All distinct palindromic sub-strings of a given string
Last Updated : 11 Mar, 2025
Given a string str of lowercase ASCII characters. The task is to find all the distinct continuous palindromic sub-strings which are present in the string str.
Examples:
Input: str = “abaaa”
Output: [ “a”, “aa”, “aaa”, “aba”, “b” ]
Explanation: All 5 distinct continuous palindromic sub-strings are listed above.
Input: str = “geek”
Output: [ “e”, “ee”, “g”, “k” ]
Explanation: All 4 distinct continuous palindromic sub-strings are listed above.
[Naive Approach – 1] – Generating All Substrings – O(n ^ 3) Time and O(n ^ 2) Space
The idea is to generate all possible substrings and find all the palindromic substrings, and use set to store all the distinct one.
C++ #include <bits/stdc++.h> using namespace std; // Function to find all distinct palindrome // substrings in a given string vector<string> palindromicSubstr(string &str) { int n = str.length(); // Create a set to store the result unordered_set<string> result; // generate all substrings for(int i = 0; i < n; i++) { // to store the substring string cur = ""; for(int j = i; j < n; j++) { cur += str[j]; // check if cur is palindrome int l = 0, r = cur.length() - 1; bool isPalindrome = true; while(l < r) { if(cur[l] != cur[r]) { isPalindrome = false; break; } l++; r--; } // if cur is palindrome, // insert it into the set if(isPalindrome) { result.insert(cur); } } } // Convert the set to a vector vector<string> res(result.begin(), result.end()); return res; } int main() { string str = "abaaa"; vector<string> result = palindromicSubstr(str); for(string s : result) cout << s << " "; return 0; } // Function to find all distinct palindrome // substrings in a given string import java.util.*; class GfG { // Function to find all distinct palindrome // substrings in a given string static ArrayList<String> palindromicSubstr(String str) { int n = str.length(); // Create a set to store the result HashSet<String> result = new HashSet<>(); // generate all substrings for (int i = 0; i < n; i++) { // to store the substring String cur = ""; for (int j = i; j < n; j++) { cur += str.charAt(j); // check if cur is palindrome int l = 0, r = cur.length() - 1; boolean isPalindrome = true; while (l < r) { if (cur.charAt(l) != cur.charAt(r)) { isPalindrome = false; break; } l++; r--; } // if cur is palindrome, // insert it into the set if (isPalindrome) { result.add(cur); } } } // Convert the set to a vector ArrayList<String> res = new ArrayList<>(result); return res; } public static void main(String[] args) { String str = "abaaa"; ArrayList<String> result = palindromicSubstr(str); for (String s : result) System.out.print(s + " "); } } # Function to find all distinct palindrome # substrings in a given string. def palindromicSubstr(str): n = len(str) # Create a set to store the result result = set() # generate all substrings for i in range(n): # to store the substring cur = "" for j in range(i, n): cur += str[j] # check if cur is palindrome l = 0 r = len(cur) - 1 isPalindrome = True while l < r: if cur[l] != cur[r]: isPalindrome = False break l += 1 r -= 1 # if cur is palindrome, # insert it into the set if isPalindrome: result.add(cur) # Convert the set to a vector res = list(result) return res if __name__ == "__main__": str = "abaaa" result = palindromicSubstr(str) for s in result: print(s, end=" ")
// Function to find all distinct palindrome // substrings in a given string using System; using System.Collections.Generic; class GfG { // Function to find all distinct palindrome // substrings in a given string static List<string> PalindromicSubstr(string str) { int n = str.Length; // Create a set to store the result HashSet<string> result = new HashSet<string>(); // generate all substrings for (int i = 0; i < n; i++) { // to store the substring string cur = ""; for (int j = i; j < n; j++) { cur += str[j]; // check if cur is palindrome int l = 0, r = cur.Length - 1; bool isPalindrome = true; while (l < r) { if (cur[l] != cur[r]) { isPalindrome = false; break; } l++; r--; } // if cur is palindrome, // insert it into the set if (isPalindrome) { result.Add(cur); } } } // Convert the set to a list List<string> res = new List<string>(result); return res; } public static void Main(string[] args) { string str = "abaaa"; List<string> result = PalindromicSubstr(str); foreach (string s in result) Console.Write(s + " "); } } // Function to find all distinct palindrome // substrings in a given string. function palindromicSubstr(str) { let n = str.length; // Create a set to store the result let result = new Set(); // generate all substrings for (let i = 0; i < n; i++) { // to store the substring let cur = ""; for (let j = i; j < n; j++) { cur += str[j]; // check if cur is palindrome let l = 0, r = cur.length - 1; let isPalindrome = true; while (l < r) { if (cur[l] !== cur[r]) { isPalindrome = false; break; } l++; r--; } // if cur is palindrome, // insert it into the set if (isPalindrome) { result.add(cur); } } } // Convert the set to a vector let res = Array.from(result); return res; } let str = "abaaa"; let result = palindromicSubstr(str); console.log(result.join(" ")); [Naive Approach – 2] – Using Dynamic Programming and Set – O(n ^ 3) Time and O(n ^ 2) Space
The idea is to use Dynamic Programming to find all palindromic substrings and set to store all the distinct one. To do so, create a 2d array dp[][] of order n * n, where dp[i][j] is True if substring s[i..j] is palindrome. Any substring s[i..j] will be palindromic if s[i] == s[j] and dp[i+1][j-1] is True. Thus run a nested loop where the outer loop marks the start and inner loop marks the end of substring.
