Count distinct substrings of a string using Rabin Karp algorithm

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;

// Driver code
int main()
{
  int t = 1;

  // store prime to reduce overflow
  long long mod = 9007199254740881;

  for(int i = 0; i < t; i++)
  {

    // string to check number of distinct substring
    string s = "abcd";

    // to store substrings
    vector<vector<long long>>l;

    // to store hash values by Rabin Karp algorithm
    unordered_map<long long,int>d;

    for(int i=0;i<s.length();i++){
      int suma = 0;
      long long pre = 0;

      // Number of input alphabets
      long long D = 256;

      for(int j=i;j<s.length();j++){

        // calculate new hash value by adding next element
        pre = (pre*D+s[j]) % mod;

        // store string length if non repeat
        if(d.find(pre) == d.end())
          l.push_back({i, j});
        d[pre] = 1;
      }
    }

    // resulting length
    cout<<l.size()<<endl;

    // resulting distinct substrings
    for(int i = 0; i < l.size(); i++)
      cout << s.substr(l[i][0],l[i][1]+1-l[i][0]) << " ";
  }
}

// This code is contributed by shinjanpatra

Java

import java.util.*;

public class Main {
    public static void main(String[] args) {
        int t = 1;
        // store prime to reduce overflow
        long mod = 9007199254740881L;

        for (int i = 0; i < t; i++) {

            // string to check number of distinct substring
            String s = "abcd";

            // to store substrings
            List<List<Integer>> l = new ArrayList<>();

            // to store hash values by Rabin Karp algorithm
            Map<Long, Integer> d = new HashMap<>();

            for (int j = 0; j < s.length(); j++) {
                long suma = 0;
                long pre = 0;

                // Number of input alphabets
                int D = 256;

                for (int k = j; k < s.length(); k++) {

                    // calculate new hash value by adding next element
                    pre = (pre*D + (long)s.charAt(k)) % mod;

                    // store string length if non repeat
                    if (!d.containsKey(pre)) {
                        List<Integer> sublist = new ArrayList<>();
                        sublist.add(j);
                        sublist.add(k);
                        l.add(sublist);
                    }
                    d.put(pre, 1);
                }
            }

            // resulting length
            System.out.println(l.size());

            // resulting distinct substrings
            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

# importing libraries
import sys
import math as mt
t = 1
# store prime to reduce overflow
mod = 9007199254740881

for ___ in range(t):

    # string to check number of distinct substring
    s = 'abcd'

    # to store substrings
    l = []

    # to store hash values by Rabin Karp algorithm
    d = {}

    for i in range(len(s)):
        suma = 0
        pre = 0

        # Number of input alphabets
        D = 256

        for j in range(i, len(s)):

            # calculate new hash value by adding next element
            pre = (pre*D+ord(s[j])) % mod

            # store string length if non repeat
            if d.get(pre, -1) == -1:
                l.append([i, j])
            d[pre] = 1

    # resulting length
    print(len(l))

    # resulting distinct substrings
    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;


    // store prime to reduce overflow
    long mod = 9007199254740881;

    for (int i = 0; i < t; i++)
    {
        // string to check number of distinct substring
        string s = "abcd";

        // to store substrings
        List<List<long>> l = new List<List<long>>();

        // to store hash values by Rabin Karp algorithm
        Dictionary<long, int> d = new Dictionary<long, int>();

        for (int j = 0; j < s.Length; j++)
        {
            int suma = 0;
            long pre = 0;

            // Number of input alphabets
            long D = 256;

            for (int k = j; k < s.Length; k++)
            {
                // calculate new hash value by adding next element
                pre = (pre * D + s[k]) % mod;

                // store string length if non repeat
                if (!d.ContainsKey(pre))
                {
                    List<long> sub = new List<long>();
                    sub.Add(j);
                    sub.Add(k);
                    l.Add(sub);
                }
                d[pre] = 1;
            }
        }

        // resulting length

        Console.WriteLine(l.Count);


        // resulting distinct substrings
        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]) + " ");
        }
    }
}
}
//This code is contributed by rudra1807raj

Javascript

<script>

let t = 1

// store prime to reduce overflow
let mod = 9007199254740881

for(let i = 0; i < t; i++){
    // string to check number of distinct substring
    let s = 'abcd'

    // to store substrings
    let l = []

    // to store hash values by Rabin Karp algorithm
    let d = new Map()

    for(let i=0;i<s.length;i++){
        let suma = 0
        let pre = 0

        // Number of input alphabets
        let D = 256

        for(let j=i;j<s.length;j++){

            // calculate new hash value by adding next element
            pre = (pre*D+s.charCodeAt(j)) % mod

            // store string length if non repeat
            if(d.has([pre, -1]) == false)
                l.push([i, j])
            d.set(pre , 1)
        }
    }

    // resulting length
    document.write(l.length,"</br>")

    // resulting distinct substrings
    for(let i = 0; i < l.length; i++)
        document.write(s.substring(l[i][0],l[i][1]+1)," ")
}

// This code is contributed by shinjanpatra

</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:

1. Define an input string “s”.
2. Create an empty unordered set “substrings” to store the distinct substrings.
3. 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.
4. 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”.
5. 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() {
    // Input string
    string s = "abcd";

    // Set to store distinct substrings
    unordered_set<string> substrings;

    // Iterate over all possible substrings and add them to the set
    for (int i = 0; i < s.size(); i++) {
        for (int j = i; j < s.size(); j++) {
            substrings.insert(s.substr(i, j - i + 1));
        }
    }
//This code is contributed rudra1807raj
    // Output the number of distinct substrings
    cout << substrings.size() << endl;
}

Java

import java.util.*;

public class GFG {
    public static void main(String[] args) {
        // Input string
        String s = "abcd";

        // Set to store distinct substrings
        Set<String> substrings = new HashSet<>();

        // Iterate over all possible substrings and add them to the set
        for (int i = 0; i < s.length(); i++) {
            for (int j = i; j < s.length(); j++) {
                substrings.add(s.substring(i, j + 1));
            }
        }

        // Output the number of distinct substrings
        System.out.println(substrings.size());
    }
}
// This code is contributed by rudra1807raj

Python

def main():
    # Input string
    s = "abcd"

    # Set to store distinct substrings
    substrings = set()

    # Iterate over all possible substrings and add them to the set
    for i in range(len(s)):
        for j in range(i, len(s)):
            substrings.add(s[i:j + 1])

    # Output the number of distinct substrings
    print(len(substrings))

if __name__ == "__main__":
    main()

C#

using System;
using System.Collections.Generic;

class GFG
{
    static void Main(string[] args)
    {
        // Input string
        string s = "abcd";

        // Set to store distinct substrings
        HashSet<string> substrings = new HashSet<string>();

        // Iterate over all possible substrings and add them to the set
        for (int i = 0; i < s.Length; i++){
           for (int j = i; j < s.Length; j++){
              substrings.Add(s.Substring(i, j - i + 1));
            }
          }
        // Output the number of distinct substrings
        Console.WriteLine(substrings.Count);
    }
}

Javascript

<script>
function countDistinctSubstrings(s) {
    const substrings = new Set();

    // Iterate over all possible substrings and add them to the set
    for (let i = 0; i < s.length; i++) {
        for (let j = i; j < s.length; j++) {
            substrings.add(s.substring(i, j + 1));
        }
    }

    // Output the number of distinct substrings
    return substrings.size;
}

// Input string
const s = "abcd";

// Get the number of distinct substrings
const distinctSubstringsCount = countDistinctSubstrings(s);
document.write(distinctSubstringsCount);
</script>

Output

10

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.