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:

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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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Output:

5


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