What is the key observation in 'Check if Strings Can be Made Equal With Operations II'?

The key observation is that characters can only be swapped if they share the same parity (even or odd index positions).

How does hash table usage help solve this problem?

Hash tables efficiently track the frequency of characters at even and odd indices, enabling quick comparison between the two strings.

Can this problem be solved without considering parity constraints?

No, the problem specifically requires that swaps only occur between characters at positions with matching parity.

What is the time complexity of this problem?

The time complexity is O(n), where n is the length of the strings, due to the need to iterate through both strings once.

Are there any common mistakes to avoid in solving this problem?

Common mistakes include forgetting to check for parity constraints in swaps or overcomplicating the solution by not focusing on efficient counting methods.

#2840

Medium

auto_awesome哈希·表·结合·string

LeetCode 题解工作台

判断通过操作能否让字符串相等 II

给你两个字符串 s1 和 s2 ，两个字符串长度都为 n ，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 ，满足 i 且 j - i 是偶数，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…

哈希表字符串排序

题目描述

给你两个字符串 s1 和 s2 ，两个字符串长度都为 n ，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 ，满足 i < j 且 j - i 是偶数，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 = "abcdba", s2 = "cabdab"
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 = "cbadba" 。
- 选择下标 i = 2 ，j = 4 ，得到字符串 s1 = "cbbdaa" 。
- 选择下标 i = 1 ，j = 5 ，得到字符串 s1 = "cabdab" = s2 。

示例 2：

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

提示：

n == s1.length == s2.length
1 <= n <= 10⁵
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：

2839. 判断通过操作能否让字符串相等 I

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class Solution:
    def checkStrings(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
Assess whether the candidate can think through the problem with the correct pattern, focusing on parity and hashing.
question_mark
Look for an understanding of string manipulation and character frequency counting for efficient solutions.
question_mark
Evaluate how the candidate handles edge cases, such as strings with mismatched characters or impossible swaps.

warning

常见陷阱

外企场景

error
Forgetting to account for the parity constraint in swaps, leading to incorrect conclusions about the possibility of equalizing the strings.
error
Overcomplicating the problem with unnecessary operations instead of focusing on character parity and frequency.
error
Not considering edge cases like when the strings are already equal or contain only one character.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Introduce a condition where swaps are allowed only between characters at specific intervals or positions, testing how well the candidate adapts the solution.
arrow_right_alt
Modify the problem to handle strings of varying lengths, where padding or trimming may be necessary before applying the operations.
arrow_right_alt
Increase the complexity by introducing constraints on the number of swaps allowed, adding a new layer of problem-solving.

help

常见问题

外企场景

继续练习

#2933 高访问员工 #3016 输入单词需要的最少按键次数 II #3035 回文字符串的最大数量