LeetCode: 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.

Solution

This is another perfect problem to practice Dynamic Programming.

My thoughts:

Firstly I oversimplified the problem. Thought it might work by iterate through s3 to match a letter either in S1 or S2. Of course it is wrong. Anti-example: s1= aa, s2=ab, s3=aaba.
Then I realized it can be solved with recursion. The implementation is as follows:

However, it failed on large test cases in leetcode.

Naturally I resorted to DP.  The typical way of solving a problem with DP is to find the state, and the optimal function. Here, the state can be considered as: dp[i][j] meaning S3[i+j] can be formed by S1[i] and S2[j] (for simplicity here string starts from 1, in the code we need to deal with that string starts from 0). With the recursion code as in lines 43~47, it is easy to come up with the optimal function:
dp[i][j]=(dp[i-1][j]==true && s1.charAt(i-1)==s3.charAt(i+j-1))
              ||(dp[i][j-1]==true && s2.charAt(j-1)==s3.charAt(i+j-1) )

Note: the initial states require great care to populate as well as the index. See the code:
To run the code in your mind, simply use s1="a", s2="b", s3="ab" as an example.