97. Interleaving String

Medium

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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 they are divided into non-empty substrings 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

Example 2:

Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc"
Output: false

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?

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