Follow the below given steps:
- Declare a Boolean 2-D array and fill it diagonally.
- Now, check for every possible length, i.e. 0, 1, 2, and so on.
- If gap=0, that means there is only one element and we put true since a single character is palindrome
- if gap = 1, we check whether extremes are same and if so we put true, else false;
- Else for any other value of gap, we check if extremes are same and dp[i+1][j-1] yields true, if so then we put true, else false.
- Each time when a true is encountered, add the string in the set data structure.
- At last, store all the strings in the set, in an array, and return it.
C++ #include <bits/stdc++.h> using namespace std; // Function to find all distinct palindrome // substrings in a given string vector<string> palindromicSubstr(string &str) { int n = str.length(); // Create a 2D array vector<vector<bool>> dp(n, vector<bool>(n, false)); // Create a set to store the result unordered_set<string> result; // Traverse the string for (int gap = 0; gap < n; gap++) { for (int i = 0, j = gap; j < n; i++, j++) { // If the gap is 0, it is a palindrome if (gap == 0) dp[i][j] = true; // If the gap is 1, check if // the characters are the same else if (gap == 1) dp[i][j] = (str[i] == str[j]); // check if the characters are the same // and the inner substring is a palindrome else dp[i][j] = (str[i] == str[j] && dp[i + 1][j - 1]); // If the substring is a palindrome // insert it into the set if (dp[i][j]) result.insert(str.substr(i, j - i + 1)); } } // Convert the set to a vector vector<string> res(result.begin(), result.end()); return res; } int main() { string str = "abaaa"; vector<string> result = palindromicSubstr(str); for(string s : result) cout << s << " "; return 0; } // Function to find all distinct palindrome // substrings in a given string import java.util.*; class GfG { // Function to find all distinct palindrome // substrings in a given string static ArrayList<String> palindromicSubstr(String str) { int n = str.length(); // Create a 2D array boolean[][] dp = new boolean[n][n]; // Create a set to store the result HashSet<String> result = new HashSet<>(); // Traverse the string for (int gap = 0; gap < n; gap++) { for (int i = 0, j = gap; j < n; i++, j++) { // If the gap is 0, it is a palindrome if (gap == 0) dp[i][j] = true; // If the gap is 1, check if // the characters are the same else if (gap == 1) dp[i][j] = (str.charAt(i) == str.charAt(j)); // check if the characters are the same // and the inner substring is a palindrome else dp[i][j] = (str.charAt(i) == str.charAt(j) && dp[i + 1][j - 1]); // If the substring is a palindrome // insert it into the set if (dp[i][j]) result.add(str.substring(i, j + 1)); } } // Convert the set to a vector ArrayList<String> res = new ArrayList<>(result); return res; } public static void main(String[] args) { String str = "abaaa"; ArrayList<String> result = palindromicSubstr(str); for (String s : result) System.out.print(s + " "); } } # Function to find all distinct palindrome # substrings in a given string. def palindromicSubstr(str): n = len(str) # Create a 2D array dp = [[False for _ in range(n)] for _ in range(n)] # Create a set to store the result result = set() # Traverse the string for gap in range(n): for i in range(0, n - gap): j = i + gap # If the gap is 0, it is a palindrome if gap == 0: dp[i][j] = True # If the gap is 1, check if # the characters are the same elif gap == 1: dp[i][j] = (str[i] == str[j]) # check if the characters are the same # and the inner substring is a palindrome else: dp[i][j] = (str[i] == str[j] and dp[i + 1][j - 1]) # If the substring is a palindrome # insert it into the set if dp[i][j]: result.add(str[i:j+1]) # Convert the set to a vector res = list(result) return res if __name__ == "__main__": str = "abaaa" result = palindromicSubstr(str) for s in result: print(s, end=" ")
