Find all distinct palindromic sub-strings of a given string

4.1

Given a string of lowercase ASCII characters, find all distinct continuous palindromic sub-strings of it.

Examples:

Input: str = "abaaa"
Output:  Below are 5 palindrome sub-strings
a
aa
aaa
aba
b


Input: str = "geek"
Output:  Below are 4 palindrome sub-strings
e
ee
g
k

Step 1: Finding all palindromes using modified Manacher’s algorithm:
Considering each character as a pivot, expand on both sides to find the length of both even and odd length palindromes centered at the pivot character under consideration and store the length in the 2 arrays (odd & even).
Time complexity for this step is O(n^2)

Step 2: Inserting all the found palindromes in a HashMap:
Insert all the palindromes found from the previous step into a HashMap. Also insert all the individual characters from the string into the HashMap (to generate distinct single letter palindromic sub-strings).
Time complexity of this step is O(n^3) assuming that the hash insert search takes O(1) time. Note that there can be at most O(n^2) palindrome sub-strings of a string. In below C++ code ordered hashmap is used where the time complexity of insert and search is O(Logn). In C++, ordered hashmap is implemented using Red Black Tree.

Step 3: Printing the distinct palindromes and number of such distinct palindromes:
The last step is to print all values stored in the HashMap (only distinct elements will be hashed due to the property of HashMap). The size of the map gives the number of distinct palindromic continuous sub-strings.

Below is the implementation of the above idea.

C/C++

// C++ program to find all distinct palindrome sub-strings
// of a given string
#include <iostream>
#include <map>
using namespace std;

// Function to print all distinct palindrome sub-strings of s
void palindromeSubStrs(string s)
{
    map<string, int> m;
    int n = s.size();

    // table for storing results (2 rows for odd-
    // and even-length palindromes
    int R[2][n+1];

    // Find all sub-string palindromes from the given input
    // string insert 'guards' to iterate easily over s
    s = "@" + s + "#";

    for (int j = 0; j <= 1; j++)
    {
        int rp = 0;   // length of 'palindrome radius'
        R[j][0] = 0;

        int i = 1;
        while (i <= n)
        {
            //  Attempt to expand palindrome centered at i
            while (s[i - rp - 1] == s[i + j + rp])
                rp++;  // Incrementing the length of palindromic
                       // radius as and when we find vaid palindrome

            // Assigning the found palindromic length to odd/even
            // length array
            R[j][i] = rp;
            int k = 1;
            while ((R[j][i - k] != rp - k) && (k < rp))
            {
                R[j][i + k] = min(R[j][i - k],rp - k);
                k++;
            }
            rp = max(rp - k,0);
            i += k;
        }
    }

    // remove 'guards'
    s = s.substr(1, n);

    // Put all obtained palindromes in a hash map to
    // find only distinct palindromess
    m[string(1, s[0])]=1;
    for (int i = 1; i <= n; i++)
    {
        for (int j = 0; j <= 1; j++)
            for (int rp = R[j][i]; rp > 0; rp--)
               m[s.substr(i - rp - 1, 2 * rp + j)]=1;
        m[string(1, s[i])]=1;
    }

    //printing all distinct palindromes from hash map
   cout << "Below are " << m.size()-1
        << " palindrome sub-strings";
   map<string, int>::iterator ii;
   for (ii = m.begin(); ii!=m.end(); ++ii)
      cout << (*ii).first << endl;
}

// Driver program
int main()
{
    palindromeSubStrs("abaaa");
    return 0;
}

Java

// Java program to find all distinct palindrome
// sub-strings of a given string
import java.util.Map;
import java.util.TreeMap;

public class GFG 
{     
    // Function to print all distinct palindrome
    // sub-strings of s
    static void palindromeSubStrs(String s)
    {
        //map<string, int> m;
        TreeMap<String , Integer> m = new TreeMap<>();
        int n = s.length();
     
        // table for storing results (2 rows for odd-
        // and even-length palindromes
        int[][] R = new int[2][n+1];
     
