# 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:
Palindromic sub-strings are “a”, “aa”, “aaa”, “aba” and “b”

Input: str = “abcd”
Output:

Approach: The solution to this problem has been discussed here using Manacherâ€™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`

## C#

 `// C# implementation of the approach ` `using` `System;` `using` `System.Collections.Generic;`   `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` `        ``Dictionary m = ``new` `Dictionary();`   `        ``for` `(i = 0; i < s.Length; i++) ` `        ``{`   `            ``// Sub-strings of length 1 are palindromes` `            ``dp[i,i] = 1;`   `            ``// Store continuous palindromic sub-strings` `            ``if``(!m.ContainsKey(s.Substring(i, 1)))` `                ``m.Add(s.Substring(i, 1), ``true``);` `        ``}`   `        ``// Store palindromes of size 2` `        ``for` `(i = 0; i < s.Length - 1; i++) ` `        ``{` `            ``if` `(s[i] == s[i + 1]) ` `            ``{` `                ``dp[i, i + 1] = 1;` `                ``if``(!m.ContainsKey(s.Substring(i, 2)))` `                    ``m.Add(s.Substring(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[st] == s[end] && ` `                    ``dp[st + 1, end - 1] == 1) ` `                ``{`   `                    ``// Set dp[start,end] = 1` `                    ``dp[st, end] = 1;` `                    ``m.Add(s.Substring(st, end + 1-st), ``true``);` `                ``}`   `                ``// Not a palindrome` `                ``else` `                    ``dp[st, end] = 0;` `            ``}` `        ``}`   `        ``// Return the count of distinct palindromes` `        ``return` `m.Count;` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `Main(String[] args)` `    ``{` `        ``String s = ``"abaaa"``;` `        ``Console.WriteLine(palindromeSubStrs(s));` `    ``}` `}`   `// This code is contributed by PrinciRaj1992`

## Javascript

 ``

Output

```5
```

Time complexity : O((n^2)logn), where n is the length of the input string. This is because we are using a nested loop to iterate over all possible substrings and check if they are palindromic.

Space complexity : O(n^2). This is because we are using a 2D array of size n x n to store the results of subproblems, and a map to store the distinct palindromic substrings.

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