Distinct palindromic sub-strings of the given string using Dynamic Programming

Given a string str of lowercase alphabets, the task is to find all distinct palindromic sub-strings of the given string.

Examples:

Input: str = “abaaa”
Output: 5
Palindromic sub-strings are “a”, “aa”, “aaa”, “aba” and “b”



Input: str = “abcd”
Output: 4

Approach: The solution to this problem has been discussed here using Manacher’s algorithm. However we can also solve it using dynamic programming.
Create an array dp[][] where dp[i][j] is set to 1 if str[i…j] is a palindrome else 0. After the array has been generated, store all the palindromic sub-strings in a map in order to get the count of distinct sub-strings.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the count
// of distinct palindromic sub-strings
// of the given string s
int palindromeSubStrs(string s)
{
  
    // To store the positions of
    // palindromic sub-strings
    int dp[s.size()][s.size()];
    int st, end, i, j, len;
  
    // Map to store the sub-strings
    map<string, bool> m;
    for (i = 0; i < s.size(); i++) {
  
        // Sub-strings of length 1 are palindromes
        dp[i][i] = 1;
  
        // Store continuous palindromic sub-strings
        m[string(s.begin() + i, s.begin() + i + 1)] = 1;
    }
  
    // Store palindromes of size 2
    for (i = 0; i < s.size() - 1; i++) {
        if (s[i] == s[i + 1]) {
            dp[i][i + 1] = 1;
            m[string(s.begin() + i, s.begin() + i + 2)] = 1;
        }
  
        // If str[i...(i+1)] is not a palindromic
        // then set dp[i][i + 1] = 0
        else {
            dp[i][i + 1] = 0;
        }
    }
  
    // Find palindromic sub-strings of length>=3
    for (len = 3; len <= s.size(); len++) {
        for (st = 0; st <= s.size() - len; st++) {
  
            // End of palindromic substring
            end = st + len - 1;
  
            // If s[start] == s[end] and
            // dp[start+1][end-1] is already palindrome
            // then s[start....end] is also a palindrome
            if (s[st] == s[end] && dp[st + 1][end - 1]) {
  
                // Set dp[start][end] = 1
                dp[st][end] = 1;
                m[string(s.begin() + st, s.begin() + end + 1)] = 1;
            }
  
            // Not a palindrome
            else
                dp[st][end] = 0;
        }
    }
  
    // Return the count of distinct palindromes
    return m.size();
}
  
// Driver code
int main()
{
    string s = "abaaa";
    cout << palindromeSubStrs(s);
  
    return 0;
}

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Java

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// Java implementation of the approach 
import java.util.HashMap;
  
class GFG 
{
  
    // Function to return the count
    // of distinct palindromic sub-strings
    // of the given string s
    static int palindromeSubStrs(String s)
    {
  
        // To store the positions of
        // palindromic sub-strings
        int[][] dp = new int[s.length()][s.length()];
        int st, end, i, len;
  
        // Map to store the sub-strings
        HashMap<String, 
                Boolean> m = new HashMap<>();
  
        for (i = 0; i < s.length(); i++) 
        {
  
            // Sub-strings of length 1 are palindromes
            dp[i][i] = 1;
  
            // Store continuous palindromic sub-strings
            m.put(s.substring(i, i + 1), true);
        }
  
        // Store palindromes of size 2
        for (i = 0; i < s.length() - 1; i++) 
        {
            if (s.charAt(i) == s.charAt(i + 1)) 
            {
                dp[i][i + 1] = 1;
                m.put(s.substring(i, i + 2), true);
            }
  
            // If str[i...(i+1)] is not a palindromic
            // then set dp[i][i + 1] = 0
            else
                dp[i][i + 1] = 0;
        }
  
        // Find palindromic sub-strings of length>=3
        for (len = 3; len <= s.length(); len++) 
        {
            for (st = 0; st <= s.length() - len; st++)
            {
  
                // End of palindromic substring
                end = st + len - 1;
  
                // If s[start] == s[end] and
                // dp[start+1][end-1] is already palindrome
                // then s[start....end] is also a palindrome
                if (s.charAt(st) == s.charAt(end) && 
                    dp[st + 1][end - 1] == 1
                {
  
                    // Set dp[start][end] = 1
                    dp[st][end] = 1;
                    m.put(s.substring(st, end + 1), true);
                }
  
                // Not a palindrome
                else
                    dp[st][end] = 0;
            }
        }
  
        // Return the count of distinct palindromes
        return m.size();
    }
  
    // Driver Code
    public static void main(String[] args)
    {
        String s = "abaaa";
        System.out.println(palindromeSubStrs(s));
    }
}
  
// This code is contributed by
// sanjeev2552

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Python3

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# Python3 implementation of the approach 
  
# import numpy lib as np
import numpy as np;
  
# Function to return the count 
# of distinct palindromic sub-strings 
# of the given string s 
def palindromeSubStrs(s) : 
  
    # To store the positions of 
    # palindromic sub-strings 
    dp = np.zeros((len(s),len(s))); 
      
    # Map to store the sub-strings 
    m = {}; 
      
    for i in range(len(s)) :
  
        # Sub-strings of length 1 are palindromes 
        dp[i][i] = 1
  
        # Store continuous palindromic sub-strings 
        m[s[i: i + 1]] = 1
      
  
    # Store palindromes of size 2 
    for i in range(len(s)- 1) : 
        if (s[i] == s[i + 1]) :
            dp[i][i + 1] = 1
            m[ s[i : i + 2]] = 1
           
  
        # If str[i...(i+1)] is not a palindromic 
        # then set dp[i][i + 1] = 0 
        else :
            dp[i][i + 1] = 0
  
    # Find palindromic sub-strings of length>=3 
    for length in range(3,len(s) + 1) : 
        for st in range(len(s) - length + 1) :
  
            # End of palindromic substring 
            end = st + length - 1
  
            # If s[start] == s[end] and 
            # dp[start+1][end-1] is already palindrome 
            # then s[start....end] is also a palindrome 
            if (s[st] == s[end] and dp[st + 1][end - 1]) :
  
                # Set dp[start][end] = 1 
                dp[st][end] = 1
                m[s[st : end + 1]] = 1
  
            # Not a palindrome 
            else :
                dp[st][end] = 0;
  
    # Return the count of distinct palindromes 
    return len(m); 
  
  
# Driver code 
if __name__ == "__main__"
  
    s = "abaaa"
    print(palindromeSubStrs(s)); 
  
# This code is contributed by AnkitRai01

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Output:

5


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Improved By : AnkitRai01, sanjeev2552