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

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-stringsaaaaaaababInput: str = "geek"Output:  Below are 4 palindrome sub-stringseeegk`

## BRUTE APPROACH

Intuition:

1. We declare a boolean 2-D array and fill it diagonally.
2. We check for every gap.ie(0,1,2,…..)
3. suppose gap=0., that means there is only one element and we put  true since  a single character is palindrome
4. if gap = 1, we check whether extremes are same and if so we put true,else false;
5. else for any other value of gap, we check if extremes are same and dp[i+1][j-1] yields true  and if so then we put true else false;
6. at every time when a true is encountered we add the string in the set data structure.
7. Atlast we return the ans

Implementation:

## Java

 `// Java program to find all distinct palindrome``// sub-strings of a given string` `import` `java.io.*;``import` `java.util.*;` `class` `GFG {``    ``static` `int` `palindromeSubStrs(String str,``                                 ``Set set)``    ``{``        ``// code here``        ``int` `n = str.length();``        ``boolean` `dp[][] = ``new` `boolean``[n][n];` `        ``for` `(``int` `gap = ``0``; gap < n; gap++) {``            ``for` `(``int` `i = ``0``, j = gap; j < n; i++, j++) {``                ``if` `(gap == ``0``)``                    ``dp[i][j] = ``true``;``                ``else` `if` `(gap == ``1``)``                    ``dp[i][j]``                        ``= str.charAt(i) == str.charAt(j);``                ``else``                    ``dp[i][j]``                        ``= (str.charAt(i) == str.charAt(j)``                           ``&& dp[i + ``1``][j - ``1``]);` `                ``if` `(dp[i][j])``                    ``set.add(str.substring(i, j + ``1``));``            ``}``        ``}``        ``return` `set.size();``    ``}``    ``public` `static` `void` `main(String[] args)``    ``{``        ``String str = ``"abaaa"``;``        ``Set set = ``new` `TreeSet<>();``        ``palindromeSubStrs(str, set);``        ``System.out.println(``            ``"No of distinct palindromic substrings are : "``            ``+ palindromeSubStrs(str, set));` `        ``for` `(String s : set)``            ``System.out.println(s);``    ``}``}``// This code is contributed by Raunak Singh`

## C#

 `// C# program to find all distinct palindrome``// sub-strings of a given string``using` `System;``using` `System.Collections.Generic;` `class` `GFG``{``    ``static` `int` `PalindromeSubStrs(``string` `str, HashSet<``string``> ``set``)``    ``{``      ``// code here``        ``int` `n = str.Length;``        ``bool``[,] dp = ``new` `bool``[n, n];` `        ``for` `(``int` `gap = 0; gap < n; gap++)``        ``{``            ``for` `(``int` `i = 0, j = gap; j < n; i++, j++)``            ``{``                ``if` `(gap == 0)``                    ``dp[i, j] = ``true``;``                ``else` `if` `(gap == 1)``                    ``dp[i, j] = str[i] == str[j];``                ``else``                    ``dp[i, j] = (str[i] == str[j] && dp[i + 1, j - 1]);` `                ``if` `(dp[i, j])``                    ``set``.Add(str.Substring(i, j - i + 1));``            ``}``        ``}``        ``return` `set``.Count;``    ``}` `    ``public` `static` `void` `Main(``string``[] args)``    ``{``        ``string` `str = ``"abaaa"``;``        ``HashSet<``string``> ``set` `= ``new` `HashSet<``string``>();``        ``PalindromeSubStrs(str, ``set``);``        ``Console.WriteLine(``            ``"No of distinct palindromic substrings are: "` `+``            ``PalindromeSubStrs(str, ``set``));` `        ``foreach` `(``string` `s ``in` `set``)``            ``Console.WriteLine(s);``    ``}``}`

Output

```No of distinct palindromic substrings are : 5
a
aa
aaa
aba
b

```

Time Complexity: O(N^2 * log(N)) since we are iterating through the matrix and while doing so we put elements in set which takes log(N) time

