# Maximum length palindromic substring such that it starts and ends with given char

• Difficulty Level : Easy
• Last Updated : 07 May, 2021

Given a string str and a character ch, the task is to find the longest palindromic sub-string of str such that it starts and ends with the given character ch.
Examples:

Input: str = “lapqooqpqpl”, ch = ‘p’
Output:
“pqooqp” is the maximum length palindromic
sub-string that starts and ends with ‘p’.
Input: str = “geeksforgeeks”, ch = ‘k’
Output:
“k” is the valid sub-string.

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Approach: For every possible index pair (i, j) such that str[i] = str[j] = ch check whether the sub-string str[i…j] is palindrome or not. For all the found palindromes, store the length of the longest palindrome found so far.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function that returns true if``// str[i...j] is a palindrome``bool` `isPalindrome(string str, ``int` `i, ``int` `j)``{``    ``while` `(i < j) {``        ``if` `(str[i] != str[j])``            ``return` `false``;``        ``i++;``        ``j--;``    ``}``    ``return` `true``;``}` `// Function to return the length of the``// longest palindromic sub-string such that``// it starts and ends with the character ch``int` `maxLenPalindrome(string str, ``int` `n, ``char` `ch)``{``    ``int` `maxLen = 0;` `    ``for` `(``int` `i = 0; i < n; i++) {` `        ``// If current character is``        ``// a valid starting index``        ``if` `(str[i] == ch) {` `            ``// Instead of finding the ending index from``            ``// the beginning, find the index from the end``            ``// This is because if the current sub-string``            ``// is a palindrome then there is no need to check``            ``// the sub-strings of smaller length and we can``            ``// skip to the next iteration of the outer loop``            ``for` `(``int` `j = n - 1; j >= i; j--) {` `                ``// If current character is``                ``// a valid ending index``                ``if` `(str[j] == ch) {` `                    ``// If str[i...j] is a palindrome then update``                    ``// the length of the maximum palindrome so far``                    ``if` `(isPalindrome(str, i, j)) {``                        ``maxLen = max(maxLen, j - i + 1);``                        ``break``;``                    ``}``                ``}``            ``}``        ``}``    ``}``    ``return` `maxLen;``}` `// Driver code``int` `main()``{``    ``string str = ``"lapqooqpqpl"``;``    ``int` `n = str.length();``    ``char` `ch = ``'p'``;` `    ``cout << maxLenPalindrome(str, n, ch);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``class` `GFG``{``    ` `    ``// Function that returns true if``    ``// str[i...j] is a palindrome``    ``static` `boolean` `isPalindrome(String str,``                               ``int` `i, ``int` `j)``    ``{``        ``while` `(i < j)``        ``{``            ``if` `(str.charAt(i) != str.charAt(j))``            ``{``                ``return` `false``;``            ``}``            ``i++;``            ``j--;``        ``}``        ``return` `true``;``    ``}` `    ``// Function to return the length of the``    ``// longest palindromic sub-string such that``    ``// it starts and ends with the character ch``    ``static` `int` `maxLenPalindrome(String str, ``int` `n, ``char` `ch)``    ``{``        ``int` `maxLen = ``0``;` `        ``for` `(``int` `i = ``0``; i < n; i++)``        ``{` `            ``// If current character is``            ``// a valid starting index``            ``if` `(str.charAt(i) == ch)``            ``{` `                ``// Instead of finding the ending index from``                ``// the beginning, find the index from the end``                ``// This is because if the current sub-string``                ``// is a palindrome then there is no need to check``                ``// the sub-strings of smaller length and we can``                ``// skip to the next iteration of the outer loop``                ``for` `(``int` `j = n - ``1``; j >= i; j--)``                ``{` `                    ``// If current character is``                    ``// a valid ending index``                    ``if` `(str.charAt(j) == ch)``                    ``{` `                        ``// If str[i...j] is a palindrome then update``                        ``// the length of the maximum palindrome so far``                        ``if` `(isPalindrome(str, i, j))``                        ``{``                            ``maxLen = Math.max(maxLen, j - i + ``1``);``                            ``break``;``                        ``}``                    ``}``                ``}``            ``}``        ``}``        ``return` `maxLen;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``String str = ``"lapqooqpqpl"``;``        ``int` `n = str.length();``        ``char` `ch = ``'p'``;` `        ``System.out.println(maxLenPalindrome(str, n, ch));``    ``}``}` `// This code is contributed by Princi Singh`

