516. Longest Palindromic Subsequence

// Recursive: Time Limit Exceeded Error
class Solution {
    public int longestPalindromeSubseq(String s) {
        int n = s.length();
        return helper(s, 0, n - 1);
    }
    
    private int helper(String s, int lo, int hi) {
        if (lo > hi) return 0;
        if (lo == hi) return 1;
        int max = 0;
        
        for (int i = lo; i < hi; i++) {
            for (int j = hi; j >= i; j--) {
                if (j - i + 1 < max) return max;
                if (i == j) {
                    max = Math.max(max, 1);
                    break;
                }
                
                if (s.charAt(i) == s.charAt(j)) {
                    max = Math.max(max, 2 + helper(s, i + 1, j - 1));
                    break;
                }
            }
        }
        
        return max;
    }
}

// Dynamic Programming
class Solution {
    public int longestPalindromeSubseq(String s) {
        int n = s.length();
        int[][] dp = new int[n][n];
        
        for (int i = n - 1; i >= 0; i--) {
            dp[i][i] = 1;
            for (int j = i + 1; j < n; j++) {
                if (s.charAt(i) == s.charAt(j)) {
                    dp[i][j] = dp[i + 1][j - 1] + 2;
                } else {
                    dp[i][j] = Math.max(dp[i + 1][j],
                                       dp[i][j - 1]);
                }
            }
        }
        
        return dp[0][n - 1];
    }
}