Space Complexity: O(N^2) since we using 2-D array

Method 1:

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++ program to find all distinct palindrome sub-strings``// of a given string``#include ``#include ``using` `namespace` `std;``  ` `// Function to print all distinct palindrome sub-strings of s``void` `palindromeSubStrs(string s)``{``    ``map 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 valid 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::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 m;``        ``TreeMap 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 valid 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 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`

## Python3

 `# 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` `range``(n``+``1``)] ``for` `x ``in` `range``(``2``)]` `    ``# Find all sub-string palindromes from the given input``    ``# string insert 'guards' to iterate easily over s``    ``s ``=` `"@"` `+` `s ``+` `"#"` `    ``for` `j ``in` `range``(``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` `range``(``1``,n):``        ``for` `j ``in` `range``(``2``):``            ``for` `rp ``in` `range``(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`

## C#

 `// C# program to find all distinct palindrome``// sub-strings of a given string``using` `System;``using` `System.Collections.Generic;` `class` `GFG``{` `    ``// Function to print all distinct palindrome``    ``// sub-strings of s``    ``public` `static` `void` `palindromeSubStrs(``string` `s)``    ``{``        ``//map m;``        ``Dictionary < ``string``,``        ``int` `> m = ``new` `Dictionary < ``string``,``        ``int` `> ();``        ``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[i - rp - 1] == s[i + j + rp])` `                ``// Incrementing the length of``                ``// palindromic radius as and``                ``// when we find valid palindrome``                ``rp++;` `                ``// 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);` `        ``// Put all obtained palindromes in a hash map to``        ``// find only distinct palindromess``        ``if` `(!m.ContainsKey(s.Substring(0, 1)))``            ``m.Add(s.Substring(0, 1), 1);``        ``else``            ``m[s.Substring(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--)``            ``{``                ``if` `(!m.ContainsKey(s.Substring(i - rp - 1, 2 * rp + j)))``                    ``m.Add(s.Substring(i - rp - 1, 2 * rp + j), 1);``                ``else``                    ``m[s.Substring(i - rp - 1, 2 * rp + j)]++;``            ``}` `            ``if` `(!m.ContainsKey(s.Substring(i, 1)))``                ``m.Add(s.Substring(i, 1), 1);``            ``else``                ``m[s.Substring(i, 1)]++;``        ``}` `        ``// printing all distinct palindromes from``        ``// hash map``        ``Console.WriteLine(``"Below are "` `+ (m.Count));` `        ``foreach``(KeyValuePair < ``string``, ``int` `> ii ``in` `m)``        ``Console.WriteLine(ii.Key);``    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main(``string``[] args)``    ``{``        ``palindromeSubStrs(``"abaaa"``);``    ``}``}` `// This code is contributed by``// sanjeev2552`

## Javascript

 ``

Output:

` Below are 5 palindrome sub-stringsaaaaaaabab `

Method 2 :

String length – N

Step 1 : Find all the palindromic sub-strings

First for every sub-string check if it is palindrome or not using dynamic programming like this – https://www.geeksforgeeks.org/count-palindrome-sub-strings-string/

Time complexity – O(N2)   and   Space complexity – O(N2)

Step 2 : Remove duplicate palindromes

For every index starting from index 0 we will use KMP algorithm and check if prefix and suffix is same and is palindrome then we will put 0 the dp array for that suffix sub-string

Time complexity O(N2)    and   Space complexity O(N) for KMP array

Step 3 : Print the distinct palindromes and number of such palindromes

For every sub-string check if it is present in dp array (i.e dp[i][j] == true) and print it.

Time complexity O(N2)   and    Space complexity O(N)

Overall Time complexity – O(N2)

Overall Space complexity – O(N2)

Below is the implementation of the above idea.

