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 _97_InterleavingString {
    /**
     * 97. Interleaving String
     * Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2.

     For example,
     Given:
     s1 = "aabcc",
     s2 = "dbbca",

     When s3 = "aadbbcbcac", return true.

     When s3 = "aadbbbaccc", return false.

     [true, true, true, false, false, false]
     [false, false, false, false, false, false]
     [false, false, false, false, false, false]
     [false, false, false, false, false, false]
     [false, false, false, false, false, false]
     [false, false, false, false, false, false]

     [true, true, true, false, false, false]
     [false, false, true, true, false, false]
     [false, false, true, true, true, false]
     [false, false, true, false, true, true]
     [false, false, true, true, true, false]
     [false, false, false, false, true, true]

     time : O(m * n)
     space : O(m * n)

     * @param s1
     * @param s2
     * @param s3
     * @return
     */
    public static boolean isInterleave(String s1, String s2, String s3) {
        if ((s1.length() + s2.length()) != s3.length()) {
            return false;
        }

        boolean[][] dp = new boolean[s2.length() + 1][s1.length() + 1];
        dp[0][0] = true;

        for (int i = 1; i < dp.length; i++) {
            dp[i][0] = dp[i - 1][0] && (s2.charAt(i - 1) == s3.charAt(i - 1));
        }
        for (int i = 1; i < dp[0].length; i++) {
            dp[0][i] = dp[0][i - 1] && (s1.charAt(i - 1) == s3.charAt(i - 1));
        }

        for (int i = 1; i < dp.length; i++) {
            for (int j = 1; j < dp[0].length; j++) {
                dp[i][j] = (dp[i - 1][j] && s2.charAt(i - 1) == s3.charAt(i + j - 1))
                        || (dp[i][j - 1] && s1.charAt(j - 1) == s3.charAt(i + j - 1));
            }
        }

        return dp[s2.length()][s1.length()];
    }
}