# 97. Interleaving String

#### 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 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`.

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**Example 1:**

![](https://assets.leetcode.com/uploads/2020/09/02/interleave.jpg)

```
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
```

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**Constraints:**

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

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

```python
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]
```