## C++

 `// C++ program to find all distinct palindrome sub-strings``// of a given string``#include ``#include ``using` `namespace` `std;` `int` `solve(string s)``{``    ``int` `n = s.size();``    ``// dp array to store whether a substring is palindrome``    ``// or not using dynamic programming we can solve this``    ``// in O(N^2)``    ``// dp[i][j] will be true (1) if substring (i, j) is``    ``// palindrome else false (0)``    ``vector > dp(n, vector<``bool``>(n, ``false``));``    ``for` `(``int` `i = 0; i < n; i++) {``        ``// base case every char is palindrome``        ``dp[i][i] = 1;``        ``// check for every substring of length 2``        ``if` `(i < n && s[i] == s[i + 1]) {``            ``dp[i][i + 1] = 1;``        ``}``    ``}``    ``// check every substring of length greater than 2 for``    ``// palindrome``    ``for` `(``int` `len = 3; len <= n; len++) {``        ``for` `(``int` `i = 0; i + len - 1 < n; i++) {``            ``if` `(s[i] == s[i + (len - 1)]``                ``&& dp[i + 1][i + (len - 1) - 1]) {``                ``dp[i][i + (len - 1)] = ``true``;``            ``}``        ``}``    ``}` `    ``//*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*``    ``// here we will apply kmp algorithm for substrings``    ``// starting from i = 0 to n-1 when we will find prefix``    ``// and suffix of a substring to be equal and it is``    ``// palindrome we will make dp[i][j] for that suffix to be``    ``// false which means it is already added in the prefix``    ``// and we should not count it anymore.``    ``vector<``int``> kmp(n, 0);``    ``for` `(``int` `i = 0; i < n; i++) {``        ``// starting kmp for every i from 0 to n-1``        ``int` `j = 0, k = 1;``        ``while` `(k + i < n) {``            ``if` `(s[j + i] == s[k + i]) {``                ``// make suffix to be false``                ``// if this suffix is palindrome then it is``                ``// already included in prefix``                ``dp[k + i - j][k + i] = ``false``;``                ``kmp[k++] = ++j;``            ``}``            ``else` `if` `(j > 0) {``                ``j = kmp[j - 1];``            ``}``            ``else` `{``                ``kmp[k++] = 0;``            ``}``        ``}``    ``}``    ``//*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*``    ``int` `count = 0;``    ``for` `(``int` `i = 0; i < n; i++) {``        ``string str;``        ``for` `(``int` `j = i; j < n; j++) {``            ``str += s[j];``            ``if` `(dp[i][j]) {``                ``// count number of resultant distinct``                ``// substrings and print  that substring``                ``count++;``                ``cout << str << ``'\n'``;``            ``}``        ``}``    ``}``    ``cout << ``"Total number of distinct palindromes is "``         ``<< count << ``'\n'``;``      ``return` `0;``}` `// Driver code starts``// This code is contributed by Aditya Anand``int` `main()``{``    ``string s1 = ``"abaaa"``, s2 = ``"aaaaaaaaaa"``;``    ``solve(s1);``    ``solve(s2);``    ``return` `0;``}``// Driver code ends`