// Function to find all distinct palindrome // substrings in a given string using System; using System.Collections.Generic; class GfG { // Function to find all distinct palindrome // substrings in a given string static List<string> PalindromicSubstr(string str) { int n = str.Length; // Create a 2D array bool[,] dp = new bool[n, n]; // Create a set to store the result HashSet<string> result = new HashSet<string>(); // Traverse the string for (int gap = 0; gap < n; gap++) { for (int i = 0, j = gap; j < n; i++, j++) { // If the gap is 0, it is a palindrome if (gap == 0) dp[i, j] = true; // If the gap is 1, check if // the characters are the same else if (gap == 1) dp[i, j] = (str[i] == str[j]); // check if the characters are the same // and the inner substring is a palindrome else dp[i, j] = (str[i] == str[j] && dp[i + 1, j - 1]); // If the substring is a palindrome // insert it into the set if (dp[i, j]) result.Add(str.Substring(i, j - i + 1)); } } // Convert the set to a list List<string> res = new List<string>(result); return res; } public static void Main(string[] args) { string str = "abaaa"; List<string> result = PalindromicSubstr(str); foreach (string s in result) Console.Write(s + " "); } } // Function to find all distinct palindrome // substrings in a given string. function palindromicSubstr(str) { let n = str.length; // Create a 2D array let dp = new Array(n); for (let i = 0; i < n; i++) { dp[i] = new Array(n).fill(false); } // Create a set to store the result let result = new Set(); // Traverse the string for (let gap = 0; gap < n; gap++) { for (let i = 0, j = gap; j < n; i++, j++) { // If the gap is 0, it is a palindrome if (gap === 0) dp[i][j] = true; // If the gap is 1, check if // the characters are the same else if (gap === 1) dp[i][j] = (str[i] === str[j]); // check if the characters are the same // and the inner substring is a palindrome else dp[i][j] = (str[i] === str[j] && dp[i + 1][j - 1]); // If the substring is a palindrome // insert it into the set if (dp[i][j]) result.add(str.substring(i, j + 1)); } } // Convert the set to a vector let res = Array.from(result); return res; } let str = "abaaa"; let result = palindromicSubstr(str); console.log(result.join(" ")); [Better Approach] – Using Manacher’s Algorithm and Hash Set – O(n ^ 2) Time and O(n ^ 2) Space
The idea is to use Manacher’s Algorithm to find all palindromic substrings and set to store all the distinct one.
Follow the below given steps:
- Consider each character as a pivot and expand on both sides to find the longest odd-length and even-length palindromes centered at that character; store these lengths in two separate arrays.
- For every pivot, generate the actual palindromic substrings using the computed lengths.
- Insert each found palindrome, along with all single characters, into a set so that only distinct palindromic substrings are kept.
- Finally, store all the palindromes stored in the set in an array and return it.
