Write a function that returns the longest palindrome in a given string

Solution 1:

You can find the the longest palindrome using Manacher's Algorithm in O(n) time! Its implementation can be found here and here.

For input String s = "HYTBCABADEFGHABCDEDCBAGHTFYW1234567887654321ZWETYGDE" it finds the correct output which is 1234567887654321.

Solution 2:

The Algo 2 may not work for all string. Here is an example of such a string "ABCDEFCBA".

Not that the string has "ABC" and "CBA" as its substring. If you reverse the original string, it will be "ABCFEDCBA". and the longest matching substring is "ABC" which is not a palindrome.

You may need to additionally check if this longest matching substring is actually a palindrome which has the running time of O(n^3).

Solution 3:

As far as I understood the problem, we can find palindromes around a center index and span our search both ways, to the right and left of the center. Given that and knowing there's no palindrome on the corners of the input, we can set the boundaries to 1 and length-1. While paying attention to the minimum and maximum boundaries of the String, we verify if the characters at the positions of the symmetrical indexes (right and left) are the same for each central position till we reach our max upper bound center.

The outer loop is O(n) (max n-2 iterations), and the inner while loop is O(n) (max around (n / 2) - 1 iterations)

Here's my Java implementation using the example provided by other users.

class LongestPalindrome {

    /**
     * @param input is a String input
     * @return The longest palindrome found in the given input.
     */
    public static String getLongestPalindrome(final String input) {
        int rightIndex = 0, leftIndex = 0;
        String currentPalindrome = "", longestPalindrome = "";
        for (int centerIndex = 1; centerIndex < input.length() - 1; centerIndex++) {
            leftIndex = centerIndex - 1;  rightIndex = centerIndex + 1;
            while (leftIndex >= 0 && rightIndex < input.length()) {
                if (input.charAt(leftIndex) != input.charAt(rightIndex)) {
                    break;
                }
                currentPalindrome = input.substring(leftIndex, rightIndex + 1);
                longestPalindrome = currentPalindrome.length() > longestPalindrome.length() ? currentPalindrome : longestPalindrome;
                leftIndex--;  rightIndex++;
            }
        }
        return longestPalindrome;
    }

    public static void main(String ... args) {
        String str = "HYTBCABADEFGHABCDEDCBAGHTFYW12345678987654321ZWETYGDE";
        String longestPali = getLongestPalindrome(str);
        System.out.println("String: " + str);
        System.out.println("Longest Palindrome: " + longestPali);
    }
}

The output of this is the following:

marcello:datastructures marcello$ javac LongestPalindrome
marcello:datastructures marcello$ java LongestPalindrome
String: HYTBCABADEFGHABCDEDCBAGHTFYW12345678987654321ZWETYGDE
Longest Palindrome: 12345678987654321

Solution 4:

with regex and ruby you can scan for short palindromes like this:

PROMPT> irb
>> s = "longtextwithranynarpalindrome"
=> "longtextwithranynarpalindrome"
>> s =~ /((\w)(\w)(\w)(\w)(\w)\6\5\4\3\2)/; p $1
nil
=> nil
>> s =~ /((\w)(\w)(\w)(\w)\w\5\4\3\2)/; p $1
nil
=> nil
>> s =~ /((\w)(\w)(\w)(\w)\5\4\3\2)/; p $1
nil
=> nil
>> s =~ /((\w)(\w)(\w)\w\4\3\2)/; p $1
"ranynar"
=> nil

Solution 5:

I have written the following Java program out of curiosity, simple and self-explanatory HTH. Thanks.

/**
 *
 * @author sanhn
 */
public class CheckPalindrome {

    private static String max_string = "";

    public static void checkSubString(String s){
        System.out.println("Got string is "+s);
        for(int i=1;i<=s.length();i++){
            StringBuilder s1 = new StringBuilder(s.substring(0,i));
            StringBuilder s2 = new StringBuilder(s.substring(0,i));
            s2.reverse();
            if(s1.toString().equals(s2.toString())){
                if(max_string.length()<=s1.length()){
                    max_string = s1.toString();
                    System.out.println("tmp max is "+max_string);
                }

            }
        }
    }

    public static void main(String[] args){
        String s="HYTBCABADEFGHABCDEDCBAGHTFYW1234567887654321ZWETYGDE";

        for(int i=0; i<s.length(); i++)
            checkSubString(s.substring(i, s.length()));

        System.out.println("Max string is "+max_string);
    }
}