Shortest Palindrome

class Solution {
public:
    string buildShortestPalindrome(string str) {
        if (str.empty()) {
            return str;
        }

        string modStr = enhancement(str);
        int len = modStr.size();
        vector<int> radius(len, 0);
        int ctr = 0, rightEdge = 0;
        int maxLen = 0;

        for (int pos = 1; pos < len - 1; ++pos) {
            int sym = 2 * ctr - pos;

            if (rightEdge > pos) {
                radius[pos] = min(rightEdge - pos, radius[sym]);
            }

            while (modStr[pos + 1 + radius[pos]] == modStr[pos - 1 - radius[pos]]) {
                ++radius[pos];
            }

            if (pos + radius[pos] > rightEdge) {
                ctr = pos;
                rightEdge = pos + radius[pos];
            }

            if (pos - radius[pos] == 1) {
                maxLen = max(maxLen, radius[pos]);
            }
        }

        string tail = str.substr(maxLen);
        reverse(tail.begin(), tail.end());
        return tail + str;
    }

private:
    string enhancement(const string& original) {
        string resultant = "^";
        for (char ch : original) {
            resultant += "#" + string(1, ch);
        }
        resultant += "#$";
        return resultant;
    }
};

The goal of the provided C++ program is to find and create the shortest palindrome from a given string by appending characters only at the beginning of the string. Here's a breakdown of how the solution achieves this:

• Initialization: The buildShortestPalindrome function starts by checking if the input string is empty and returns it immediately if true.

• String enhancement: The enhancement function modifies the original string to facilitate the palindrome checking process. This is accomplished by inserting special characters ("^", "#", and "$") around each character of the string. This modified string helps in handling edge cases and simplifies the palindrome expansion logic.

• Use of Manacher’s Algorithm: This algorithm is applied on the enhanced string. It allows calculating the longest palindromic substring centered at each position in the string efficiently.

• Calculation Process:

  • The solution initializes variables to keep track of the center (ctr), the right edge of the current palindrome (rightEdge), and radius, which stores the radius (half the length) of the palindrome centered at each position.
  • As the algorithm iterates through the length of the modified string, it expands potential palindromes and updates the rightEdge and ctr based on the expansions.
  • Specifically, it checks if expanding the palindrome at the current position will exceed the current rightEdge. If it does, the center and right edge are updated.
  • Additionally, the palindrome centered at each position and expanding till the beginning of the string is tracked using maxLen.

• Constructing the shortest palindrome:

  • The substring from position maxLen to the end of the original string is used to identify the characters to be reversed and prepended to the original string to create the palindrome.
  • After reversing and prepending this substring, the updated string is returned as the shortest possible palindrome that can be formed by adding characters only at the beginning.

This method provides a proficient way to determine the shortest palindrome by leveraging efficient symmetry checks through Manacher’s Algorithm and is ideal for strings where straightforward palindromic expansions are computationally expensive.

class Solution {

    public String findShortestPalindrome(String inputStr) {
        if (inputStr == null || inputStr.isEmpty()) {
            return inputStr;
        }

        String augmented = insertSpecialCharacters(inputStr);
        int[] pLengths = new int[augmented.length()];
        int mid = 0, right = 0;
        int longestPalindromeStartsAt0 = 0;

        for (int i = 1; i < augmented.length() - 1; i++) {
            int mirror = 2 * mid - i;

            if (right > i) {
                pLengths[i] = Math.min(
                    right - i,
                    pLengths[mirror]
                );
            }

            while (
                augmented.charAt(i + 1 + pLengths[i]) ==
                augmented.charAt(i - 1 - pLengths[i])
            ) {
                pLengths[i]++;
            }

            if (i + pLengths[i] > right) {
                mid = i;
                right = i + pLengths[i];
            }

            if (i - pLengths[i] == 1) {
                longestPalindromeStartsAt0 = Math.max(
                    longestPalindromeStartsAt0,
                    pLengths[i]
                );
            }
        }

        StringBuilder suffixReverse = new StringBuilder(
            inputStr.substring(longestPalindromeStartsAt0)
        ).reverse();
        return suffixReverse.append(inputStr).toString();
    }

    private String insertSpecialCharacters(String str) {
        StringBuilder resultBuilder = new StringBuilder("^");
        for (char ch : str.toCharArray()) {
            resultBuilder.append("#").append(ch);
        }
        return resultBuilder.append("#$").toString();
    }
}

To solve the problem of finding the shortest palindrome by appending characters to the beginning of a given string in Java, the provided solution implements the following steps:

• First, check if the input string is null or empty. If true, return the input string as there's no further processing required.