C++ #include <bits/stdc++.h> using namespace std; // Function to find all distinct palindrome // substrings in a given string vector<string> palindromicSubstr(string &str) { int n = str.length(); // Create a set to store the result unordered_set<string> result; // table for storing results of // odd even length subarrays vector<vector<int>> dp(2, vector<int>(n + 1, 0)); // store all the palindromic substrings // of length 1 in the set for (int i = 0; i < n; i++) result.insert(str.substr(i, 1)); // insert 'guards' to iterate easily over str str = "@" + str + "#"; for (int j = 0; j <= 1; j++) { // length of 'palindrome radius' int rp = 0; dp[j][0] = 0; int i = 1; while (i <= n) { // expand palindrome centered at i while (str[i - rp - 1] == str[i + j + rp]) rp++; // Assign the found palindromic length // to odd/even length array dp[j][i] = rp; int k = 1; while ((dp[j][i - k] != rp - k) && (k < rp)) { dp[j][i + k] = min(dp[j][i - k], rp - k); k++; } // reset the length of palindromic radius rp = max(rp - k, 0); i += k; } } // remove 'guards' str = str.substr(1, n); // store the results in a set for (int i = 1; i <= n; i++) { for (int j = 0; j <= 1; j++) { for (int rp = dp[j][i]; rp > 0; rp--) { result.insert(str.substr(i - rp - 1, 2 * rp + j)); } } } // Convert the set to a vector vector<string> res(result.begin(), result.end()); return res; } int main() { string str = "abaaa"; vector<string> result = palindromicSubstr(str); for(string str : result) cout << str << " "; return 0; } // Function to find all distinct palindrome // substrings in a given string import java.util.*; class GfG { // Function to find all distinct palindrome // substrings in a given string static ArrayList<String> palindromicSubstr(String str) { int n = str.length(); // Create a set to store the result HashSet<String> result = new HashSet<>(); // table for storing results of // odd even length subarrays int[][] dp = new int[2][n + 1]; // store all the palindromic substrings // of length 1 in the set for (int i = 0; i < n; i++) result.add(str.substring(i, i + 1)); // insert 'guards' to iterate easily over str str = "@" + str + "#"; for (int j = 0; j <= 1; j++) { // length of 'palindrome radius' int rp = 0; dp[j][0] = 0; int i = 1; while (i <= n) { // expand palindrome centered at i while (str.charAt(i - rp - 1) == str.charAt(i + j + rp)) rp++; // Assign the found palindromic length // to odd/even length array dp[j][i] = rp; int k = 1; while ((dp[j][i - k] != rp - k) && (k < rp)) { dp[j][i + k] = Math.min(dp[j][i - k], rp - k); k++; } // reset the length of palindromic radius rp = Math.max(rp - k, 0); i += k; } } // remove 'guards' str = str.substring(1, n + 1); // store the results in a set for (int i = 1; i <= n; i++) { for (int j = 0; j <= 1; j++) { for (int rp = dp[j][i]; rp > 0; rp--) { result.add(str.substring(i - rp - 1, i - rp - 1 + 2 * rp + j)); } } } // Convert the set to a vector ArrayList<String> res = new ArrayList<>(result); return res; } public static void main(String[] args) { String str = "abaaa"; ArrayList<String> result = palindromicSubstr(str); for (String s : result) System.out.print(s + " "); } } # Function to find all distinct palindrome # substrings in a given string. def palindromicSubstr(str): n = len(str) # Create a set to store the result result = set() # table for storing results of # odd even length subarrays dp = [[0]*(n+1) for _ in range(2)] # store all the palindromic substrings # of length 1 in the set for i in range(n): result.add(str[i:i+1]) # insert 'guards' to iterate easily over str str = "@" + str + "#" for j in range(2): # length of 'palindrome radius' rp = 0 dp[j][0] = 0 i = 1 while i <= n: # expand palindrome centered at i while str[i - rp - 1] == str[i + j + rp]: rp += 1 # Assign the found palindromic length # to odd/even length array dp[j][i] = rp k = 1 while (dp[j][i - k] != rp - k) and (k < rp): dp[j][i + k] = min(dp[j][i - k], rp - k) k += 1 # reset the length of palindromic radius rp = max(rp - k, 0) i += k # remove 'guards' str = str[1:n+1] # store the results in a set for i in range(1, n+1): for j in range(2): for rp in range(dp[j][i], 0, -1): result.add(str[i - rp - 1: i - rp - 1 + 2 * rp + j]) # Convert the set to a vector res = list(result) return res if __name__ == "__main__": str = "abaaa" result = palindromicSubstr(str) for s in result: print(s, end=" ")
