#205

Easy

LeetCode Problem Workspace

Isomorphic Strings

Determine if two strings are isomorphic by checking if one string can be transformed into the other using character replacements.

hash table string

Problem Statement

Given two strings, s and t, you need to check if they are isomorphic. The strings are isomorphic if each character in s can be replaced by a character in t, maintaining the character order. Importantly, no two characters in s can map to the same character in t, but a character can map to itself.

For example, the strings 'egg' and 'add' are isomorphic because the character 'e' maps to 'a' and 'g' maps to 'd'. However, 'foo' and 'bar' are not isomorphic, as the 'o' in 'foo' would need to map to both 'a' and 'r'.

Examples

Example 1

Input: s = "egg", t = "add"

Output: true

The strings s and t can be made identical by:

Example 2

Input: s = "foo", t = "bar"

Output: false

The strings s and t can not be made identical as 'o' needs to be mapped to both 'a' and 'r' .

Example 3

Input: s = "paper", t = "title"

Output: true

Example details omitted.

Constraints

1 <= s.length <= 5 * 104
t.length == s.length
s and t consist of any valid ascii character.

Solution Approach

Use Hash Maps for Character Mappings

Utilize two hash maps to track the character mappings between the two strings. As you iterate through the strings, check if each character from string s can be mapped to a unique character in t. If any mapping conflict occurs, return false.

Check for Bijective Mappings

Ensure that the mapping between characters is bijective. This means that not only should every character in s map to a unique character in t, but also vice versa. This guarantees that no two characters from s map to the same character in t.

Ensure Order of Characters is Preserved

As you process the strings, confirm that the order of characters is maintained. The order should not change while performing replacements, ensuring the strings remain structurally identical after the transformation.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the length of the input strings, as you must iterate through each character of both strings once. Using hash maps allows for constant time lookups, so the overall time complexity is O(n), where n is the length of the strings. The space complexity is also O(n) due to the additional hash maps used for character mappings.

What Interviewers Usually Probe

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

Common Pitfalls or Variants

Common pitfalls

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

Follow-up variants

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

FAQ

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.

terminal

Solution

Solution 1: Hash Table or Array

We can use two hash tables or arrays $d_1$ and $d_2$ to record the character mapping relationship between $s$ and $t$.

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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

Solution 2

#### Python3

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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

Continue Practicing

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