Length of Longest Palindrome Substring

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++

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// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to obtain the length of
// the longest palindromic substring
int 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 Code
int main()
{
    // Given string
    string str = "forgeeksskeegfor";
 
    // Function Call
    cout << longestPalSubstr(str);
    return 0;
}

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Java

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// Java program for the above approach
import java.io.*;
 
class GFG{
  
// Function to obtain the length of
// the longest palindromic substring
static 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 Code
public static void main (String[] args)
{
     
    // Given string
    String str = "forgeeksskeegfor";
  
    // Function call
    System.out.print(longestPalSubstr(str));
}
}
 
// This code is contributed by code_hunt

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Python3

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# Python3 program for the above approach
 
# Function to obtain the length of
# the longest palindromic substring
def 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 string
str = "forgeeksskeegfor"
  
# Function call
print(longestPalSubstr(str))
 
# This code is contributed by code_hunt

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C#

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// C# program for the above approach 
using System;
 
class GFG{
  
// Function to obtain the length of
// the longest palindromic substring
static 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 Code
public static void Main ()
{
     
    // Given string
    string str = "forgeeksskeegfor";
  
    // Function call
    Console.Write(longestPalSubstr(str));
}
}
 
// This code is contributed by code_hunt

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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: 

  1. Maintain a boolean table[N][N] that is filled in a bottom-up manner.
  2. The value of table[i][j] is true if the substring is a palindrome, otherwise false.
  3. 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.
  4. Otherwise, the value of table[i][j] is update as false.

Below is the illustration for the string “geeks”

Below is the implementation of the above approach:

C++

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// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the length of
// the longest palindromic substring
int 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 Code
int main()
{
    // Given string str
    string str = "forgeeksskeegfor";
 
    // Function Call
    cout << longestPalSubstr(str);
 
    return 0;
}

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Java

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// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to find the length of
// the longest palindromic subString
static 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 Code
public static void main(String[] args)
{
     
    // Given String str
    String str = "forgeeksskeegfor";
 
    // Function Call
    System.out.print(longestPalSubstr(str));
}
}
 
// This code is contributed by Amit Katiyar

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C#

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// C# program for
// the above approach
using System;
class GFG{
 
// Function to find the length of
// the longest palindromic subString
static 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 Code
public static void Main(String[] args)
{
  // Given String str
  String str = "forgeeksskeegfor";
 
  // Function Call
  Console.Write(longestPalSubstr(str));
}
}
 
// This code is contributed by Rajput-Ji

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Python3

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# Python program for the above approach
 
 
# Function to find the length of
# the longest palindromic subString
def 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 Code
if __name__ == '__main__':
    # Given String str
    str = "forgeeksskeegfor";
 
    # Function Call
    print(longestPalSubstr(str));
 
# This code is contributed by 29AjayKumar

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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: 

  1. Add the special characters in the given string S as explained above and let its length be N.
  2. 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.
  3. 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).
  4. Now, check if there are some characters after r such that the palindrome becomes ever longer.
  5. If (i + d[i]) is greater than r, update r = (i + d[i]) and center as i.
  6. 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:

Python3

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# Python program for the above approach
 
# Function that placed '#' intermediately
# before and after each character
def 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 substring
def 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 str
str = "forgeeksskeegfor"
 
# Function Call
print(Manacher(str))

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Output: 

10









 

Time Complexity: O(N), where N is the length of the given string. 
Auxiliary Space: O(N) 

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