Given a string S of length N, the task is to find the length of the longest palindromic substring from a given string.
Examples:
Input: S = “abcbab”
Output: 5
Explanation:
string “abcba” is the longest substring that is a palindrome which is of length 5.Input: S = “abcdaa”
Output: 2
Explanation:
string “aa” is the longest substring that is a palindrome which is of length 2.
Naive Approach: The simplest approach to solve the problem is to generate all possible substrings of the given string and print the length of the longest substring which is a palindrome.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to obtain the length of// the longest palindromic substringint longestPalSubstr(string str){ // Length of given string int n = str.size(); // Stores the maximum length int maxLength = 1, start = 0; // Iterate over the string for (int i = 0; i < str.length(); i++) { // Iterate over the string for (int j = i; j < str.length(); j++) { int flag = 1; // Check for palindrome for (int k = 0; k < (j - i + 1) / 2; k++) if (str[i + k] != str[j - k]) flag = 0; // If string [i, j - i + 1] // is palindromic if (flag && (j - i + 1) > maxLength) { start = i; maxLength = j - i + 1; } } } // Return length of LPS return maxLength;}// Driver Codeint main(){ // Given string string str = "forneveropenskeegfor"; // Function Call cout << longestPalSubstr(str); return 0;} |
Java
// Java program for the above approachimport java.io.*;class GFG{ // Function to obtain the length of// the longest palindromic substringstatic int longestPalSubstr(String str){ // Length of given string int n = str.length(); // Stores the maximum length int maxLength = 1, start = 0; // Iterate over the string for(int i = 0; i < str.length(); i++) { // Iterate over the string for(int j = i; j < str.length(); j++) { int flag = 1; // Check for palindrome for(int k = 0; k < (j - i + 1) / 2; k++) if (str.charAt(i + k) != str.charAt(j - k)) flag = 0; // If string [i, j - i + 1] // is palindromic if (flag != 0 && (j - i + 1) > maxLength) { start = i; maxLength = j - i + 1; } } } // Return length of LPS return maxLength;} // Driver Codepublic static void main (String[] args) { // Given string String str = "forneveropenskeegfor"; // Function call System.out.print(longestPalSubstr(str));}}// This code is contributed by code_hunt |
Python3
# Python3 program for the above approach# Function to obtain the length of# the longest palindromic substringdef longestPalSubstr(str): # Length of given string n = len(str) # Stores the maximum length maxLength = 1 start = 0 # Iterate over the string for i in range(len(str)): # Iterate over the string for j in range(i, len(str), 1): flag = 1 # Check for palindrome for k in range((j - i + 1) // 2): if (str[i + k] != str[j - k]): flag = 0 # If string [i, j - i + 1] # is palindromic if (flag != 0 and (j - i + 1) > maxLength): start = i maxLength = j - i + 1 # Return length of LPS return maxLength# Driver Code# Given stringstr = "forneveropenskeegfor" # Function callprint(longestPalSubstr(str))# This code is contributed by code_hunt |
C#
// C# program for the above approach using System;class GFG{ // Function to obtain the length of// the longest palindromic substringstatic int longestPalSubstr(string str){ // Length of given string int n = str.Length; // Stores the maximum length int maxLength = 1, start = 0; // Iterate over the string for(int i = 0; i < str.Length; i++) { // Iterate over the string for(int j = i; j < str.Length; j++) { int flag = 1; // Check for palindrome for(int k = 0; k < (j - i + 1) / 2; k++) if (str[i + k] != str[j - k]) flag = 0; // If string [i, j - i + 1] // is palindromic if (flag != 0 && (j - i + 1) > maxLength) { start = i; maxLength = j - i + 1; } } } // Return length of LPS return maxLength;} // Driver Codepublic static void Main () { // Given string string str = "forneveropenskeegfor"; // Function call Console.Write(longestPalSubstr(str));}}// This code is contributed by code_hunt |
Javascript
<script>// JavaScript program for the above approach// Function to obtain the length of// the longest palindromic substringfunction longestPalSubstr(str){ // Length of given string var n = str.length; // Stores the maximum length var maxLength = 1, start = 0; // Iterate over the string for (var i = 0; i < str.length; i++) { // Iterate over the string for (var j = i; j < str.length; j++) { var flag = 1; // Check for palindrome for (var k = 0; k < (j - i + 1) / 2; k++) if (str[i + k] != str[j - k]) flag = 0; // If string [i, j - i + 1] // is palindromic if (flag && (j - i + 1) > maxLength) { start = i; maxLength = j - i + 1; } } } // Return length of LPS return maxLength;}// Driver Code// Given stringvar str = "forneveropenskeegfor";// Function Calldocument.write( longestPalSubstr(str));</script> |
Output:
10
Time Complexity: O(N3), where N is the length of the given string.
