What is the time complexity of Isomorphic Strings?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Isomorphic Strings?

This problem uses Hash Table, String. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Isomorphic Strings?

Insight 1: Using two hash maps (or dictionaries) is crucial for efficiently tracking the mapping between characters in both strings. One map tracks the mapping from `s` to `t`, and the other tracks the reverse mapping. Insight 2: The crucial condition for isomorphism is that each character in `s` must map to only one character in `t`, and vice-versa. The maps help enforce this one-to-one correspondence.

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Isomorphic Strings - LeetCode 205 Solution

Q: Which companies ask LeetCode 205. Isomorphic Strings in interviews?

This problem is frequently asked at Barclays, EPAM Systems, Google, HashedIn, Infosys and 8 more companies.

Isomorphic Strings - Complete Solution Guide

Isomorphic Strings is LeetCode problem 205, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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' . Ex

Detailed Explanation

The problem asks whether two strings, `s` and `t`, are isomorphic. Isomorphic strings mean that there's a one-to-one mapping between the characters of `s` and the characters of `t`, preserving the order. This mapping doesn't need to be a direct character-for-character match; it's about consistent replacements. For example, 'egg' and 'add' are isomorphic because 'e' maps to 'a' and 'g' maps to 'd' consistently. However, 'foo' and 'bar' are not, because 'o' would need to map to both 'a' and 'r', violating the one-to-one rule. The input consists of two strings, `s` and `t`, and the output is a boolean value (true if isomorphic, false otherwise). The constraints specify that the lengths of both strings are equal and are within a certain range.

Solution Approach

The solution uses two hash maps (dictionaries in Python) to track the mappings between characters in `s` and `t`. It iterates through both strings simultaneously. For each pair of characters, it checks if the mappings are consistent. If a character in `s` already has a mapping, it checks if that mapping matches the corresponding character in `t`. Similarly, it checks the reverse mapping from `t` to `s`. If any inconsistency is found (a character maps to two different characters or a conflict in reverse mapping), it returns `false`. Otherwise, if all characters are mapped consistently and the lengths are equal, it returns `true`.

Step-by-Step Algorithm

Step 1: Check if the lengths of strings `s` and `t` are equal. If not, return `false`.
Step 2: Initialize two hash maps, `s_map` and `t_map`. `s_map` will map characters from `s` to `t`, and `t_map` will map characters from `t` to `s`.
Step 3: Iterate through the strings using an index `i`. For each pair of characters `s[i]` and `t[i]`:
Step 4: Check if `s[i]` is in `s_map`. If it is, check if `s_map[s[i]]` equals `t[i]`. If not, return `false`.
Step 5: Check if `t[i]` is in `t_map`. If it is, check if `t_map[t[i]]` equals `s[i]`. If not, return `false`.
Step 6: If neither `s[i]` nor `t[i]` is in their respective maps, add the mapping to both maps: `s_map[s[i]] = t[i]` and `t_map[t[i]] = s[i]`.
Step 7: After the loop, if no inconsistencies were found, return `true`.

Key Insights

Insight 1: Using two hash maps (or dictionaries) is crucial for efficiently tracking the mapping between characters in both strings. One map tracks the mapping from `s` to `t`, and the other tracks the reverse mapping.
Insight 2: The crucial condition for isomorphism is that each character in `s` must map to only one character in `t`, and vice-versa. The maps help enforce this one-to-one correspondence.
Insight 3: Checking the lengths of the strings upfront is a simple but essential optimization. If the lengths are different, they cannot possibly be isomorphic.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Hash Table, String.

Companies

Asked at: Barclays, EPAM Systems, Google, HashedIn, Infosys, LinkedIn, Oracle, Remitly, Salesforce, Tinkoff, Yandex, Zoho, eBay.

