What is the primary technique for solving the Interleaving String problem?

The primary technique for solving this problem is dynamic programming, specifically using a state transition approach to build up solutions incrementally.

How do we handle empty strings in the Interleaving String problem?

If both s1 and s2 are empty, s3 must also be empty for the result to be true. Any other combination where s3’s length doesn’t match the sum of s1 and s2 lengths will return false.

Why is dynamic programming used in this problem?

Dynamic programming is used because it allows you to solve overlapping subproblems efficiently by breaking down the problem into smaller, manageable states, avoiding redundant calculations.

What are common mistakes when solving the Interleaving String problem?

Common mistakes include not properly managing the state transitions in the DP table, failing to account for edge cases, or misunderstanding the interleaving condition.

How can GhostInterview help me with the Interleaving String problem?

GhostInterview helps by providing a clear breakdown of the problem-solving process, offering interactive examples, and guiding you through the steps needed to apply dynamic programming techniques.

#97

Medium

auto_awesome状态·转移·动态规划

LeetCode 题解工作台

交错字符串

给定三个字符串 s1 、 s2 、 s3 ，请你帮忙验证 s3 是否是由 s1 和 s2 交错组成的。两个字符串 s 和 t 交错的定义与过程如下，其中每个字符串都会被分割成若干非空子字符串： s = s 1 + s 2 + ... + s n t = t 1 + t 2 + ... +…

字符串动态规划

题目描述

给定三个字符串 s1、s2、s3，请你帮忙验证 s3 是否是由 s1 和 s2 交错组成的。

两个字符串 s 和 t 交错的定义与过程如下，其中每个字符串都会被分割成若干非空子字符串：

s = s₁ + s₂ + ... + s_n
t = t₁ + t₂ + ... + t_m
|n - m| <= 1
交错是 s₁ + t₁ + s₂ + t₂ + s₃ + t₃ + ... 或者 t₁ + s₁ + t₂ + s₂ + t₃ + s₃ + ...

注意：a + b 意味着字符串 a 和 b 连接。

示例 1：

输入：s1 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac"
输出：true

示例 2：

输入：s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc"
输出：false

示例 3：

输入：s1 = "", s2 = "", s3 = ""
输出：true

提示：

0 <= s1.length, s2.length <= 100
0 <= s3.length <= 200
s1、s2、和 s3 都由小写英文字母组成

进阶：您能否仅使用 O(s2.length) 额外的内存空间来解决它?

lightbulb

解题思路

方法一：记忆化搜索

我们记字符串 $s_1$ 的长度为 $m$ ，字符串 $s_2$ 的长度为 $n$ ，如果 $m + n \neq |s_3|$ ，那么 $s_3$ 一定不是 $s_1$ 和 $s_2$ 的交错字符串，返回 false。

接下来，我们设计一个函数 $dfs(i, j)$ ，表示从 $s_1$ 的第 $i$ 个字符和 $s_2$ 的第 $j$ 个字符开始，能否交错组成 $s_3$ 的剩余部分。那么答案就是 $dfs(0, 0)$ 。

函数 $dfs(i, j)$ 的计算过程如下：

如果 $i \geq m$ 并且 $j \geq n$ ，那么说明 $s_1$ 和 $s_2$ 都已经遍历完毕，返回 true。

如果 $i < m$ 并且 $s_1[i] = s_3[i + j]$ ，那么说明 $s_1[i]$ 这个字符是 $s_3[i + j]$ 中的一部分，因此递归地调用 $dfs(i + 1, j)$ 判断 $s_1$ 的下一个字符能否和 $s_2$ 的当前字符匹配，如果能匹配成功，就返回 true。

同理，如果 $j < n$ 并且 $s_2[j] = s_3[i + j]$ ，那么说明 $s_2[j]$ 这个字符是 $s_3[i + j]$ 中的一部分，因此递归地调用 $dfs(i, j + 1)$ 判断 $s_2$ 的下一个字符能否和 $s_1$ 的当前字符匹配，如果能匹配成功，就返回 true。

否则，返回 false。

为了避免重复计算，我们可以使用记忆化搜索。

时间复杂度 $O(m \times n)$ ，空间复杂度 $O(m \times n)$ 。其中 $m$ 和 $n$ 分别是字符串 $s_1$ 和 $s_2$ 的长度。

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class Solution:
    def isInterleave(self, s1: str, s2: str, s3: str) -> bool:
        @cache
        def dfs(i: int, j: int) -> bool:
            if i >= m and j >= n:
                return True
            k = i + j
            if i < m and s1[i] == s3[k] and dfs(i + 1, j):
                return True
            if j < n and s2[j] == s3[k] and dfs(i, j + 1):
                return True
            return False

        m, n = len(s1), len(s2)
        if m + n != len(s3):
            return False
        return dfs(0, 0)

speed

复杂度分析

指标	值
时间	Depends on the final approach
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
The candidate should explain how to fill the DP table, especially how the state transitions work for each character comparison.
question_mark
A correct answer will involve explaining the transition between s1, s2, and s3 and handling edge cases like empty strings.
question_mark
The interviewer should watch for the candidate's understanding of dynamic programming and how it can optimize the problem-solving process.

warning

常见陷阱

外企场景

error
Misunderstanding the interleaving concept and attempting to rearrange characters rather than preserving the order from both strings.
error
Failing to account for edge cases, such as when one or both strings are empty.
error
Not properly filling the DP table or skipping necessary state transitions during the solution process.

swap_horiz

进阶变体

外企场景

arrow_right_alt
The problem could be modified by changing the constraints, such as limiting the maximum lengths of s1 and s2, which would require optimizations in the solution approach.
arrow_right_alt
Another variant could involve allowing characters from s1 and s2 to overlap, requiring more sophisticated DP or backtracking techniques.
arrow_right_alt
A variant could also introduce additional constraints like including special characters or uppercase letters, which would require adaptations in handling input.

help

常见问题

外企场景

继续练习

#91 解码方法 #87 扰乱字符串 #115 不同的子序列