class Solution:
    def isInterleave(self, s1: str, s2: str, s3: str) -> bool:
        # 使用dp矩阵
        len1 = len(s1)
        len2 = len(s2)
        len3 = len(s3)

        # 如果s3的长度和s1+s2的不等，就不行
        if len1 + len2 != len3:
            return False
        
        # dp数组全是False
        dp = [[False]*(len2+1) for _ in range(len1+1)]
        # 对于两个数组都是空的情况
        dp[0][0] = True

        # 对于第一行，则判断s2是否能构成s3

        for j in range(1,len2+1):
            if dp[0][j-1] == False:
                break
            else:
                dp[0][j] = (dp[0][j-1] and s2[j-1]==s3[j-1])
        
        # 对于第一列，则判断s1是否能构成s3

        for i in range(1,len1+1):
            if dp[i-1][0] == False:
                break
            else:
                dp[i][0] = (dp[i-1][0] and s1[i-1]==s3[i-1])
        
        # 对于其他位置的情况，依次遍历，从左到右，从上到下
        for i in range(1,len1+1):
            for j in range(1,len2+1):
                dp[i][j] = (dp[i][j-1] and s2[j-1]==s3[i+j-1]) or (dp[i-1][j] and s1[i-1]==s3[i+j-1])
        
        return dp[-1][-1]