Auxiliary Space: O(N)
Dynamic Programming Approach: The above approach can be optimized by storing results of Overlapping Subproblems. The idea is similar to this post. Below are the steps:
- Maintain a boolean table[N][N] that is filled in a bottom-up manner.
- The value of table[i][j] is true if the substring is a palindrome, otherwise false.
- To calculate table[i][j], check the value of table[i + 1][j – 1], if the value is true and str[i] is same as str[j], then update table[i][j] true.
- Otherwise, the value of table[i][j] is update as false.
Below is the illustration for the string “neveropen”:
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the length of// the longest palindromic substringint longestPalSubstr(string str){ // Length of string str int n = str.size(); // Stores the dp states bool table[n][n]; // Initialise table[][] as false memset(table, 0, sizeof(table)); // All substrings of length 1 // are palindromes int maxLength = 1; for (int i = 0; i < n; ++i) table[i][i] = true; // Check for sub-string of length 2 int start = 0; for (int i = 0; i < n - 1; ++i) { // If adjacent character are same if (str[i] == str[i + 1]) { // Update table[i][i + 1] table[i][i + 1] = true; start = i; maxLength = 2; } } // Check for lengths greater than 2 // k is length of substring for (int k = 3; k <= n; ++k) { // Fix the starting index for (int i = 0; i < n - k + 1; ++i) { // Ending index of substring // of length k int j = i + k - 1; // Check for palindromic // substring str[i, j] if (table[i + 1][j - 1] && str[i] == str[j]) { // Mark true table[i][j] = true; // Update the maximum length if (k > maxLength) { start = i; maxLength = k; } } } } // Return length of LPS return maxLength;}// Driver Codeint main(){ // Given string str string str = "forneveropenskeegfor"; // Function Call cout << longestPalSubstr(str); return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{// Function to find the length of// the longest palindromic subStringstatic int longestPalSubstr(String str){ // Length of String str int n = str.length(); // Stores the dp states boolean [][]table = new boolean[n][n]; // All subStrings of length 1 // are palindromes int maxLength = 1; for(int i = 0; i < n; ++i) table[i][i] = true; // Check for sub-String of length 2 int start = 0; for(int i = 0; i < n - 1; ++i) { // If adjacent character are same if (str.charAt(i) == str.charAt(i + 1)) { // Update table[i][i + 1] table[i][i + 1] = true; start = i; maxLength = 2; } } // Check for lengths greater than 2 // k is length of subString for(int k = 3; k <= n; ++k) { // Fix the starting index for(int i = 0; i < n - k + 1; ++i) { // Ending index of subString // of length k int j = i + k - 1; // Check for palindromic // subString str[i, j] if (table[i + 1][j - 1] && str.charAt(i) == str.charAt(j)) { // Mark true table[i][j] = true; // Update the maximum length if (k > maxLength) { start = i; maxLength = k; } } } } // Return length of LPS return maxLength;}// Driver Codepublic static void main(String[] args){ // Given String str String str = "forneveropenskeegfor"; // Function Call System.out.print(longestPalSubstr(str));}}// This code is contributed by Amit Katiyar |
C#