// Function to find all distinct palindrome // substrings in a given string using System; using System.Collections.Generic; class GfG { // Function to find all distinct palindrome // substrings in a given string static List<string> PalindromicSubstr(string str) { int n = str.Length; // Create a set to store the result HashSet<string> result = new HashSet<string>(); // table for storing results of // odd even length subarrays int[,] dp = new int[2, n + 1]; // store all the palindromic substrings // of length 1 in the set for (int i = 0; i < n; i++) result.Add(str.Substring(i, 1)); // insert 'guards' to iterate easily over str str = "@" + str + "#"; for (int j = 0; j <= 1; j++) { // length of 'palindrome radius' int rp = 0; dp[j, 0] = 0; int i = 1; while (i <= n) { // expand palindrome centered at i while (str[i - rp - 1] == str[i + j + rp]) rp++; // Assign the found palindromic length // to odd/even length array dp[j, i] = rp; int k = 1; while ((dp[j, i - k] != rp - k) && (k < rp)) { dp[j, i + k] = Math.Min(dp[j, i - k], rp - k); k++; } // reset the length of palindromic radius rp = Math.Max(rp - k, 0); i += k; } } // remove 'guards' str = str.Substring(1, n); // store the results in a set for (int i = 1; i <= n; i++) { for (int j = 0; j <= 1; j++) { for (int rp = dp[j, i]; rp > 0; rp--) { result.Add(str.Substring(i - rp - 1, 2 * rp + j)); } } } // Convert the set to a list List<string> res = new List<string>(result); return res; } public static void Main(string[] args) { string str = "abaaa"; List<string> result = PalindromicSubstr(str); foreach (string s in result) Console.Write(s + " "); } } // Function to find all distinct palindrome // substrings in a given string. function palindromicSubstr(str) { let n = str.length; // Create a set to store the result let result = new Set(); // table for storing results of // odd even length subarrays let dp = [new Array(n + 1).fill(0), new Array(n + 1).fill(0)]; // store all the palindromic substrings // of length 1 in the set for (let i = 0; i < n; i++) result.add(str.substring(i, i + 1)); // insert 'guards' to iterate easily over str str = "@" + str + "#"; for (let j = 0; j <= 1; j++) { // length of 'palindrome radius' let rp = 0; dp[j][0] = 0; let i = 1; while (i <= n) { // expand palindrome centered at i while (str[i - rp - 1] === str[i + j + rp]) rp++; // Assign the found palindromic length // to odd/even length array dp[j][i] = rp; let k = 1; while ((dp[j][i - k] !== rp - k) && (k < rp)) { dp[j][i + k] = Math.min(dp[j][i - k], rp - k); k++; } // reset the length of palindromic radius rp = Math.max(rp - k, 0); i += k; } } // remove 'guards' str = str.substring(1, n + 1); // store the results in a set for (let i = 1; i <= n; i++) { for (let j = 0; j <= 1; j++) { for (let rp = dp[j][i]; rp > 0; rp--) { result.add(str.substring(i - rp - 1, i - rp - 1 + 2 * rp + j)); } } } // Convert the set to a vector let res = Array.from(result); return res; } let str = "abaaa"; let result = palindromicSubstr(str); console.log(result.join(" ")); [Expected Approach] – Using Dynamic Programming and KMP Algorithm – O(n^2) Time and O(n^2) Space
The idea is to identify all palindromic substrings in a given string using a dynamic programming method, then eliminate duplicates by leveraging the KMP algorithm, and finally print the distinct palindromes along with their count.
Follow the below given steps:
- First, for every possible substring, use dynamic programming to check if it is a palindrome.
- Next, for every index starting from 0, apply the KMP algorithm to compare prefixes and suffixes. If a substring is both a palindrome and its prefix matches its suffix, mark it (for example, by setting the corresponding dp value to false) so that duplicate occurrences are removed.
- Finally, iterate over all substrings and print those that remain marked as palindromic in the dp array; the number of such entries gives the total count of distinct palindromic substrings.