## Java

 `// Java program to find all distinct palindrome sub-strings``// of a given string``import` `java.util.*;` `class` `GFG``{``  ` `  ``// Driver code starts``  ``public` `static` `void` `main(String[] args)``  ``{``    ``String s1 = ``"abaaa"``, s2 = ``"aaaaaaaaaa"``;``    ``solve(s1);``    ``solve(s2);``  ``}` `  ``public` `static` `int` `solve(String s)``  ``{``    ``int` `n = s.length();``    ` `    ``// dp array to store whether a substring is``    ``// palindrome or not using dynamic programming we``    ``// can solve this in O(N^2) dp[i][j] will be true``    ``// (1) if substring (i, j) is palindrome else false``    ``// (0)``    ``boolean``[][] dp = ``new` `boolean``[n][n];``    ``for` `(``int` `i = ``0``; i < n; i++)``    ``{``      ` `      ``// base case every char is palindrome``      ``dp[i][i] = ``true``;``      ` `      ``// check for every substring of length 2``      ``if` `(i < n - ``1``          ``&& s.charAt(i) == s.charAt(i + ``1``)) {``        ``dp[i][i + ``1``] = ``true``;``      ``}``    ``}``    ` `    ``// check every substring of length greater than 2``    ``// for palindrome``    ``for` `(``int` `len = ``3``; len <= n; len++) {``      ``for` `(``int` `i = ``0``; i + len - ``1` `< n; i++) {``        ``if` `(s.charAt(i) == s.charAt(i + (len - ``1``))``            ``&& dp[i + ``1``][i + (len - ``1``) - ``1``]) {``          ``dp[i][i + (len - ``1``)] = ``true``;``        ``}``      ``}``    ``}` `    ``//*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*``    ``// here we will apply kmp algorithm for substrings``    ``// starting from i = 0 to n-1 when we will find``    ``// prefix and suffix of a substring to be equal and``    ``// it is palindrome we will make dp[i][j] for that``    ``// suffix to be false which means it is already``    ``// added in the prefix and we should not count it``    ``// anymore.``    ``int``[] kmp = ``new` `int``[n];``    ``for` `(``int` `i = ``0``; i < n; i++)``    ``{``      ` `      ``// starting kmp for every i from 0 to n-1``      ``int` `j = ``0``, k = ``1``;``      ``while` `(k + i < n) {``        ``if` `(s.charAt(j + i) == s.charAt(k + i))``        ``{``          ` `          ``// make suffix to be false``          ``// if this suffix is palindrome then it``          ``// is already included in prefix``          ``dp[k + i - j][k + i] = ``false``;``          ``kmp[k++] = ++j;``        ``}``        ``else` `if` `(j > ``0``) {``          ``j = kmp[j - ``1``];``        ``}``        ``else` `{``          ``kmp[k++] = ``0``;``        ``}``      ``}``    ``}``    ``//*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*``    ``int` `count = ``0``;``    ``for` `(``int` `i = ``0``; i < n; i++) {``      ``String str = ``""``;``      ``for` `(``int` `j = i; j < n; j++) {``        ``str += s.charAt(j);``        ``if` `(dp[i][j])``        ``{``          ` `          ``// count number of resultant distinct``          ``// substrings and print  that substring``          ``count++;``          ``System.out.println(str);``        ``}``      ``}``    ``}``    ``System.out.println(``      ``"Total number of distinct palindromes is "``      ``+ count);``    ``return` `0``;``  ``}``}` `// This code is contributed by kdheeraj.`