// C# program for // the above approachusing System;class GFG{// Function to find the length of// the longest palindromic subStringstatic int longestPalSubstr(String str){ // Length of String str int n = str.Length; // Stores the dp states bool [,]table = new bool[n, n]; // All subStrings of length 1 // are palindromes int maxLength = 1; for(int i = 0; i < n; ++i) table[i, i] = true; // Check for sub-String // of length 2 int start = 0; for(int i = 0; i < n - 1; ++i) { // If adjacent character are same if (str[i] == str[i + 1]) { // Update table[i,i + 1] table[i, i + 1] = true; start = i; maxLength = 2; } } // Check for lengths greater than 2 // k is length of subString for(int k = 3; k <= n; ++k) { // Fix the starting index for(int i = 0; i < n - k + 1; ++i) { // Ending index of subString // of length k int j = i + k - 1; // Check for palindromic // subString str[i, j] if (table[i + 1, j - 1] && str[i] == str[j]) { // Mark true table[i, j] = true; // Update the maximum length if (k > maxLength) { start = i; maxLength = k; } } } } // Return length of LPS return maxLength;}// Driver Codepublic static void Main(String[] args){ // Given String str String str = "forneveropenskeegfor"; // Function Call Console.Write(longestPalSubstr(str));}}// This code is contributed by Rajput-Ji |
Python3
# Python program for the above approach# Function to find the length of# the longest palindromic subStringdef longestPalSubstr(str): # Length of String str n = len(str); # Stores the dp states table = [[False for i in range(n)] for j in range(n)]; # All subStrings of length 1 # are palindromes maxLength = 1; for i in range(n): table[i][i] = True; # Check for sub-String of length 2 start = 0; for i in range(n - 1): # If adjacent character are same if (str[i] == str[i + 1]): # Update table[i][i + 1] table[i][i + 1] = True; start = i; maxLength = 2; # Check for lengths greater than 2 # k is length of subString for k in range(3, n + 1): # Fix the starting index for i in range(n - k + 1): # Ending index of subString # of length k j = i + k - 1; # Check for palindromic # subString str[i, j] if (table[i + 1][j - 1] and str[i] == str[j]): # Mark True table[i][j] = True; # Update the maximum length if (k > maxLength): start = i; maxLength = k; # Return length of LPS return maxLength;# Driver Codeif __name__ == '__main__': # Given String str str = "forneveropenskeegfor"; # Function Call print(longestPalSubstr(str));# This code is contributed by 29AjayKumar |
Javascript
<script>// javascript program for the above approach // Function to find the length of // the longest palindromic subString function longestPalSubstr(str) { // Length of String str var n = str.length; // Stores the dp states var table = Array(n).fill().map(()=>Array(n).fill(false)); // All subStrings of length 1 // are palindromes var maxLength = 1; for (var i = 0; i < n; ++i) table[i][i] = true; // Check for sub-String of length 2 var start = 0; for (i = 0; i < n - 1; ++i) { // If adjacent character are same if (str.charAt(i) == str.charAt(i + 1)) { // Update table[i][i + 1] table[i][i + 1] = true; start = i; maxLength = 2; } } // Check for lengths greater than 2 // k is length of subString for (k = 3; k <= n; ++k) { // Fix the starting index for (i = 0; i < n - k + 1; ++i) { // Ending index of subString // of length k var j = i + k - 1; // Check for palindromic // subString str[i, j] if (table[i + 1][j - 1] && str.charAt(i) == str.charAt(j)) { // Mark true table[i][j] = true; // Update the maximum length if (k > maxLength) { start = i; maxLength = k; } } } } // Return length of LPS return maxLength; } // Driver Code // Given String str var str = "forneveropenskeegfor"; // Function Call document.write(longestPalSubstr(str));// This code is contributed by umadevi9616</script> |
Output:
10
Time Complexity: O(N2), where N is the length of the given string.
Auxiliary Space: O(N)
Efficient Approach: To optimize the above approach, the idea is to use Manacher’s Algorithm. By using this algorithm, for each character c, the longest palindromic substring that has c as its center can be found whose length is odd. But the longest palindromic substring can also have an even length which does not have any center. Therefore, some special characters can be added between each character.
