How do you solve Isomorphic Strings on LeetCode?

Isomorphic Strings (LeetCode #205) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Isomorphic strings require a one-to-one mapping between characters.

What is the brute force solution for Isomorphic Strings?

The brute force approach for Isomorphic Strings is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible character mapping between the two strings. This means we will try to create a mapping for each character in the first string to the characters in the second string, which can be inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Isomorphic Strings?

Isomorphic Strings uses the following concepts: Hash Table, String. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Isomorphic Strings?

Always clarify the problem constraints and requirements before jumping into coding. Think about edge cases, such as strings with the same characters but different lengths. Practice explaining your thought process as you code, as communication is key in interviews.

What are common mistakes when solving Isomorphic Strings?

Failing to check both directions of mapping (s to t and t to s). Not considering cases where the same character appears multiple times.

#205

Isomorphic Strings

Easy

Hash TableStringHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

We can use two hash maps to track the mappings from characters in s to t and from t to s. This way, we ensure that each character maps uniquely and consistently.

⚙️

Algorithm

5 steps

1Step 1: Initialize two hash maps: one for mapping characters from s to t and another for t to s.
2Step 2: Iterate through the characters of both strings simultaneously.
3Step 3: For each character pair, check if the current character from s is already mapped to a different character in t or vice versa. If so, return false.
4Step 4: If the mappings are valid, update the hash maps.
5Step 5: If the loop completes without conflicts, return true.

solution.py12 lines

1# Full working Python code
2
3def isIsomorphic(s, t):
4    map_s_to_t = {}
5    map_t_to_s = {}
6    for char_s, char_t in zip(s, t):
7        if (char_s in map_s_to_t and map_s_to_t[char_s] != char_t) or 
8           (char_t in map_t_to_s and map_t_to_s[char_t] != char_s):
9            return False
10        map_s_to_t[char_s] = char_t
11        map_t_to_s[char_t] = char_s
12    return True

ℹ

Complexity note: The time complexity is linear because we only traverse the strings once. The space complexity is also linear due to the hash maps storing the mappings.

1Isomorphic strings require a one-to-one mapping between characters.
2Using hash maps allows efficient checking and updating of character mappings.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.