/*
 Author:     King, wangjingui@outlook.com
 Date:       Jan 3, 2015
 Problem:    Interleaving String
 Difficulty: Medium
 Source:     https://oj.leetcode.com/problems/interleaving-string/
 Notes:
 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.

 Solution: 1. dp. O(MN) time & space. 'dp[i][j] == true' means that there is at least one way to construct 
              the string s3[0...i+j-1] by interleaving s1[0...j-1] and s2[0...i-1].
 */

public class Solution {
    public boolean isInterleave_1(String s1, String s2, String s3) {
        int l1 = s1.length(), l2 = s2.length(), l3 = s3.length();
        if (l1 == 0) return s2.compareTo(s3) == 0;
        if (l2 == 0) return s1.compareTo(s3) == 0;
        if (l1 + l2 != l3) return false;
        boolean[][] dp = new boolean[l1+1][l2+1];
        dp[0][0] = true;
        for (int i = 1; i <= l1; ++i) {
            dp[i][0] = dp[i-1][0] && (s1.charAt(i-1) == s3.charAt(i-1));
        }
        for (int j = 1; j <= l2; ++j) {
            dp[0][j] = dp[0][j-1] && (s2.charAt(j-1) == s3.charAt(j-1));
        }
        for (int i = 1; i <= l1; ++i) {
            for (int j = 1; j <= l2; ++j) {
                if (s1.charAt(i - 1) == s3.charAt(i+j-1)) dp[i][j] = dp[i-1][j];
                if (s2.charAt(j - 1) == s3.charAt(i+j-1)) dp[i][j] = dp[i][j] | dp[i][j-1];
            }
        }
        return dp[l1][l2];
    }
    public boolean isInterleave(String s1, String s2, String s3) {
        int l1 = s1.length(), l2 = s2.length(), l3 = s3.length();
        if (l1 == 0) return s2.compareTo(s3) == 0;
        if (l2 == 0) return s1.compareTo(s3) == 0;
        if (l1 + l2 != l3) return false;
        boolean[] dp = new boolean[l2+1];
        dp[0] = true;
        for (int j = 1; j <= l2; ++j) {
            dp[j] = dp[j-1] && (s2.charAt(j-1) == s3.charAt(j-1));
        }
        for (int i = 1; i <= l1; ++i) {
            dp[0] = dp[0] && (s1.charAt(i-1) == s3.charAt(i-1));
            for (int j = 1; j <= l2; ++j) {
                boolean before = dp[j]; dp[j] = false;
                if (s1.charAt(i - 1) == s3.charAt(i+j-1)) dp[j] = before;
                if (s2.charAt(j - 1) == s3.charAt(i+j-1)) dp[j] = dp[j] | dp[j-1];
            }
        }
        return dp[l2];
    }
}