97.Interleaving String
===

###### tags: `Medium` `String` `DP`

[97. Interleaving String](https://leetcode.com/problems/interleaving-string/)

### 題目描述

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`.

### 範例

**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?

### 解答

#### TypeScript

`dp[i][j]` 表示 `s1` 的前 `i` 個字元和 `s2` 的前 `j` 個字元是否可以交錯組合成 `s3` 的前 `i + j` 個字元。

從下面兩個方向達到目前狀態 `(i,j)`：

1. 從 `(i - 1, j)` 轉移：
- 如果 `dp[i - 1][j] == true`，代表著 `s1` 的前 `i - 1` 個字元和 `s2` 的前 `j` 個字元可以交錯組合成 `s3` 的前 `i + j - 1` 個字元。
- 然後我們只需檢查 `s1[i - 1]` 是否等於 `s3[i + j - 1]`。如果相等，則 `dp[i][j]` 也為 `true`
2. 從 `i, j - 1` 轉移：
- 如果 `dp[i, j - 1] == true`，代表著 `s1` 的前 `i` 個字元和 `s2` 的前 `j - 1` 個字元可以交錯組合成 `s3` 的前 `i + j - 1` 個字元。
- 一樣檢查 `s2[j - 1]` 是否等於 `s3[i + j - 1]`。如果相等，則 `dp[i][j]` 也為 `true`

function isInterleave(s1: string, s2: string, s3: string): boolean {
  if (s1.length + s2.length !== s3.length) return false;

  const dp: boolean[][] = new Array(s1.length + 1)
    .fill(0)
    .map(() => new Array(s2.length + 1).fill(false));
  dp[0][0] = true;

  // s2 = ''
  for (let i = 1; i <= s1.length; i++) {
    dp[i][0] = dp[i - 1][0] && s1[i - 1] === s3[i - 1];
  }

  // s1 = ''
  for (let j = 1; j <= s2.length; j++) {
    dp[0][j] = dp[0][j - 1] && s2[j - 1] === s3[j - 1];
  }

  for (let i = 1; i <= s1.length; i++) {
    for (let j = 1; j <= s2.length; j++) {
      dp[i][j] =
        (dp[i - 1][j] && s1[i - 1] === s3[i + j - 1]) ||
        (dp[i][j - 1] && s2[j - 1] === s3[i + j - 1]);
    }
  }

  return dp[s1.length][s2.length];
}

> [name=sheep][time=Fri, 25 Aug 2023]

#### Python

class Solution:
    def isInterleave(self, s1: str, s2: str, s3: str) -> bool:
        if len(s1) + len(s2) != len(s3):
            return False

        @cache
        def dfs(i, j):
            if i == 0 and j == 0:
                return True
            if i == 0:
                return s2[j - 1] == s3[j - 1] and dfs(i, j - 1)
            if j == 0:
                return s1[i - 1] == s3[i - 1] and dfs(i - 1, j)
            return (s1[i - 1] == s3[i + j - 1] and dfs(i - 1, j)) \
                or (s2[j - 1] == s3[i + j - 1] and dfs(i, j - 1))

        return dfs(len(s1), len(s2))

> [name=Yen-Chi Chen][time=Fri, Aug 25, 2023]

### Reference

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