## Python3

 `# Python3 program to find all distinct palindrome sub-strings``# of a given string``def` `solve(s):``    ``n ``=` `len``(s)``    ` `    ``# dp array to store whether a substring is palindrome``    ``# or not using dynamic programming we can solve this``    ``# in O(N^2)``    ``# dp[i][j] will be true (1) if substring (i, j) is``    ``# palindrome else false (0)``    ``dp ``=` `[[``False` `for` `j ``in` `range``(n)] ``for` `i ``in` `range``(n)]``    ``for` `i ``in` `range``(n):``      ` `        ``# base case every char is palindrome``        ``dp[i][i] ``=` `True``        ` `        ``# check for every substring of length 2``        ``if` `i < n``-``1` `and` `s[i] ``=``=` `s[i``+``1``]:``            ``dp[i][i``+``1``] ``=` `True``            ` `    ``# check every substring of length greater than 2 for``    ``# palindrome``    ``for` `lenk ``in` `range``(``3``, n``+``1``):``        ``for` `i ``in` `range``(n``-``lenk``+``1``):``            ``if` `s[i] ``=``=` `s[i``+``lenk``-``1``] ``and` `dp[i``+``1``][i``+``lenk``-``2``]:``                ``dp[i][i``+``lenk``-``1``] ``=` `True``                ` `    ``# here we will apply kmp algorithm for substrings``    ``# starting from i = 0 to n-1 when we will find prefix``    ``# and suffix of a substring to be equal and it is``    ``# palindrome we will make dp[i][j] for that suffix to be``    ``# false which means it is already added in the prefix``    ``# and we should not count it anymore.``    ``kmp ``=` `[``0``]``*``n``    ``for` `i ``in` `range``(n):``      ` `        ``# starting kmp for every i from 0 to n-1``        ``j, k ``=` `0``, ``1``        ``while` `k``+``i < n:``            ``if` `s[j``+``i] ``=``=` `s[k``+``i]:``              ` `                ``# make suffix to be false``                ``# if this suffix is palindrome then it is``                ``# already included in prefix``                ``dp[k``+``i``-``j][k``+``i] ``=` `False``                ``kmp[k] ``=` `j``+``1``                ``k ``+``=` `1``                ``j ``+``=` `1``            ``elif` `j > ``0``:``                ``j ``=` `kmp[j``-``1``]``            ``else``:``                ``kmp[k] ``=` `0``                ``k ``+``=` `1``    ``count ``=` `0``    ``for` `i ``in` `range``(n):``        ``str` `=` `""``        ``for` `j ``in` `range``(i, n):``            ``str` `+``=` `s[j]``            ``if` `dp[i][j]:``                ``# count number of resultant distinct``                ``# substrings and print  that substring``                ``count ``+``=` `1``                ``print``(``str``)``    ``print``(``"Total number of distinct palindromes is"``, count)``    ``return` `0` `# Driver code starts` `if` `__name__ ``=``=` `"__main__"``:``    ``s1 ``=` `"abaaa"``    ``s2 ``=` `"aaaaaaaaaa"``    ``solve(s1)``    ``solve(s2)` `    ``# This code is contributed by lokeshpotta20.`

## C#

 `using` `System;``class` `GFG``{``  ``static` `int` `solve(``string` `s)``  ``{``    ``int` `n = s.Length;` `    ``// dp array to store whether a substring is palindrome``    ``// or not using dynamic programming we can solve this``    ``// in O(N^2)``    ``// dp[i][j] will be true (1) if substring (i, j) is``    ``// palindrome else false (0)``    ``bool``[][] dp = ``new` `bool``[n][];``    ``for` `(``int` `i = 0; i < n; i++)``    ``{``      ``dp[i] = ``new` `bool``[n];` `      ``// base case every char is palindrome``      ``dp[i][i] = ``true``;` `      ``// check for every substring of length 2``      ``if` `(i < n - 1 && s[i] == s[i + 1])``      ``{``        ``dp[i][i + 1] = ``true``;``      ``}``    ``}` `    ``// check every substring of length greater than 2 for``    ``// palindrome``    ``for` `(``int` `len = 3; len <= n; len++)``    ``{``      ``for` `(``int` `i = 0; i + len - 1 < n; i++)``      ``{``        ``if` `(s[i] == s[i + len - 1] && dp[i + 1][i + len - 2])``        ``{``          ``dp[i][i + len - 1] = ``true``;``        ``}``      ``}``    ``}` `    ``//--------------------------------*``    ``// here we will apply kmp algorithm for substrings``    ``// starting from i = 0 to n-1 when we will find prefix``    ``// and suffix of a substring to be equal and it is``    ``// palindrome we will make dp[i][j] for that suffix to be``    ``// false which means it is already added in the prefix``    ``// and we should not count it anymore.``    ``int``[] kmp = ``new` `int``[n];``    ``for` `(``int` `i = 0; i < n; i++)``    ``{` `      ``// starting kmp for every i from 0 to n-1``      ``int` `j = 0, k = 1;``      ``while` `(k + i < n)``      ``{``        ``if` `(s[j + i] == s[k + i])``        ``{` `          ``// make suffix to be false``          ``// if this suffix is palindrome then it is``          ``// already included in prefix``          ``dp[k + i - j][k + i] = ``false``;``          ``kmp[k++] = ++j;``        ``}``        ``else` `if` `(j > 0)``        ``{``          ``j = kmp[j - 1];``        ``}``        ``else``        ``{``          ``kmp[k++] = 0;``        ``}``      ``}``    ``}``    ``//---------------------------------``    ``int` `count = 0;``    ``for` `(``int` `i = 0; i < n; i++)``    ``{``      ``string` `str = ``""``;``      ``for` `(``int` `j = i; j < n; j++)``      ``{``        ``str += s[j];``        ``if` `(dp[i][j])``        ``{` `          ``// count number of resultant distinct``          ``// substrings and print that substring``          ``count++;``          ``Console.WriteLine(str);``        ``}``      ``}` `    ``}``    ``Console.WriteLine(``"Total number of distinct palindromes is "``+ count);` `    ``return` `count;``  ``}` `  ``// Driver code starts``  ``static` `void` `Main(``string``[] args)``  ``{``    ``string` `s1 = ``"abaaa"``;``    ``string` `s2 = ``"aaaaaaaaaa"``;``    ``solve(s1);``    ``solve(s2);``  ``}` `  ``// Driver code ends``}` `// This code is contributed by imruhrbf8.`

