Is Isomorphic Strings easy or hard?

Isomorphic Strings is categorized as an Easy problem on LeetCode with an acceptance rate around 48%. The logic is straightforward once you recognize the need for a one-to-one character mapping, making it a common early practice problem for hash tables and string manipulation.

Isomorphic Strings Python/Java solution

In Python, dictionaries are used to store character mappings and sets track used characters. In Java, HashMap and HashSet serve the same role. Both implementations iterate once over the strings, validate mappings, and return false immediately if a conflict appears.

How to solve Isomorphic Strings in O(n)?

Traverse both strings simultaneously and maintain a mapping from characters in the first string to characters in the second. If a character already has a mapping, verify it matches the current character in the other string. Also track used target characters to prevent duplicates. This single pass with hash lookups gives O(n) time.

What is the best approach for Isomorphic Strings?

The best approach uses a one-to-one character mapping with hash maps. Iterate through both strings and map each character from s to t while ensuring no two characters map to the same target. Each lookup and insertion is O(1), so the full algorithm runs in O(n) time with O(k) extra space for unique characters.

Is Isomorphic Strings asked at Google/Amazon/Meta?

Isomorphic Strings is a common entry-level string and hash table problem frequently seen in interviews at companies like Amazon, Google, and Microsoft. It tests understanding of bijective mappings, hash maps, and careful constraint validation during iteration.

What data structure is used in Isomorphic Strings?

The primary data structure is a hash table (or dictionary) to store character mappings. Many solutions also use a set or a second hash map to ensure each target character is mapped only once. These structures provide O(1) lookups and updates.

What is the time complexity of Isomorphic Strings?

The optimal solution runs in O(n) time where n is the length of the strings. Each character pair is processed once and validated using constant-time hash table lookups. Space complexity is typically O(k), where k is the number of unique characters encountered.

Description Solution Video 10:07

205. Isomorphic Strings

Easy48.8% acceptance15 companies

Given two strings s and t, determine if they are isomorphic.

Two strings s and t are isomorphic if the characters in s can be replaced to get t.

All occurrences of a character must be replaced with another character while preserving the order of characters. No two characters may map to the same character, but a character may map to itself.

Example 1:

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

Output: true

Explanation:

The strings s and t can be made identical by: