What are isomorphic strings?

Isomorphic strings are two strings where each character from one string can be replaced by another character to form the second string, preserving the order of characters.

What is the optimal approach to solve the Isomorphic Strings problem?

The optimal approach involves using two hash maps to track character mappings and ensure bijective mapping between the strings.

How do hash tables help in solving Isomorphic Strings?

Hash tables allow for efficient mapping and lookups of character transformations, ensuring that each character in one string maps uniquely to a character in the other string.

What is the time complexity of the Isomorphic Strings problem?

The time complexity is O(n), where n is the length of the input strings, as we process each string once.

Can you explain the failure modes of Isomorphic Strings?

Failure modes occur when there is a conflict in character mapping, such as when two characters from string s map to the same character in t, or when the order of characters is disrupted.

#205

Easy

auto_awesome哈希·表·结合·string

LeetCode 题解工作台

同构字符串

给定两个字符串 s 和 t ，判断它们是否是同构的。如果 s 中的字符可以按某种映射关系替换得到 t ，那么这两个字符串是同构的。每个出现的字符都应当映射到另一个字符，同时不改变字符的顺序。不同字符不能映射到同一个字符上，相同字符只能映射到同一个字符上，字符可以映射到自己本身。示例 1：输入…

哈希表字符串

题目描述

给定两个字符串 s 和 t ，判断它们是否是同构的。

如果 s 中的字符可以按某种映射关系替换得到 t ，那么这两个字符串是同构的。

每个出现的字符都应当映射到另一个字符，同时不改变字符的顺序。不同字符不能映射到同一个字符上，相同字符只能映射到同一个字符上，字符可以映射到自己本身。

示例 1：

输入：s = "egg", t = "add"

输出：true

解释：

字符串 s 和 t 可以通过以下方式变得相同：

将 'e' 映射为 'a'。
将 'g' 映射为 'd'。

示例 2：

输入：s = "f11", t = "b23"

输出：false

解释：

字符串 s 和 t 无法变得相同，因为 '1' 需要同时映射到 '2' 和 '3'。

示例 3：

输入：s = "paper", t = "title"

输出：true

提示：

1 <= s.length <= 5 * 10⁴
t.length == s.length
s 和 t 由任意有效的 ASCII 字符组成

lightbulb

解题思路

方法一：哈希表或数组

我们可以用两个哈希表或数组 $d_1$ 和 $d_2$ 记录 $s$ 和 $t$ 中字符的映射关系。

遍历 $s$ 和 $t$ ，如果 $d_1$ 和 $d_2$ 中对应的字符映射关系不同，则返回 false，否则更新 $d_1$ 和 $d_2$ 中对应的字符映射关系。遍历结束，说明 $s$ 和 $t$ 是同构的，返回 true。

时间复杂度 $O(n)$ ，空间复杂度 $O(C)$ 。其中 $n$ 为字符串 $s$ 的长度；而 $C$ 为字符集大小，本题中 $C = 256$ 。

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class Solution:
    def isIsomorphic(self, s: str, t: str) -> bool:
        d1 = {}
        d2 = {}
        for a, b in zip(s, t):
            if (a in d1 and d1[a] != b) or (b in d2 and d2[b] != a):
                return False
            d1[a] = b
            d2[b] = a
        return True

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
Check if the candidate correctly identifies the need for two hash maps to track character mappings.
question_mark
Evaluate the candidate's ability to ensure bijective character mappings without conflicts.
question_mark
See if the candidate can explain the importance of maintaining the character order during transformation.

warning

常见陷阱

外企场景

error
Not checking for bijective mapping, which can cause mapping conflicts where multiple characters map to the same character.
error
Forgetting to handle cases where characters from one string map to themselves, such as when 'a' maps to 'a'.
error
Relying on string comparison without using hash maps, which can lead to inefficiencies and incorrect results.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Consider strings of varying lengths, ensuring the solution works for edge cases such as empty strings or strings with one character.
arrow_right_alt
Handle cases where the strings are of equal length but contain a mix of uppercase and lowercase characters.
arrow_right_alt
Test the solution with strings containing non-alphabetical ASCII characters to ensure robustness.

help

常见问题

外企场景

继续练习

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