C++ #include <bits/stdc++.h> using namespace std; // Function to find all distinct palindrome // substrings in a given string vector<string> palindromicSubstr(string &str) { int n = str.length(); // Create a 2D array vector<vector<bool>> 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 && str[i] == str[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 (str[i] == str[i + (len - 1)] && dp[i + 1][i + (len - 1) - 1]) { dp[i][i + (len - 1)] = true; } } } // create an array of size n // to operate the kmp algorithm 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 (str[j + i] == str[k + i]) { // make suffix to be false, if this suffix is // palindrome then it is 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; } } } // Create an array to store the result vector<string> result; for (int i = 0; i < n; i++) { // to store the current string string cur; for (int j = i; j < n; j++) { cur += str[j]; if (dp[i][j]) { result.push_back(cur); } } } return result; } int main() { string str = "abaaa"; vector<string> result = palindromicSubstr(str); for(string s : result) cout << s << " "; return 0; } // Function to find all distinct palindrome // substrings in a given string import java.util.*; class GfG { // Function to find all distinct palindrome // substrings in a given string static ArrayList<String> palindromicSubstr(String str) { int n = str.length(); // Create a 2D array 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 && str.charAt(i) == str.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 (str.charAt(i) == str.charAt(i + len - 1) && dp[i + 1][i + len - 2]) { dp[i][i + len - 1] = true; } } } // create an array of size n // to operate the kmp algorithm int[] kmp = new int[n]; Arrays.fill(kmp, 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 (str.charAt(j + i) == str.charAt(k + i)) { // make suffix to be false, if this suffix is // palindrome then it is 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; } } } // Create an array to store the result ArrayList<String> result = new ArrayList<>(); for (int i = 0; i < n; i++) { // to store the current string String cur = ""; for (int j = i; j < n; j++) { cur += str.charAt(j); if (dp[i][j]) { result.add(cur); } } } return result; } public static void main(String[] args) { String str = "abaaa"; ArrayList<String> result = palindromicSubstr(str); for (String s : result) System.out.print(s + " "); } } # Function to find all distinct palindrome # substrings in a given string. def palindromicSubstr(str): n = len(str) # Create a 2D array dp = [[False for _ in range(n)] for _ 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 str[i] == str[i + 1]: dp[i][i + 1] = True # check every substring of length # greater than 2 for palindrome for length in range(3, n + 1): for i in range(n - length + 1): j = i + length - 1 if str[i] == str[j] and dp[i + 1][j - 1]: dp[i][j] = True # create an array of size n # to operate the kmp algorithm kmp = [0] * n for i in range(n): # starting kmp for every i from 0 to n-1 j = 0 k = 1 while k + i < n: if str[i + j] == str[i + k]: # make suffix to be false, if this suffix is # palindrome then it is included in prefix dp[i + k - j][i + k] = False kmp[k] = j + 1 j += 1 k += 1 elif j > 0: j = kmp[j - 1] else: kmp[k] = 0 k += 1 # Create an array to store the result result = [] for i in range(n): # to store the current string cur = "" for j in range(i, n): cur += str[j] if dp[i][j]: result.append(cur) return result if __name__ == "__main__": str = "abaaa" result = palindromicSubstr(str) for s in result: print(s, end=" ")
// Function to find all distinct palindrome // substrings in a given string using System; using System.Collections.Generic; class GfG { // Function to find all distinct palindrome // substrings in a given string. static List<string> palindromicSubstr(ref string str) { int n = str.Length; // Create a 2D array bool[,] dp = new bool[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 && str[i] == str[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++) { int j = i + len - 1; if (str[i] == str[j] && dp[i + 1, j - 1]) dp[i, j] = true; } } // create an array of size n // to operate the kmp algorithm int[] kmp = new int[n]; for (int i = 0; i < n; i++) { kmp[i] = 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 (str[i + j] == str[i + k]) { // make suffix to be false, if this suffix is // palindrome then it is included in prefix dp[i + k - j, i + k] = false; kmp[k] = ++j; k++; } else if (j > 0) { j = kmp[j - 1]; } else { kmp[k] = 0; k++; } } } // Create an array to store the result List<string> result = new List<string>(); for (int i = 0; i < n; i++) { // to store the current string string cur = ""; for (int j = i; j < n; j++) { cur += str[j]; if (dp[i, j]) result.Add(cur); } } return result; } static void Main() { string str = "abaaa"; List<string> result = palindromicSubstr(ref str); foreach (string s in result) Console.Write(s + " "); } } // Function to find all distinct palindrome // substrings in a given string. function palindromicSubstr(str) { let n = str.length; // Create a 2D array let dp = new Array(n); for (let i = 0; i < n; i++) { dp[i] = new Array(n).fill(false); } for (let 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 && str[i] === str[i + 1]) dp[i][i + 1] = true; } // 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++) { let j = i + len - 1; if (str[i] === str[j] && dp[i + 1][j - 1]) dp[i][j] = true; } } // create an array of size n // to operate the kmp algorithm 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 (str[i + j] === str[i + k]) { // make suffix to be false, if this suffix is // palindrome then it is included in prefix dp[i + k - j][i + k] = false; kmp[k] = ++j; k++; } else if (j > 0) { j = kmp[j - 1]; } else { kmp[k] = 0; k++; } } } // Create an array to store the result let result = []; for (let i = 0; i < n; i++) { // to store the current string let cur = ""; for (let j = i; j < n; j++) { cur += str[j]; if (dp[i][j]) result.push(cur); } } return result; } let str = "abaaa"; let result = palindromicSubstr(str); console.log(result.join(" ")); [Alternate Approach] – Using Center Expansion and Set – O(n ^ 2) Time and O(n ^ 2) Space
The idea is to use center expansion technique to find all even and odd length palindromic substrings separately, and then store them in a set to find the unique palindromic substrings.
