Palindrome Partitioning

Given a string s, partition s such that every substringof the partition is a palindrome. Return all possible palindrome partitioning of s.

LeetCode Problem - 131

class Solution {
    // Method to partition the input string into all possible palindromic substrings
    public List<List<String>> partition(String s) {
        // List to store the result of all possible palindromic partitions
        List<List<String>> result = new ArrayList<>();
        // List to store the current partition being constructed
        List<String> currentPartition = new ArrayList<>();

        // Start the backtracking process from the first character
        backtrack(s, 0, currentPartition, result);

        // Return the final result list
        return result;
    }

    // Helper method to perform backtracking
    private void backtrack(String s, int start, List<String> currentPartition, List<List<String>> result) {
        // If we've reached the end of the string, add the current partition to the result list
        if (start == s.length()) {
            result.add(new ArrayList<>(currentPartition));
            return;
        }

        // Iterate over possible end indices for substrings starting from 'start'
        for (int end = start; end < s.length(); end++) {
            // If the substring s[start:end+1] is a palindrome, proceed with the current partition
            if (isPalindrome(s, start, end)) {
                // Add the palindromic substring to the current partition
                currentPartition.add(s.substring(start, end + 1));
                // Recur to continue partitioning the remaining substring
                backtrack(s, end + 1, currentPartition, result);
                // Backtrack by removing the last added substring
                currentPartition.remove(currentPartition.size() - 1);
            }
        }
    }

    // Helper method to check if a substring s[left:right+1] is a palindrome
    private boolean isPalindrome(String s, int left, int right) {
        // Compare characters from the ends towards the center
        while (left < right) {
            // If characters don't match, it's not a palindrome
            if (s.charAt(left) != s.charAt(right)) {
                return false;
            }
            left++;
            right--;
        }
        // If all characters match, it's a palindrome
        return true;
    }
}