What is the main approach for solving the 'Check if Strings Can be Made Equal With Operations I' problem?

The main approach involves checking if one string is a rotation or permutation of the other, and performing swaps between indices to make the strings equal.

How can I optimize the brute force approach for this problem?

Optimization can be achieved by recognizing patterns such as string rotations or matching the character sets of both strings before attempting swaps.

Can I use a recursive approach to solve this problem?

Yes, you could use recursion to attempt all possible swaps and check if the strings become equal.

Are there any performance concerns with brute-forcing this problem?

Since the strings are small (length 4), brute force is manageable. However, for larger strings, brute force would be inefficient, and optimized methods should be used.

What are the most common mistakes when solving the 'Check if Strings Can be Made Equal With Operations I' problem?

The most common mistakes include overcomplicating the solution, not leveraging the small input size for brute-force, and not considering string rotations or permutations.

#2839

Easy

auto_awesomeString-driven solution strategy

LeetCode 题解工作台

判断通过操作能否让字符串相等 I

给你两个字符串 s1 和 s2 ，两个字符串的长度都为 4 ，且只包含小写英文字母。你可以对两个字符串中的任意一个执行以下操作任意次：选择两个下标 i 和 j 且满足 j - i = 2 ，然后交换这个字符串中两个下标对应的字符。如果你可以让字符串 s1 和 s2 相等，那么返…

字符串

题目描述

给你两个字符串 s1 和 s2 ，两个字符串的长度都为 4 ，且只包含小写英文字母。

你可以对两个字符串中的 任意一个 执行以下操作任意次：

选择两个下标 i 和 j 且满足 j - i = 2 ，然后交换这个字符串中两个下标对应的字符。

如果你可以让字符串 s1 和 s2 相等，那么返回 true ，否则返回 false 。

示例 1：

输入：s1 = "abcd", s2 = "cdab"
输出：true
解释： 我们可以对 s1 执行以下操作：
- 选择下标 i = 0 ，j = 2 ，得到字符串 s1 = "cbad" 。
- 选择下标 i = 1 ，j = 3 ，得到字符串 s1 = "cdab" = s2 。

示例 2：

输入：s1 = "abcd", s2 = "dacb"
输出：false
解释：无法让两个字符串相等。

提示：

s1.length == s2.length == 4
s1 和 s2 只包含小写英文字母。

lightbulb

解题思路

方法一：计数

我们观察题目中的操作，可以发现，如果字符串的两个下标 $i$ 和 $j$ 的奇偶性相同，那么它们可以通过交换改变顺序。

因此，我们可以统计两个字符串中奇数下标的字符的出现次数，以及偶数下标的字符的出现次数，如果两个字符串的统计结果相同，那么我们就可以通过操作使得两个字符串相等。

时间复杂度 $O(n + |\Sigma|)$ ，空间复杂度 $O(|\Sigma|)$ 。其中 $n$ 是字符串的长度，而 $\Sigma$ 是字符集。

相似题目：

2840. 判断通过操作能否让字符串相等 II

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class Solution:
    def canBeEqual(self, s1: str, s2: str) -> bool:
        return Counter(s1[::2]) == Counter(s2[::2]) and Counter(s1[1::2]) == Counter(
            s2[1::2]
        )

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
Candidate understands the string-driven solution approach.
question_mark
Candidate suggests efficient ways to handle small input sizes.
question_mark
Candidate proposes checking string rotations or character set equivalence as a strategy.

warning

常见陷阱

外企场景

error
Overcomplicating the problem by using unnecessary algorithms or data structures.
error
Failing to consider small input sizes which allow for brute-force solutions.
error
Not checking for rotations or permutations of strings which could offer a simpler solution.

swap_horiz

进阶变体

外企场景

arrow_right_alt
If strings are of larger length, brute force would not be feasible, and more efficient algorithms are required.
arrow_right_alt
A variant might include using recursion to repeatedly swap characters until a solution is found.
arrow_right_alt
Testing performance with larger strings as input can provide insight into optimal solution techniques.

help

常见问题

外企场景

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