C++ #include <bits/stdc++.h> using namespace std; // Function to find all distinct palindrome // substrings in a given string vector<string> palindromicSubstr(string &str) { int n = str.length(); // Create a set to store the result unordered_set<string> result; // Check for odd length palindromes for (int i = 0; i < n; i++) { int left = i, right = i; while (left >= 0 && right < n && str[left] == str[right]) { // Add the palindrome substring result.insert(str.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 && str[left] == str[right]) { // Add the palindrome substring result.insert(str.substr(left, right - left + 1)); left--, right++; } } // Convert the set to a vector vector<string> res(result.begin(), result.end()); return res; } int main() { string str = "abaaa"; vector<string> result = palindromicSubstr(str); for(string str : result) cout << str << " "; return 0; } // Function to find all distinct palindrome // substrings in a given string import java.util.*; class GfG { // Function to find all distinct palindrome // substrings in a given string static ArrayList<String> palindromicSubstr(String str) { int n = str.length(); // Create a set to store the result HashSet<String> result = new HashSet<>(); // Check for odd length palindromes for (int i = 0; i < n; i++) { int left = i, right = i; while (left >= 0 && right < n && str.charAt(left) == str.charAt(right)) { // Add the palindrome substring result.add(str.substring(left, right + 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 && str.charAt(left) == str.charAt(right)) { // Add the palindrome substring result.add(str.substring(left, right + 1)); left--; right++; } } // Convert the set to a vector ArrayList<String> res = new ArrayList<>(result); return res; } public static void main(String[] args) { String str = "abaaa"; ArrayList<String> result = palindromicSubstr(str); for (String s : result) System.out.print(s + " "); } } from __future__ import print_function # Function to find all distinct palindrome # substrings in a given string. def palindromicSubstr(str): n = len(str) # Create a set to store the result result = set() # Check for odd length palindromes for i in range(n): left = i right = i while left >= 0 and right < n and str[left] == str[right]: # Add the palindrome substring result.add(str[left:right+1]) left -= 1 right += 1 # Check for even length palindromes for i in range(n - 1): left = i right = i + 1 while left >= 0 and right < n and str[left] == str[right]: # Add the palindrome substring result.add(str[left:right+1]) left -= 1 right += 1 # Convert the set to a vector res = list(result) return res if __name__ == "__main__": str = "abaaa" result = palindromicSubstr(str) for s in result: print(s, end=" ")
// Function to find all distinct palindrome // substrings in a given string using System; using System.Collections.Generic; using System.Linq; class GfG { // Function to find all distinct palindrome // substrings in a given string static List<string> PalindromicSubstr(string str) { int n = str.Length; // Create a set to store the result HashSet<string> result = new HashSet<string>(); // Check for odd length palindromes for (int i = 0; i < n; i++) { int left = i, right = i; while (left >= 0 && right < n && str[left] == str[right]) { // Add the palindrome substring result.Add(str.Substring(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 && str[left] == str[right]) { // Add the palindrome substring result.Add(str.Substring(left, right - left + 1)); left--; right++; } } // Convert the set to a list List<string> res = result.ToList(); return res; } public static void Main(string[] args) { string str = "abaaa"; List<string> result = PalindromicSubstr(str); foreach (string s in result) Console.Write(s + " "); } } // Function to find all distinct palindrome // substrings in a given string. function palindromicSubstr(str) { let n = str.length; // Create a set to store the result let result = new Set(); // Check for odd length palindromes for (let i = 0; i < n; i++) { let left = i, right = i; while (left >= 0 && right < n && str[left] === str[right]) { // Add the palindrome substring result.add(str.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 && str[left] === str[right]) { // Add the palindrome substring result.add(str.substring(left, right + 1)); left--; right++; } } // Convert the set to a vector let res = Array.from(result); return res; } let str = "abaaa"; let result = palindromicSubstr(str); console.log(result.join(" ")); Related Article:
Count All Palindrome Sub-Strings in a String
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