For example, if the given string is “abababc” then it will become “$#a#b#a#b#a#b#c#@”. Now, notice that in this case, for each character c, the longest palindromic substring with the center c will have an odd length.
Below are the steps:
- Add the special characters in the given string S as explained above and let its length be N.
- Initialize an array d[], center, and r with 0 where d[i] stores the length of the left part of the palindrome where S[i] is the center, r denotes the rightmost visited boundary and center denotes the current index of character which is the center of this rightmost boundary.
- While traversing the string S, for each index i, if i is smaller than r then its answer has previously been calculated and d[i] can be set equals to answer for the mirror of character at i with the center which can be calculated as (2*center – i).
- Now, check if there are some characters after r such that the palindrome becomes ever longer.
- If (i + d[i]) is greater than r, update r = (i + d[i]) and center as i.
- After finding the longest palindrome for every character c as the center, print the maximum value of (2*d[i] + 1)/2 where 0 ? i < N because d[i] only stores the left part of the palindrome.
Below is the implementation for the above approach:
C++14
// C++ program for the above approach:#include <bits/stdc++.h>using namespace std;// Function that placed '#' intermediately// before and after each characterstring UpdatedString(string s){ string newString = "#"; // Traverse the string for(auto ch : s){ newString += ch; newString += "#"; } // Return the string return newString;}// Function that finds the length of// the longest palindromic substringint Manacher(string s){ // Update the string s = UpdatedString(s); // Stores the longest proper prefix // which is also a suffix int LPS[s.length()] = {}; int C = 0; int R = 0; for (int i = 0 ; i < s.length() ; i++){ int imir = 2 * C - i; // Find the minimum length of // the palindrome if (R > i){ LPS[i] = min(R-i, LPS[imir]); } else{ // Find the actual length of // the palindrome LPS[i] = 0; } // Exception Handling while( ((i + 1 + LPS[i]) < s.length()) && ((i - 1 - LPS[i]) >= 0) && s[i + 1 + LPS[i]] == s[i - 1 - LPS[i]]){ LPS[i] += 1; } // Update C and R if (i + LPS[i] > R){ C = i; R = i + LPS[i]; } } int r = 0, c = -1; for(int i = 0 ; i < s.length() ; i++){ r = max(r, LPS[i]); if(r == LPS[i]){ c = i; } } // Return the length r return r;}// Driver codeint main(){ // Given string str string str = "forneveropenskeegfor"; // Function Call cout << Manacher(str) << endl;}// This code is contributed by subhamgoyal2014. |
Java
// Java code for the above approachimport java.util.Arrays;class GFG { // Function that placed '#' intermediately // before and after each character static String UpdatedString(String s) { String newString = "#"; // Traverse the string for (char ch : s.toCharArray()) { newString += ch; newString += "#"; } // Return the string return newString; } // Function that finds the length of // the longest palindromic substring static int Manacher(String s) { // Update the string s = UpdatedString(s); // Stores the longest proper prefix // which is also a suffix int[] LPS = new int[s.length()]; int C = 0; int R = 0; for (int i = 0; i < s.length(); i++) { int imir = 2 * C - i; // Find the minimum length of // the palindrome if (R > i) { LPS[i] = Math.min(R - i, LPS[imir]); } else { // Find the actual length of // the palindrome LPS[i] = 0; } // Exception Handling while (((i + 1 + LPS[i]) < s.length()) && ((i - 1 - LPS[i]) >= 0) && (s.charAt(i + 1 + LPS[i]) == s.charAt(i - 1 - LPS[i]))) { LPS[i] += 1; } // Update C and R if (i + LPS[i] > R) { C = i; R = i + LPS[i]; } } int r = 0; for (int i = 0; i < s.length(); i++) { r = Math.max(r, LPS[i]); } // Return the length r return r; } // Driver code public static void main(String[] args) { // Given string str String str = "forneveropenskeegfor"; // Function Call System.out.println(Manacher(str)); }}// This code is contributed by lokeshpotta20. |
