Implement the solution to the Palindrome Partitioning problem using dynamic programming.?

The Palindrome Partitioning problem can be stated as follows:

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

For example, given the string s = "aab", the possible palindrome partitionings are:

- ["aa", "b"]
- ["a", "a", "b"]

To solve this problem, we can use dynamic programming. We can use a 2D boolean array to store whether a substring from i to j is a palindrome. Then we can use backtracking to generate all possible palindrome partitionings.

Here’s the Java code:

public List<List<String>> partition(String s) {
    int n = s.length();
    boolean[][] isPalindrome = new boolean[n][n];

    // fill isPalindrome table
    for (int i = n - 1; i >= 0; i--) {
        for (int j = i; j < n; j++) {
            if (s.charAt(i) == s.charAt(j) && (j - i <= 2 || isPalindrome[i + 1][j - 1])) {
                isPalindrome[i][j] = true;
            }
        }
    }

    List<List<String>> result = new ArrayList<>();
    dfs(result, new ArrayList<>(), s, isPalindrome, 0);
    return result;
}

private void dfs(List<List<String>> result, List<String> currentList, String s, boolean[][] isPalindrome, int start) {
    if (start == s.length()) {
        result.add(new ArrayList<>(currentList));
        return;
    }

    for (int end = start; end < s.length(); end++) {
        if (isPalindrome[start][end]) {
            currentList.add(s.substring(start, end + 1));
            dfs(result, currentList, s, isPalindrome, end + 1);
            currentList.remove(currentList.size() - 1);
        }
    }
}

The time complexity of this solution is O(n²) to fill the isPalindrome table, and the space complexity is also O(n²) to store the table. The backtracking part has a time complexity of O(n * 2ⁿ), since in the worst-case scenario, every substring is a palindrome and every partition is a single character, resulting in 2ⁿ possible partitions. However, in practice, the number of possible partitions is much smaller than 2ⁿ.