What is the time complexity of Valid Anagram?

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

What data structures are used to solve Valid Anagram?

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

What are the key insights for solving Valid Anagram?

Insight 1: Anagrams have the same number of each character. This means we can count the occurrences of each character in both strings and compare the counts. Insight 2: Using a hash table (or array in this case, since characters are limited) is efficient for counting character frequencies. Alternatively, sorting both strings allows for direct comparison.

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Valid Anagram - LeetCode 242 Solution

Q: How to solve Valid Anagram?

The provided solutions use different approaches to solve the problem. The Python solution elegantly uses the `sorted()` function to sort both strings and compares the sorted results. The Java and C solutions use an array to count character frequencies, ensuring constant-time access to counts. The C++ solution uses maps to achieve the same result, albeit with slightly higher overhead.

Q: Which companies ask LeetCode 242. Valid Anagram in interviews?

This problem is frequently asked at Accenture, Affirm, American Express, Atlassian, BlackRock and 25 more companies.

Valid Anagram - Complete Solution Guide

Valid Anagram is LeetCode problem 242, 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 , return true if t is an anagram of s , and false otherwise. Example 1: Input: s = "anagram", t = "nagaram" Output: true Example 2: Input: s = "rat", t = "car" Output: false Constraints: 1 <= s.length, t.length <= 5 * 10 4 s and t consist of lowercase English letters. Follow up: What if the inputs contain Unicode characters? How would you adapt your solution to such a case?

Detailed Explanation

The problem asks whether two given strings, `s` and `t`, are anagrams of each other. Anagrams are words or phrases formed by rearranging the letters of another word or phrase. For example, "anagram" and "nagaram" are anagrams. The input consists of two strings, `s` and `t`, containing only lowercase English letters. The output is a boolean value: `true` if `t` is an anagram of `s`, and `false` otherwise. The strings can have lengths between 1 and 50,000 characters.

Solution Approach

The provided solutions use different approaches to solve the problem. The Python solution elegantly uses the `sorted()` function to sort both strings and compares the sorted results. The Java and C solutions use an array to count character frequencies, ensuring constant-time access to counts. The C++ solution uses maps to achieve the same result, albeit with slightly higher overhead.

Step-by-Step Algorithm

Step 1: Check if the lengths of strings `s` and `t` are different. If they are, return `false` because anagrams must have the same length.
Step 2: (For Java, C, and implicitly Python): Iterate through the characters of string `s`. For each character, increment its corresponding count in the character count array/map.
Step 3: (For Java, C): Iterate through the characters of string `t`. For each character, decrement its corresponding count in the character count array.
Step 4: (For Java, C): Iterate through the character count array/map. If any count is not zero, return `false`. Otherwise, return `true`.
Step 5: (For Python): Sort strings `s` and `t` using the `sorted()` function. Compare the sorted strings. If they are equal, return `true`; otherwise, return `false`.
Step 6: (For C++): Use two maps to store the character counts for `s` and `t`. Compare the maps for equality. If equal, return `true`, otherwise `false`.

Key Insights

Insight 1: Anagrams have the same number of each character. This means we can count the occurrences of each character in both strings and compare the counts.
Insight 2: Using a hash table (or array in this case, since characters are limited) is efficient for counting character frequencies. Alternatively, sorting both strings allows for direct comparison.
Insight 3: A quick check for unequal string lengths can immediately eliminate many non-anagram cases.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(n)

Topics

This problem involves: Hash Table, String, Sorting.

Companies

Asked at: Accenture, Affirm, American Express, Atlassian, BlackRock, Capgemini, Cognizant, ConsultAdd, Disney, EPAM Systems, Fidelity, Goldman Sachs, Google, Infosys, Mastercard, Nagarro, Nokia, Oracle, PayPal, Rokt, Siemens, Tech Mahindra, Tesla, Visa, Walmart Labs, Wipro, Yandex, Yelp, persistent systems, tcs.

Valid Anagram | LeetCode 242

Easy

Hash Table String Sorting

Time: O(n log n)Space: O(n)

Solution Code

Available in:

class Solution {
    public boolean isAnagram(String s, String t) {
        if (s.length() != t.length()) {
            return false;
        }
        int[] charCount = new int[26];
        for (int i = 0; i < s.length(); i++) {
            charCount[s.charAt(i) - 'a']++;
            charCount[t.charAt(i) - 'a']--;
        }
        for (int count : charCount) {
            if (count != 0) {
                return false;
            }
        }
        return true;
    }
}

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Problem Info

Description

Problem 242: Valid Anagram - Detailed explanation and solution approaches.

Topics

Hash Table String Sorting

Companies

AccentureAffirmAmerican ExpressAtlassianBlackRockCapgeminiCognizantConsultAdd

Stats

Difficulty:Easy

Languages:4

Companies:30

Problem #:242

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

6 steps

1
Step 1: Check if the lengths of strings `s` and `t` are different. If they are, return `false` because anagrams must have the same length.
2
Step 2: (For Java, C, and implicitly Python): Iterate through the characters of string `s`. For each character, increment its corresponding count in the character count array/map.
3
Step 3: (For Java, C): Iterate through the characters of string `t`. For each character, decrement its corresponding count in the character count array.
4
Step 4: (For Java, C): Iterate through the character count array/map. If any count is not zero, return `false`. Otherwise, return `true`.
5
Step 5: (For Python): Sort strings `s` and `t` using the `sorted()` function. Compare the sorted strings. If they are equal, return `true`; otherwise, return `false`.
6
Step 6: (For C++): Use two maps to store the character counts for `s` and `t`. Compare the maps for equality. If equal, return `true`, otherwise `false`.

💡

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 log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The Python solution's time complexity is dominated by the sorting algorithm, which typically has O(n log n) time complexity. The Java, C, and C++ solutions iterate through the strings only once, resulting in O(n) time complexity. The space complexity of the Java and C solutions is constant because they use a fixed-size array to store character counts. The Python solution uses space proportional to the length of the input strings during sorting, and C++ uses space proportional to the number of unique characters.

📖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 defaultdict

In Python, a `collections.defaultdict(int)` could replace the explicit array/map approach to count characters, simplifying the code.

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

Trade-offs: This approach is slightly more concise than the array/map approach in Python but has the same time and space complexity.

✨

Key Insights

1Insight 1: Anagrams have the same number of each character. This means we can count the occurrences of each character in both strings and compare the counts.
2Insight 2: Using a hash table (or array in this case, since characters are limited) is efficient for counting character frequencies. Alternatively, sorting both strings allows for direct comparison.
3Insight 3: A quick check for unequal string lengths can immediately eliminate many non-anagram cases.

📚

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!

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: Empty strings. The solution handles this correctly (either by sorting or by checking the lengths).
Edge case 2: Strings with only one character. The solutions correctly identify these as anagrams or not based on character equality.
Edge case 3: Strings with repeated characters. The solutions handle repeated characters gracefully by using character counts.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the case where the lengths of the two strings are different. Anagrams must have the same length.

✗Common mistake 2: Incorrectly handling character cases (e.g., not converting all characters to lowercase before comparison).

✅

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: Detecting plagiarism. Comparing the frequency of words or characters in two texts can help determine if one is a re-arrangement of the other.

🏦Real-world scenario 2: Searching for anagrams in games or puzzles. This algorithm can efficiently find all anagrams of a given word in a dictionary.

💡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

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ Watch Out

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

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: August 2, 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

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

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

❓

Frequently Asked Questions

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Valid Anagram"?

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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