Python3
# Python program for the above approach# Function that placed '#' intermediately# before and after each characterdef UpdatedString(string): newString = ['#']# Traverse the string for char in string: newString += [char, '#']# Return the string return ''.join(newString)# Function that finds the length of# the longest palindromic substringdef Manacher(string): # Update the string string = UpdatedString(string) # Stores the longest proper prefix # which is also a suffix LPS = [0 for _ in range(len(string))] C = 0 R = 0 for i in range(len(string)): imir = 2 * C - i # Find the minimum length of # the palindrome if R > i: LPS[i] = min(R-i, LPS[imir]) else: # Find the actual length of # the palindrome LPS[i] = 0 # Exception Handling try: while string[i + 1 + LPS[i]] \ == string[i - 1 - LPS[i]]: LPS[i] += 1 except: pass # Update C and R if i + LPS[i] > R: C = i R = i + LPS[i] r, c = max(LPS), LPS.index(max(LPS)) # Return the length r return r# Driver code# Given string strstr = "forneveropenskeegfor"# Function Callprint(Manacher(str)) |
C#
// C# program to implement above approachusing System;using System.Collections;using System.Collections.Generic;class GFG{ // Function that placed '#' intermediately // before and after each character static string UpdatedString(string s){ string newString = "#"; // Traverse the string foreach(char ch in s){ newString += ch; newString += "#"; } // Return the string return newString; } // Function that finds the length of // the longest palindromic substring static int Manacher(string s){ // Update the string s = UpdatedString(s); // Stores the longest proper prefix // which is also a suffix int[] LPS = new int[s.Length]; int C = 0; int R = 0; for (int i = 0 ; i < s.Length ; i++){ int imir = 2 * C - i; // Find the minimum length of // the palindrome if (R > i){ LPS[i] = Math.Min(R-i, LPS[imir]); } else{ // Find the actual length of // the palindrome LPS[i] = 0; } // Exception Handling while( ((i + 1 + LPS[i]) < s.Length) && ((i - 1 - LPS[i]) >= 0) && s[i + 1 + LPS[i]] == s[i - 1 - LPS[i]]){ LPS[i] += 1; } // Update C and R if (i + LPS[i] > R){ C = i; R = i + LPS[i]; } } int r = 0; for(int i = 0 ; i < s.Length ; i++){ r = Math.Max(r, LPS[i]); } // Return the length r return r; } // Driver code public static void Main(string[] args){ // Given string str string str = "forneveropenskeegfor"; // Function Call Console.WriteLine(Manacher(str)); }}// This code is contributed by entertain2022. |
Javascript
//Javascript code for the above approach// Function that placed '#' intermediately// before and after each characterfunction UpdatedString(string) { let newString = ['#'];// Traverse the string for (let char of string) { newString.push(char, '#'); }// Return the string return newString.join('');}// Function that finds the length of// the longest palindromic substringfunction Manacher(string) { // Update the string string = UpdatedString(string); // Stores the longest proper prefix // which is also a suffix let LPS = new Array(string.length).fill(0); let C = 0; let R = 0; for (let i = 0; i < string.length; i++) { let imir = 2 * C - i; // Find the minimum length of // the palindrome if (R > i) { LPS[i] = Math.min(R-i, LPS[imir]); } else { // Find the actual length of // the palindrome LPS[i] = 0; } // Exception Handling try { while (string[i + 1 + LPS[i]] === string[i - 1 - LPS[i]]) { LPS[i] += 1; } } catch (err) { // pass } // Update C and R if (i + LPS[i] > R) { C = i; R = i + LPS[i]; } } let r = Math.max(...LPS); let c = LPS.indexOf(r); // Return the length r return r;}// Driver code// Given string strlet str = "forneveropenskeegfor";// Function Callconsole.log(Manacher(str)); |
Output:
10
Time Complexity: O(N), where N is the length of the given string.
Auxiliary Space: O(N)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