## Javascript

 `// Javascript program to find all distinct palindrome sub-strings``// of a given string``function` `solve(s)``{``    ``let n = s.length;``    ` `    ``// dp array to store whether a substring is palindrome``    ``// or not using dynamic programming we can solve this``    ``// in O(N^2)``    ``// dp[i][j] will be true (1) if substring (i, j) is``    ``// palindrome else false (0)``    ``let dp = ``new` `Array(n);``    ``for``(let i = 0; i < n; i++)``        ``dp[i] = ``new` `Array(n).fill(0);``    ``for` `(let i = 0; i < n; i++)``    ``{``    ` `        ``// base case every char is palindrome``        ``dp[i][i] = 1;``        ` `        ``// check for every substring of length 2``        ``if` `(i < n && s[i] == s[i + 1]) {``            ``dp[i][i + 1] = 1;``        ``}``    ``}``    ` `    ``// check every substring of length greater than 2 for``    ``// palindrome``    ``for` `(let len = 3; len <= n; len++) {``        ``for` `(let i = 0; i + len - 1 < n; i++) {``            ``if` `(s[i] == s[i + (len - 1)]``                ``&& dp[i + 1][i + (len - 1) - 1]) {``                ``dp[i][i + (len - 1)] = ``true``;``            ``}``        ``}``    ``}` `    ``//*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*``    ``// here we will apply kmp algorithm for substrings``    ``// starting from i = 0 to n-1 when we will find prefix``    ``// and suffix of a substring to be equal and it is``    ``// palindrome we will make dp[i][j] for that suffix to be``    ``// false which means it is already added in the prefix``    ``// and we should not count it anymore.``    ``let kmp=``new` `Array(n).fill(0);``    ``for` `(let i = 0; i < n; i++)``    ``{` `        ``// starting kmp for every i from 0 to n-1``        ``let j = 0, k = 1;``        ``while` `(k + i < n)``        ``{``            ``if` `(s[j + i] == s[k + i])``            ``{``            ` `                ``// make suffix to be false``                ``// if this suffix is palindrome then it is``                ``// already included in prefix``                ``dp[k + i - j][k + i] = ``false``;``                ``kmp[k++] = ++j;``            ``}``            ``else` `if` `(j > 0) {``                ``j = kmp[j - 1];``            ``}``            ``else` `{``                ``kmp[k++] = 0;``            ``}``        ``}``    ``}``    ` `    ``//*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*``    ``let count = 0;``    ``for` `(let i = 0; i < n; i++) {``        ``let str=``""``;``        ``for` `(let j = i; j < n; j++) {``            ``str += s[j];``            ``if` `(dp[i][j])``            ``{``            ` `                ``// count number of resultant distinct``                ``// substrings and print  that substring``                ``count++;``                ``console.log(str);``            ``}``        ``}``    ``}``    ``console.log(``"Total number of distinct palindromes is "``        ``+ count);``}` `// Driver code starts``let s1 = ``"abaaa"``, s2 = ``"aaaaaaaaaa"``;``solve(s1);``solve(s2);` `// This code is contributed by ratiagrawal.`

