> ## Documentation Index
> Fetch the complete documentation index at: https://leetcode-py.wisl.dev/llms.txt
> Use this file to discover all available pages before exploring further.

> ## Agent Instructions
> leetcode-py is a Python LeetCode practice environment generator with one CLI: lcpy. It is not a service or platform.
> Each problem is a directory under leetcode/ with README.md, solution.py, test_solution.py, helpers.py, and playground.ipynb. lcpy gen creates them from JSON templates bundled with the package.
> Examples are backed by tests; copy them verbatim.

# Interleaving String Python Solution with Tests

> Tested Python solution for LeetCode 97 with 18 pytest cases. Generate a practice environment with lcpy.

LeetCode 97, Medium. Topics: String, Dynamic Programming. [View on LeetCode](https://leetcode.com/problems/interleaving-string/description/).

Generate this problem as a practice environment: tested reference solution, 18 [parametrized pytest cases](/practice/testing), and a playground notebook:

```bash theme={"theme":{"light":"github-light","dark":"github-dark"}}
lcpy gen -n 97   # by problem number
lcpy gen -s interleaving_string   # by problem name
```

## Problem

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`.

### Examples

![Example 1](https://assets.leetcode.com/uploads/2020/09/02/interleave.jpg)

```
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 "aadbbcbcac".
```

```
Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc"
Output: false
Explanation: Notice how it is hard to find a viable interleaving because s3 must preserve the character order of s1 and s2.
```

```
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.

## Solution

Reference implementation from [solution.py on GitHub](https://github.com/wislertt/leetcode-py/blob/main/leetcode/interleaving_string/solution.py), full suite in [test\_solution.py](https://github.com/wislertt/leetcode-py/blob/main/leetcode/interleaving_string/test_solution.py):

```python theme={"theme":{"light":"github-light","dark":"github-dark"}}
class Solution:
    # Time: O(m * n)
    # Space: O(n) using a single rolling row
    def is_interleave(self, s1: str, s2: str, s3: str) -> bool:
        m, n = len(s1), len(s2)
        if m + n != len(s3):
            return False
        if n > m:
            # Ensure s1 is the longer string so the rolling row stays minimal
            return self.is_interleave(s2, s1, s3)

        # dp[j] = True if s3[:i+j] is an interleaving of s1[:i] and s2[:j]
        dp = [False] * (n + 1)
        for i in range(m + 1):
            for j in range(n + 1):
                if i == 0 and j == 0:
                    dp[j] = True
                elif i == 0:
                    dp[j] = dp[j - 1] and s2[j - 1] == s3[j - 1]
                elif j == 0:
                    dp[j] = dp[j] and s1[i - 1] == s3[i - 1]
                else:
                    dp[j] = (dp[j] and s1[i - 1] == s3[i + j - 1]) or (
                        dp[j - 1] and s2[j - 1] == s3[i + j - 1]
                    )
        return dp[n]
```

## Complexity

| Time      | Space                           |
| --------- | ------------------------------- |
| O(m \* n) | O(n) using a single rolling row |

## Tags

[NeetCode 150](/catalog/neetcode-150), [NeetCode 250](/catalog/neetcode-250), [NeetCode All](/catalog/neetcode).