Dada una string str y un carácter ch , la tarea es encontrar la substring palindrómica más larga de str tal que comience y termine con el carácter ch dado .
Ejemplos:
Entrada: str = “lapqooqpqpl”, ch = ‘p’
Salida: 6
“pqooqp” es la substring palindrómica de longitud máxima
que comienza y termina con ‘p’.
Entrada: str = “geeksforgeeks”, ch = ‘k’
Salida: 1
“k” es la substring válida.
Enfoque: para cada posible par de índices (i, j) tal que str[i] = str[j] = ch , compruebe si la substring str[i…j] es palíndromo o no. Para todos los palíndromos encontrados, almacene la longitud del palíndromo más largo encontrado hasta el momento.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
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
<script>
// JavaScript implementation of the approach
// Function that returns true if
// str[i...j] is a palindrome
function isPalindrome(str, i, 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
function maxLenPalindrome(str, n, ch) {
var maxLen = 0;
for (var 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 (var 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
var str = "lapqooqpqpl";
var n = str.length;
var ch = "p";
document.write(maxLenPalindrome(str, n, ch));
</script>
Producción:
6
Time Complexity: O(n3) Space Complexity: O(1)