Length of Longest Palindrome Substring - GeeksforGeeks

28 Dec, 2021
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: 5Explanation: string “abcba” is the longest substring that is a palindrome which is of length 5.
Input: S = “abcdaa”Output: 2Explanation: 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++
Java
Python3
C#
Javascript
// 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 = "forgeeksskeegfor";     // Function Call    cout << longestPalSubstr(str);    return 0;}
// 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 = "forgeeksskeegfor";      // Function call    System.out.print(longestPalSubstr(str));}} // This code is contributed by code_hunt
# 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 = "forgeeksskeegfor"  # Function callprint(longestPalSubstr(str)) # This code is contributed by code_hunt
// 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 = "forgeeksskeegfor";      // Function call    Console.Write(longestPalSubstr(str));}} // This code is contributed by code_hunt
<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 = "forgeeksskeegfor"; // Function Calldocument.write( longestPalSubstr(str));  </script>
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.
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 “geeks”: 
Below is the implementation of the above approach:
C++
Java
C#
Python3
Javascript
// 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 = "forgeeksskeegfor";     // Function Call    cout << longestPalSubstr(str);     return 0;}
// 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 = "forgeeksskeegfor";     // Function Call    System.out.print(longestPalSubstr(str));}} // This code is contributed by Amit Katiyar
// 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 = "forgeeksskeegfor";   // Function Call  Console.Write(longestPalSubstr(str));}} // This code is contributed by Rajput-Ji
# 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 = "forgeeksskeegfor";     # Function Call    print(longestPalSubstr(str)); # This code is contributed by 29AjayKumar
<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 = "forgeeksskeegfor";         // Function Call        document.write(longestPalSubstr(str)); // This code is contributed by umadevi9616</script>
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.
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:
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 = "forgeeksskeegfor" # Function Callprint(Manacher(str))
10
 
Time Complexity: O(N), where N is the length of the given string. Auxiliary Space: O(N) 
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