Isomorphic Strings | LeetCode 205

Easy

Hash Table String

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public boolean isIsomorphic(String s, String t) {
        if (s.length() != t.length()) {
            return false;
        }

        java.util.HashMap<Character, Character> mapST = new java.util.HashMap<>();
        java.util.HashMap<Character, Character> mapTS = new java.util.HashMap<>();

        for (int i = 0; i < s.length(); i++) {
            char charS = s.charAt(i);
            char charT = t.charAt(i);

            if (mapST.containsKey(charS)) {
                if (mapST.get(charS) != charT) {
                    return false;
                }
            } else {
                if (mapTS.containsKey(charT)) {
                    return false;
                }
                mapST.put(charS, charT);
                mapTS.put(charT, charS);
            }
        }

        return true;
    }
}

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Problem Info

Description

Problem 205: Isomorphic Strings - Detailed explanation and solution approaches.

Topics

Hash Table String

Companies

BarclaysEPAM SystemsGoogleHashedInInfosysLinkedInOracleRemitly

Stats

Difficulty:Easy

Languages:4

Companies:13

Problem #:205

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Check if the lengths of strings `s` and `t` are equal. If not, return `false`.
2
Step 2: Initialize two hash maps, `s_map` and `t_map`. `s_map` will map characters from `s` to `t`, and `t_map` will map characters from `t` to `s`.
3
Step 3: Iterate through the strings using an index `i`. For each pair of characters `s[i]` and `t[i]`:
4
Step 4: Check if `s[i]` is in `s_map`. If it is, check if `s_map[s[i]]` equals `t[i]`. If not, return `false`.
5
Step 5: Check if `t[i]` is in `t_map`. If it is, check if `t_map[t[i]]` equals `s[i]`. If not, return `false`.
6
Step 6: If neither `s[i]` nor `t[i]` is in their respective maps, add the mapping to both maps: `s_map[s[i]] = t[i]` and `t_map[t[i]] = s[i]`.
7
Step 7: After the loop, if no inconsistencies were found, return `true`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The dominant operation is the single iteration through the strings. The hash map lookups and insertions are considered O(1) on average for a good hash function.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Using a single hash map

It's possible to solve this using only one hash map. The key would be a character from `s`, and the value would be the corresponding character from `t`. You would check for conflicts during insertion.

Time: O(n)Space: O(1)

Trade-offs: Using a single map might slightly reduce space usage, but the double-map approach provides a more readable and arguably clearer solution for this specific problem.

✨

Key Insights

1Insight 1: Using two hash maps (or dictionaries) is crucial for efficiently tracking the mapping between characters in both strings. One map tracks the mapping from `s` to `t`, and the other tracks the reverse mapping.
2Insight 2: The crucial condition for isomorphism is that each character in `s` must map to only one character in `t`, and vice-versa. The maps help enforce this one-to-one correspondence.
3Insight 3: Checking the lengths of the strings upfront is a simple but essential optimization. If the lengths are different, they cannot possibly be isomorphic.

📚

Key Terms Explained

For Students

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Empty strings. The solution correctly handles this case because the length check will pass, and the loop won't execute. It will return `true` because empty strings are considered isomorphic.
Edge case 2: Strings with repeated characters. The solution efficiently handles this by using the hash maps to detect inconsistencies in the mappings.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to check the reverse mapping. Only checking `s_map` can lead to incorrect results because it doesn't guarantee a one-to-one mapping.

✗Common mistake 2: Inefficiently using nested loops. A single pass through the strings is much more efficient than comparing each character to every other character.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Data encryption/decryption. Isomorphic mappings could be a simplified representation of a substitution cipher. Checking for isomorphism could be a way to verify the consistency of the cipher.

🏦Real-world scenario 2: Code optimization. In compiler design or code analysis, determining isomorphic structures can be a step in identifying optimization opportunities.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

Hash tables trade space for time. Before using one, consider if the problem constraints allow the extra memory. In interviews, always mention this trade-off.

🎯 Interview Tip

Use this problem to demonstrate clean code practices. Variable naming and code organization matter.

⚠️ Watch Out

Don't forget to handle key collisions and null/undefined keys. Consider what happens with duplicate values.

👨‍💻 From Experience

This is one of the first problems I solved when learning Hash Table. It's a great foundation for more complex problems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: June 26, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 24

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 17

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 10

❓

Frequently Asked Questions

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Isomorphic Strings"?

For this easy-level Hash Table problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether string techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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