## Python3

 `# Python implementation of the approach` `# Function that returns true if``# str[i...j] is a palindrome``def` `isPalindrome(``str``, i, j):``    ``while` `(i < j):``        ``if` `(``str``[i] !``=` `str``[j]):``            ``return` `False``;``        ``i``+``=``1``;``        ``j``-``=``1``;``    ``return` `True``;`  `# Function to return the length of the``# longest palindromic sub-string such that``# it starts and ends with the character ch``def` `maxLenPalindrome(``str``, n, ch):``    ``maxLen ``=` `0``;` `    ``for` `i ``in` `range``(n):` `        ``# If current character is``        ``# a valid starting index``        ``if` `(``str``[i] ``=``=` `ch):` `            ``# Instead of finding the ending index from``            ``# the beginning, find the index from the end``            ``# This is because if the current sub-string``            ``# is a palindrome then there is no need to check``            ``# the sub-strings of smaller length and we can``            ``# skip to the next iteration of the outer loop``            ``for` `j ``in` `range``(n``-``1``,i``+``1``,``-``1``):``                ``# If current character is``                ``# a valid ending index``                ``if` `(``str``[j] ``=``=` `ch):` `                    ``# If str[i...j] is a palindrome then update``                    ``# the length of the maximum palindrome so far``                    ``if` `(isPalindrome(``str``, i, j)):``                        ``maxLen ``=` `max``(maxLen, j ``-` `i ``+` `1``);``                        ``break``;` `    ``return` `maxLen;` `# Driver code``str` `=` `"lapqooqpqpl"``;``n ``=` `len``(``str``);``ch ``=` `'p'``;` `print``(maxLenPalindrome(``str``, n, ch));``    ` `# This code is contributed by 29AjayKumar`

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG``{``    ` `    ``// Function that returns true if``    ``// str[i...j] is a palindrome``    ``static` `bool` `isPalindrome(``string` `str,``                            ``int` `i, ``int` `j)``    ``{``        ``while` `(i < j)``        ``{``            ``if` `(str[i] != str[j])``            ``{``                ``return` `false``;``            ``}``            ``i++;``            ``j--;``        ``}``        ``return` `true``;``    ``}` `    ``// Function to return the length of the``    ``// longest palindromic sub-string such that``    ``// it starts and ends with the character ch``    ``static` `int` `maxLenPalindrome(``string` `str, ``int` `n, ``char` `ch)``    ``{``        ``int` `maxLen = 0;` `        ``for` `(``int` `i = 0; i < n; i++)``        ``{` `            ``// If current character is``            ``// a valid starting index``            ``if` `(str[i] == ch)``            ``{` `                ``// Instead of finding the ending index from``                ``// the beginning, find the index from the end``                ``// This is because if the current sub-string``                ``// is a palindrome then there is no need to check``                ``// the sub-strings of smaller length and we can``                ``// skip to the next iteration of the outer loop``                ``for` `(``int` `j = n - 1; j >= i; j--)``                ``{` `                    ``// If current character is``                    ``// a valid ending index``                    ``if` `(str[j] == ch)``                    ``{` `                        ``// If str[i...j] is a palindrome then update``                        ``// the length of the maximum palindrome so far``                        ``if` `(isPalindrome(str, i, j))``                        ``{``                            ``maxLen = Math.Max(maxLen, j - i + 1);``                            ``break``;``                        ``}``                    ``}``                ``}``            ``}``        ``}``        ``return` `maxLen;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``string` `str = ``"lapqooqpqpl"``;``        ``int` `n = str.Length;``        ``char` `ch = ``'p'``;` `        ``Console.WriteLine(maxLenPalindrome(str, n, ch));``    ``}``}` `// This code is contributed by AnkitRai01`

## Javascript

 ``
Output:
`6`

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