# 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++

 `// C++ implementation of the approach ` `#include ` `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 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; ` `} `

## Java

 `// 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 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 `

## Python3

 `# 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 `

Output:

```5
```

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Improved By : AnkitRai01, sanjeev2552