Output

```a
aba
b
aa
aaa
Total number of distinct palindromes is 5
a
aa
aaa
aaaa
aaaaa
aaaaaa
aaaaaaa
aaaaaaaa
aaaaaaaaa
aaaaaaaaaa
Total number of distinct palindromes is 10

```

Approach:

This code uses an unordered set to store distinct palindromes and avoids using dynamic programming or the KMP algorithm. It first checks for odd-length palindromes and then even-length palindromes by expanding from each character as a center. Finally, it prints all distinct palindromes and the total count.

## C++

 `#include ``#include ``using` `namespace` `std;` `void` `findAllPalindromes(``const` `string& s) {``    ``int` `n = s.size();``    ``unordered_set palindromes;` `    ``// Check for odd length palindromes``    ``for` `(``int` `i = 0; i < n; i++) {``        ``int` `left = i, right = i;``        ``while` `(left >= 0 && right < n && s[left] == s[right]) {``            ``palindromes.insert(s.substr(left, right - left + 1));``            ``left--;``            ``right++;``        ``}``    ``}` `    ``// Check for even length palindromes``    ``for` `(``int` `i = 0; i < n - 1; i++) {``        ``int` `left = i, right = i + 1;``        ``while` `(left >= 0 && right < n && s[left] == s[right]) {``            ``palindromes.insert(s.substr(left, right - left + 1));``            ``left--;``            ``right++;``        ``}``    ``}` `    ``// Print all distinct palindromes``    ``for` `(``const` `auto``& palindrome : palindromes) {``        ``cout << palindrome << endl;``    ``}` `    ``cout << ``"Total number of distinct palindromes is "` `<< palindromes.size() << endl;``}` `int` `main() {``    ``string s1 = ``"abaaa"``, s2 = ``"aaaaaaaaaa"``;``    ``findAllPalindromes(s1);``    ``findAllPalindromes(s2);``    ``return` `0;``}`

## Javascript

 `function` `findAllPalindromes(s) {``    ``let n = s.length;``    ``let palindromes = ``new` `Set();` `    ``// Check for odd length palindromes``    ``for` `(let i = 0; i < n; i++) {``        ``let left = i, right = i;``        ``while` `(left >= 0 && right < n && s[left] == s[right]) {``            ``palindromes.add(s.substring(left, right + 1));``            ``left--;``            ``right++;``        ``}``    ``}` `    ``// Check for even length palindromes``    ``for` `(let i = 0; i < n - 1; i++) {``        ``let left = i, right = i + 1;``        ``while` `(left >= 0 && right < n && s[left] == s[right]) {``            ``palindromes.add(s.substring(left, right + 1));``            ``left--;``            ``right++;``        ``}``    ``}` `    ``// Prlet all distinct palindromes``    ``for` `(let palindrome of palindromes) {``        ``console.log(palindrome);``    ``}` `    ``console.log(``"Total number of distinct palindromes is "` `+ palindromes.size);` `}` `let s1 = ``"abaaa"``, s2 = ``"aaaaaaaaaa"``;``findAllPalindromes(s1);``findAllPalindromes(s2);`

Output

```aa
aaa
aba
b
a
Total number of distinct palindromes is 5
aaaaaaaaaa
aaaaaaaa
aaaaaa
aaaa
aaaaaaa
aa
aaaaa
aaa
aaaaaaaaa
a
Total number of distinct palindromes is 10

```

Time complexity: O(N^2)
Space complexity: O(N)

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