• Transform the input string by inserting special characters between each character and at the beginning and end. This transformation helps in avoiding bound checking during the palindrome expansion.

• Initialize an array to store the lengths of the palindromes centered at each character of the transformed string.

• Use a two-pointers technique, 'mid' to track the center of the current rightmost palindrome, and 'right' to track its right boundary.

• For each character in the transformed string, calculate the potential length of the palindrome. Reflect around the center value if the current position 'i' is within an existing palindrome, otherwise expand outward.

• Adjust 'mid' and 'right' according to the expansions.

• Keep track of the longest palindrome that starts at the original string beginning.

• At the end, combine the reversed substring which is not part of the palindrome from the input string with the original string to form the shortest palindrome.

Consider the performance implications when handling large strings due to the transformation and palindrome check operations, each of which might increase complexity. However, the algorithm efficiently finds the shortest palindrome by leveraging properties of symmetric expansions using mirrored indices and dynamic adjustment of centers.

class Solution:
    # Main function to compute shortest palindrome
    def shortestPal(self, input_str: str) -> str:
        # Returns the original string if empty
        if not input_str:
            return input_str
        
        # Converts the input string into a format to manage palindromes correctly
        processed_input = self._preprocess(input_str)
        size = len(processed_input)
        radius = [0] * size
        pivot = 0
        right_edge = 0
        longest = 0

        # Evaluate each character in the converted string
        for idx in range(1, size - 1):
            mirror = 2 * pivot - idx

            # Utilize existing values to avoid redundant operations
            if right_edge > idx:
                radius[idx] = min(
                    right_edge - idx, radius[mirror]
                )

            # Try to expand palindrome around current point
            while (
                processed_input[idx + 1 + radius[idx]]
                == processed_input[idx - 1 - radius[idx]]
            ):
                radius[idx] += 1

            # Update the pivot and right edge if the current palindrome extends further
            if idx + radius[idx] > right_edge:
                pivot = idx
                right_edge = idx + radius[idx]

            # Keep the largest palindrome start from the very first position
            if idx - radius[idx] == 1:
                longest = max(
                    longest, radius[idx]
                )

        # Create the smallest palindrome by appending reversed excess to the beginning
        non_palindrome_part = input_str[longest:][::-1]
        return non_palindrome_part + input_str

    # Helper function to augment the input for palindrome processing
    def _preprocess(self, input_str: str) -> str:
        return "^" + "#" + "#".join(input_str) + "#$"

This solution describes a Python method for finding the shortest palindrome that can be created by appending characters to the beginning of a given string.

• Implement a class Solution with two primary methods: shortestPal and _preprocess.
• In shortestPal:
  • Start by checking if the input string is empty. If true, simply return it.
  • Use _preprocess to transform the input string to include separators (#) and boundary markers (^ and $), which aid in palindrome identification.
  • Use an algorithm reminiscent of the Manacher's algorithm to efficiently identify palindromes. It involves an array radius to store the radii of palindromes centered at each character in the processed string.
  • Utilize two pointer technique with pivot and right_edge to track the center and the right boundary of the most recently found palindrome.
  • For each position, possibly extend the palindrome and update pivot and right_edge accordingly.
  • Track the longest palindrome that starts from the very first character.
• At the end, obtain the part of the original input that was not included in the palindrome, reverse it, and prepend it to the original input to form the shortest palindrome.

By using an augmented string and a tailored approach leveraging properties of palindromes, the implementation efficiently computes the shortest palindrome from the input.