next shift (or shift s+1)
d: Number of characters in the alphabet
q: A prime number
h: d^(m-1)
The idea is similar as we evaluate a mathematical expression. For example, we have a string of “1234” let us compute the value of the substring “12” as 12 and we want to compute the value of the substring “123” this can be calculated as ((12)*10+3)=123, similar logic is applied here.
C++
#include <bits/stdc++.h>
using namespace std;
int main()
{
int t = 1;
long long mod = 9007199254740881;
for(int i = 0; i < t; i++)
{
string s = "abcd";
vector<vector<long long>>l;
unordered_map<long long,int>d;
for(int i=0;i<s.length();i++){
int suma = 0;
long long pre = 0;
long long D = 256;
for(int j=i;j<s.length();j++){
pre = (pre*D+s[j]) % mod;
if(d.find(pre) == d.end())
l.push_back({i, j});
d[pre] = 1;
}
}
cout<<l.size()<<endl;
for(int i = 0; i < l.size(); i++)
cout << s.substr(l[i][0],l[i][1]+1-l[i][0]) << " ";
}
}
|
Java
import java.util.*;
public class Main {
public static void main(String[] args) {
int t = 1;
long mod = 9007199254740881L;
for (int i = 0; i < t; i++) {
String s = "abcd";
List<List<Integer>> l = new ArrayList<>();
Map<Long, Integer> d = new HashMap<>();
for (int j = 0; j < s.length(); j++) {
long suma = 0;
long pre = 0;
int D = 256;
for (int k = j; k < s.length(); k++) {
pre = (pre*D + (long)s.charAt(k)) % mod;
if (!d.containsKey(pre)) {
List<Integer> sublist = new ArrayList<>();
sublist.add(j);
sublist.add(k);
l.add(sublist);
}
d.put(pre, 1);
}
}
System.out.println(l.size());
for (int j = 0; j < l.size(); j++) {
int start = l.get(j).get(0);
int end = l.get(j).get(1);
System.out.print(s.substring(start, end+1) + " ");
}
}
}
}
|
Python3
import sys
import math as mt
t = 1
mod = 9007199254740881
for ___ in range(t):
s = 'abcd'
l = []
d = {}
for i in range(len(s)):
suma = 0
pre = 0
D = 256
for j in range(i, len(s)):
pre = (pre*D+ord(s[j])) % mod
if d.get(pre, -1) == -1:
l.append([i, j])
d[pre] = 1
print(len(l))
for i in range(len(l)):
print(s[l[i][0]:l[i][1]+1], end=" ")
|
C#
using System;
using System.Collections.Generic;
class GFG {
static void Main()
{
int t = 1;
long mod = 9007199254740881;
for (int i = 0; i < t; i++)
{
string s = "abcd";
List<List<long>> l = new List<List<long>>();
Dictionary<long, int> d = new Dictionary<long, int>();
for (int j = 0; j < s.Length; j++)
{
int suma = 0;
long pre = 0;
long D = 256;
for (int k = j; k < s.Length; k++)
{
pre = (pre * D + s[k]) % mod;
if (!d.ContainsKey(pre))
{
List<long> sub = new List<long>();
sub.Add(j);
sub.Add(k);
l.Add(sub);
}
d[pre] = 1;
}
}
Console.WriteLine(l.Count);
for (int j = 0; j < l.Count; j++)
{
Console.Write(s.Substring((int)l[j][0], (int)l[j][1] + 1 - (int)l[j][0]) + " ");
}
}
}
}
|
Javascript
<script>
let t = 1
let mod = 9007199254740881
for(let i = 0; i < t; i++){
let s = 'abcd'
let l = []
let d = new Map()
for(let i=0;i<s.length;i++){
let suma = 0
let pre = 0
let D = 256
for(let j=i;j<s.length;j++){
pre = (pre*D+s.charCodeAt(j)) % mod
if(d.has([pre, -1]) == false)
l.push([i, j])
d.set(pre , 1)
}
}
document.write(l.length,"</br>")
for(let i = 0; i < l.length; i++)
document.write(s.substring(l[i][0],l[i][1]+1)," ")
}
</script>
|
Output
10
a ab abc abcd b bc bcd c cd d
Time Complexity: O(N2), N is the length of the string
Auxiliary Space: O(N*2) => O(N)
Another Approach:
- Define an input string “s”.
- Create an empty unordered set “substrings” to store the distinct substrings.
- Use two nested loops to iterate over all possible substrings of “s”. The outer loop iterates over the starting index of each substring.
The inner loop iterates over the ending index of each substring, starting from the starting index.
- Use the “substr” method of the input string “s” to extract each substring and insert it into the “substrings” set.
The “substr” method takes two arguments: the starting index of the substring and the length of the substring.
The length of the substring is computed as “j – i + 1”.
- After all substrings have been added to the set, output the size of the set to get the number of distinct substrings.
Below is the implementation of the above approach:
C++
#include <iostream>
#include <unordered_set>
#include <string>
using namespace std;
int main() {
string s = "abcd";
unordered_set<string> substrings;
for (int i = 0; i < s.size(); i++) {
for (int j = i; j < s.size(); j++) {
substrings.insert(s.substr(i, j - i + 1));
}
}
cout << substrings.size() << endl;
}
|
Java
import java.util.*;
public class GFG {
public static void main(String[] args) {
String s = "abcd";
Set<String> substrings = new HashSet<>();
for (int i = 0; i < s.length(); i++) {
for (int j = i; j < s.length(); j++) {
substrings.add(s.substring(i, j + 1));
}
}
System.out.println(substrings.size());
}
}
|
Python
def main():
s = "abcd"
substrings = set()
for i in range(len(s)):
for j in range(i, len(s)):
substrings.add(s[i:j + 1])
print(len(substrings))
if __name__ == "__main__":
main()
|
C#
using System;
using System.Collections.Generic;
class GFG
{
static void Main(string[] args)
{
string s = "abcd";
HashSet<string> substrings = new HashSet<string>();
for (int i = 0; i < s.Length; i++){
for (int j = i; j < s.Length; j++){
substrings.Add(s.Substring(i, j - i + 1));
}
}
Console.WriteLine(substrings.Count);
}
}
|
Javascript
<script>
function countDistinctSubstrings(s) {
const substrings = new Set();
for (let i = 0; i < s.length; i++) {
for (let j = i; j < s.length; j++) {
substrings.add(s.substring(i, j + 1));
}
}
return substrings.size;
}
const s = "abcd";
const distinctSubstringsCount = countDistinctSubstrings(s);
document.write(distinctSubstringsCount);
</script>
|
Time complexity: O(n^3), where n is the length of the input string “s”.
Auxiliary Space: O(n^2), where n is the length of the input string “s”. The space complexity is dominated by the number of distinct substrings that are stored in the unordered_set.
Please Login to comment...