Interleaving string

class Solution:
    def isInterleave(self, s1: str, s2: str, s3: str) -> bool:
        if len(s1) + len(s2) != len(s3):
            return False
        m, n = len(s1), len(s2)
        dp = [[False] * (n + 1) for _ in range(m + 1)]
        dp[m][n] = True
        for i in range(m, -1, -1):
            for j in range(n, -1, -1):
                if i < len(s1) and s1[i] == s3[i + j] and dp[i + 1][j]:
                    dp[i][j] = True

                if j < len(s2) and s2[j] == s3[i + j] and dp[i][j + 1]:
                    dp[i][j] = True

        return dp[0][0]

        # @cache
        # def dfs(i, j):
        #     if i == len(s1) and j == len(s2):
        #         return True

        #     if i < len(s1) and s1[i] == s3[i + j]:
        #         if dfs(i + 1, j):
        #             return True

        #     if j < len(s2) and s2[j] == s3[i + j]:
        #         if dfs(i, j + 1):
        #             return True

        #     return False

        # return dfs(0, 0)

Interleaving String

Difficulty: Medium

Given strings s1, s2, and s3, find whether s3 is formed by an interleaving of s1 and s2.

An interleaving of two strings s and t is a configuration where s and t are divided into n and m substrings respectively, such that:

  • s = s1 + s2 + ... + sn
  • t = t1 + t2 + ... + tm
  • |n - m| <= 1
  • The interleaving is s1 + t1 + s2 + t2 + s3 + t3 + ... or t1 + s1 + t2 + s2 + t3 + s3 + ...

Note: a + b is the concatenation of strings a and b.

 

Example 1:

Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac"
Output: true
Explanation: One way to obtain s3 is:
Split s1 into s1 = "aa" + "bc" + "c", and s2 into s2 = "dbbc" + "a".
Interleaving the two splits, we get "aa" + "dbbc" + "bc" + "a" + "c" = "aadbbcbcac".
Since s3 can be obtained by interleaving s1 and s2, we return true.

Example 2:

Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc"
Output: false
Explanation: Notice how it is impossible to interleave s2 with any other string to obtain s3.

Example 3:

Input: s1 = "", s2 = "", s3 = ""
Output: true

 

Constraints:

  • 0 <= s1.length, s2.length <= 100
  • 0 <= s3.length <= 200
  • s1, s2, and s3 consist of lowercase English letters.

 

Follow up: Could you solve it using only O(s2.length) additional memory space?