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

Question

e.g "ccddcc" in the string "abaccddccefe"

I thought of a solution but it runs in O(n^2) time

Algo 1:

Steps: Its a brute force method

1. Have 2 for loops
   for i = 1 to i less than array.length -1
   for j=i+1 to j less than array.length
2. This way you can get substring of every possible combination from the array
3. Have a palindrome function which checks if a string is palindrome
4. so for every substring (i,j) call this function, if it is a palindrome store it in a string variable
5. If you find next palindrome substring and if it is greater than the current one, replace it with current one.
6. Finally your string variable will have the answer

Issues: 1. This algo runs in O(n^2) time.

Algo 2:

1. Reverse the string and store it in diferent array
2. Now find the largest matching substring between both the array
3. But this too runs in O(n^2) time

Can you guys think of an algo which runs in a better time. If possible O(n) time

Answer 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.

Answer 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).

Answer 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