        // Find all sub-string palindromes from the 
        // given input string insert 'guards' to 
        // iterate easily over s
        s = "@" + s + "#";
     
        for (int j = 0; j <= 1; j++)
        {
            int rp = 0;   // length of 'palindrome radius'
            R[j][0] = 0;
     
            int i = 1;
            while (i <= n)
            {
                //  Attempt to expand palindrome centered 
                // at i
                while (s.charAt(i - rp - 1) == s.charAt(i + 
                                                j + rp))
                    rp++;  // Incrementing the length of
                           // palindromic radius as and 
                           // when we find vaid palindrome
     
                // Assigning the found palindromic length
                // to odd/even length array
                R[j][i] = rp;
                int k = 1;
                while ((R[j][i - k] != rp - k) && (k < rp))
                {
                    R[j][i + k] = Math.min(R[j][i - k], 
                                              rp - k);
                    k++;
                }
                rp = Math.max(rp - k,0);
                i += k;
            }
        }
     
        // remove 'guards'
        s = s.substring(1, s.length()-1);
     
        // Put all obtained palindromes in a hash map to
        // find only distinct palindromess
        m.put(s.substring(0,1), 1);
        for (int i = 1; i < n; i++)
        {
            for (int j = 0; j <= 1; j++)
                for (int rp = R[j][i]; rp > 0; rp--)
                   m.put(s.substring(i - rp - 1,  i - rp - 1 
                                       + 2 * rp + j), 1);
            m.put(s.substring(i, i + 1), 1);
        }
     
        // printing all distinct palindromes from 
        // hash map
       System.out.println("Below are " + (m.size())
                           + " palindrome sub-strings");
       
       for (Map.Entry<String, Integer> ii:m.entrySet())
          System.out.println(ii.getKey());
    }
     
    // Driver program
    public static void main(String args[])
    {
        palindromeSubStrs("abaaa");
    }
}
// This code is contributed by Sumit Ghosh

Python

# Python program Find all distinct palindromic sub-strings
# of a given string

# Function to print all distinct palindrome sub-strings of s
def palindromeSubStrs(s):
    m = dict()
    n = len(s)

    # table for storing results (2 rows for odd-
    # and even-length palindromes
    R = [[0 for x in xrange(n+1)] for x in xrange(2)]

    # Find all sub-string palindromes from the given input
    # string insert 'guards' to iterate easily over s
    s = "@" + s + "#"

    for j in xrange(2):
        rp = 0    # length of 'palindrome radius'
        R[j][0] = 0

        i = 1
        while i <= n:

            # Attempt to expand palindrome centered at i
            while s[i - rp - 1] == s[i + j + rp]:
                rp += 1 # Incrementing the length of palindromic
                        # radius as and when we find valid palindrome

            # Assigning the found palindromic length to odd/even
            # length array
            R[j][i] = rp
            k = 1
            while (R[j][i - k] != rp - k) and (k < rp):
                R[j][i+k] = min(R[j][i-k], rp - k)
                k += 1
            rp = max(rp - k, 0)
            i += k

    # remove guards
    s = s[1:len(s)-1]

    # Put all obtained palindromes in a hash map to
    # find only distinct palindrome
    m[s[0]] = 1
    for i in xrange(1,n):
        for j in xrange(2):
            for rp in xrange(R[j][i],0,-1):
                m[s[i - rp - 1 : i - rp - 1 + 2 * rp + j]] = 1
        m[s[i]] = 1

    # printing all distinct palindromes from hash map
    print "Below are " + str(len(m)) + " pali sub-strings"
    for i in m:
        print i

# Driver program
palindromeSubStrs("abaaa")
# This code is contributed by BHAVYA JAIN and ROHIT SIKKA


Output:
 Below are 5 palindrome sub-strings
a
aa
aaa
aba
b 

Similar Problem:
Count All Palindrome Sub-Strings in a String

This article is contributed by Vignesh Narayanan and Sowmya Sampath. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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