# 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++

 `// C++ program for the above approach` `#include ` `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;` `}`

## Java

 `// 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`

## Python3

 `# 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`

## C#

 `// 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`

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++

 `// C++ program for the above approach`   `#include ` `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;` `}`

## Java

 `// 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`

## C#

 `// 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`

## Python3

 `# 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`

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

 `# 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``))`

Output:

```10

```

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.