Isomorphic Strings

LeetCode 205 | Difficulty: Easy

Easy

Problem Description

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:

- Mapping `'e'` to `'a'`.

- Mapping `'g'` to `'d'`.

Example 2:

Input: s = "f11", t = "b23"

Output: false

Explanation:

The strings s and t can not be made identical as '1' needs to be mapped to both '2' and '3'.

Example 3:

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

Output: true

Constraints:

- `1 <= s.length <= 5 * 10^4`

- `t.length == s.length`

- `s` and `t` consist of any valid ascii character.

Topics: Hash Table, String

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Approach

Hash Map

Use a hash map for O(1) average lookups. Store seen values, frequencies, or indices. The key question: what should I store as key, and what as value?

When to use

Need fast lookups, counting frequencies, finding complements/pairs.

String Processing

Consider character frequency counts, two-pointer approaches, or building strings efficiently. For pattern matching, think about KMP or rolling hash. For palindromes, expand from center or use DP.

When to use

Anagram detection, palindrome checking, string transformation, pattern matching.

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Solutions

Solution 1: C# (Best: 139 ms)

Metric	Value
Runtime	139 ms
Memory	37.6 MB
Date	2022-02-16

Solution

public class Solution {
    public bool IsIsomorphic(string s, string t) {
        Dictionary<char,char> keyValuePairs = new Dictionary<char,char>();
            for (int i = 0; i < s.Length; i++)
            {
                if(keyValuePairs.ContainsKey(s[i]))
                {
                    if (keyValuePairs[s[i]] == t[i])
                        continue;
              
                    else
                        return false;
                    
                }
                if (keyValuePairs.ContainsValue(t[i]))
                    return false;
                keyValuePairs.Add(s[i], t[i]);
            
            }
            return true;
    }
}

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Complexity Analysis

Approach	Time	Space
Hash Map	$O(n)$	$O(n)$

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Interview Tips

Key Points

• Start by clarifying edge cases: empty input, single element, all duplicates.
• Hash map gives O(1) lookup — think about what to use as key vs value.