What is the best approach to solve Check If Two String Arrays are Equivalent?

Concatenate all strings in each array and compare the resulting strings, or use a two-pointer streaming method to reduce space.

Does the order of array elements affect equivalence?

Yes, elements must be concatenated in the original order; changing the order can make equivalent strings appear different.

Can empty strings in the arrays cause issues?

Empty strings are valid and do not affect concatenation results, but they should not be skipped in pointer traversal approaches.

How to handle very large arrays efficiently?

Use a two-pointer approach to compare characters without forming full concatenated strings, reducing space complexity.

Is this problem pattern considered Array plus String?

Yes, the core pattern involves combining array elements into strings and comparing them, which is the Array plus String pattern.

#1662

Easy

auto_awesome数组·string

LeetCode 题解工作台

检查两个字符串数组是否相等

给你两个字符串数组 word1 和 word2 。如果两个数组表示的字符串相同，返回 true ；否则，返回 false 。数组表示的字符串是由数组中的所有元素按顺序连接形成的字符串。示例 1：输入： word1 = ["ab", "c"], word2 = ["a", "bc"] 输出…

数组字符串

题目描述

给你两个字符串数组 word1 和 word2 。如果两个数组表示的字符串相同，返回 true ；否则，返回 false 。

数组表示的字符串 是由数组中的所有元素 按顺序 连接形成的字符串。

示例 1：

输入：word1 = ["ab", "c"], word2 = ["a", "bc"]
输出：true
解释：
word1 表示的字符串为 "ab" + "c" -> "abc"
word2 表示的字符串为 "a" + "bc" -> "abc"
两个字符串相同，返回 true

示例 2：

输入：word1 = ["a", "cb"], word2 = ["ab", "c"]
输出：false

示例 3：

输入：word1  = ["abc", "d", "defg"], word2 = ["abcddefg"]
输出：true

提示：

1 <= word1.length, word2.length <= 10³
1 <= word1[i].length, word2[i].length <= 10³
1 <= sum(word1[i].length), sum(word2[i].length) <= 10³
word1[i] 和 word2[i] 由小写字母组成

lightbulb

解题思路

方法一：字符串拼接

将两个数组中的字符串拼接成两个字符串，然后比较两个字符串是否相等。

时间复杂度 $O(m)$ ，空间复杂度 $O(m)$ 。其中 $m$ 为数组中字符串的总长度。

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class Solution:
    def arrayStringsAreEqual(self, word1: List[str], word2: List[str]) -> bool:
        return ''.join(word1) == ''.join(word2)

speed

复杂度分析

指标	值
时间	complexity is (\mathcal{O}(N)) where N is the total number of characters across both arrays, because every character must be examined. Space complexity is (\mathcal{O}(N)) if concatenating the arrays fully, or (\mathcal{O}(1)) with the two-pointer approach.
空间	\mathcal{O}(N)

psychology

面试官常问的追问

外企场景

question_mark
Ask about handling arrays of different lengths and empty strings.
question_mark
Probe whether the candidate can optimize space using streaming comparison.
question_mark
Check understanding of subtle off-by-one errors when joining array elements.

warning

常见陷阱

外企场景

error
Assuming arrays of equal length guarantee equivalent strings.
error
Neglecting to preserve the order of elements when concatenating.
error
Using inefficient nested loops instead of direct concatenation or pointer traversal.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Check equivalence ignoring case sensitivity.
arrow_right_alt
Verify equivalence allowing wildcard characters in one array.
arrow_right_alt
Compare arrays where elements may be reversed before concatenation.

help

常见问题

外企场景

继续练习

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