Leetcode (Python): 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:

We use dynamic programming, with the matrix isInterleaved whose element(i,j) indicates if string s3[0,...,i+j-1] can be formed by interleaving s1[0,...,i-1] and s2[0,...,j-1].
Then we can just calculate the element (i,j) by:

isInterleaved[i][j] = (isInterleaved[i-1][j] && s1.charAt(i-1)==s3.charAt(i+j-1)) || (isInterleaved[i][j-1] && s2.charAt(j-1)==s3.charAt(i+j-1))

Instead of using $O(n*m)$ space to keep the whole matrix, we can just keep the previousRow and the actual row, reducing to $O(m)$ space

class Solution:
    # @return a boolean
    def isInterleave(self, s1, s2, s3):
        isInterleaved = []
        if len(s3) != len(s1)+len(s2):
            return False
        for i in range(0,len(s1)+1):
            row = []
            for j in range(0,len(s2)+1):
                if i == 0 and  j==0:
                    row.append(True)
                else:
                    temp = False
                    if j>0: 
                        temp = row[j-1] and s3[i+j-1] == s2[j-1]
                    if i>0: 
                        temp = temp or (previousRow[j] and s3[i+j-1] == s1[i-1])
                    row.append(temp)
            previousRow = row
        return row[len(s2)]