package leetcode_1To300;

/**
 * 本代码来自 Cspiration，由 @Cspiration 提供
 * 题目来源：http://leetcode.com
 * - Cspiration 致力于在 CS 领域内帮助中国人找到工作，让更多海外国人受益
 * - 现有课程：Leetcode Java 版本视频讲解（1-900题）（上）（中）（下）三部
 * - 算法基础知识（上）（下）两部；题型技巧讲解（上）（下）两部
 * - 节省刷题时间，效率提高2-3倍，初学者轻松一天10题，入门者轻松一天20题
 * - 讲师：Edward Shi
 * - 官方网站：https://cspiration.com
 * - 版权所有，转发请注明出处
 */
public class _5_LongestPalindromicSubstring {
    /**
     * Given a string s, find the longest palindromic substring in s.
     * You may assume that the maximum length of s is 1000.

     Example:

     Input: "babad"

     Output: "bab"

     Note: "aba" is also a valid answer.
     Example:

     Input: "cbbd"

     Output: "bb"


     * @param s
     * @return
     */

    // time : O(n^2) space : O(n^2);
    public String longestPalindrome(String s) {
        if (s == null || s.length() == 0) return s;
        String res = "";
        boolean[][] dp = new boolean[s.length()][s.length()];
        int max = 0;
        for (int j = 0; j < s.length(); j++) {
            for (int i = 0; i <= j; i++) {
                dp[i][j] = s.charAt(i) == s.charAt(j) && ((j - i <= 2) || dp[i + 1][j - 1]);
                if (dp[i][j]) {
                    if (j - i + 1 > max) {
                        max = j - i + 1;
                        res = s.substring(i, j + 1);
                    }
                }
            }
        }
        return res;
    }

    String res = "";
    // time : O(n^2) space : O(1)
    public String longestPalindrome2(String s) {
        if (s == null || s.length() == 0) return s;
        for (int i = 0; i < s.length(); i++) {
            helper(s, i, i);
            helper(s, i, i + 1);
        }
        return res;
    }
    public void helper(String s, int left, int right) {
        while (left >= 0 && right < s.length() && s.charAt(left) == s.charAt(right)) {
            left--;
            right++;
        }
        String cur = s.substring(left + 1, right);
        if (cur.length() > res.length()) {
            res = cur;
        }
    }
}