Solving Anagrams with a Hashmap

Table of Contents

1. Introduction
2. Problem Statement
3. What is an Anagram?
4. Examples of Anagrams
5. Constraints
6. Solution Approach
7. Explanation of the Code
8. Time and Space Complexity
9. Follow-up Question
   • Adapting the Solution for Unicode Characters
10. Conclusion

Introduction

In this article, we will discuss the problem of determining whether two strings are anagrams of each other. We will explore the concept of anagrams, provide examples, discuss the constraints of the problem, and present a solution approach using hashing. Additionally, we will analyze the time and space complexity of the solution, and address a follow-up question related to handling Unicode characters.

Problem Statement

The problem is to determine whether two given strings, s and t, are anagrams of each other. An anagram is a word or phrase formed by rearranging the letters of another word or phrase. If t can be obtained by rearranging the letters of s, the function should return True; otherwise, it should return False.

What is an Anagram?

An anagram is a word or phrase formed by rearranging the letters of another word or phrase. For example, "abcd" and "bcad" are anagrams because rearranging the letters of "abcd" can result in "bcad".

Examples of Anagrams

• Example 1:
  • Input: s = "listen", t = "silent"
  • Output: True
• Example 2:
  • Input: s = "abc", t = "def"
  • Output: False

Constraints

• The length of the strings, s and t, will be between 1 and 10^4.
• The strings, s and t, consist only of lowercase English letters.

Solution Approach

The solution approach involves using hashing to store the frequency of characters in the strings. By comparing the frequency of characters in both strings, we can determine if they are anagrams. If the frequency of characters is the same for both strings, they are anagrams; otherwise, they are not.

Explanation of the Code

The code uses an array of size 26 to represent the frequency of lowercase characters. The frequency of each character in the first string, s, is incremented in the array, while the frequency of each character in the second string, t, is decremented. If the resulting array consists of 26 zeros, the strings are anagrams. Otherwise, they are not.

Time and Space Complexity

The time complexity of the solution is O(s + t), where s and t are the lengths of the strings. The space complexity is O(26) or O(1), as we only need to store the frequency of characters in an array of size 26.

Follow-up Question

What if the input contains Unicode characters? How would you adapt the solution to such a case?

To handle Unicode characters, the size of the array used to store the frequency of characters should be increased to cover the range of Unicode character codes. Assuming Unicode characters have sky values between 0 and 511, the array size should be 512. The rest of the algorithm remains the same.

Conclusion

In this article, we discussed the problem of determining whether two strings are anagrams. We explored the concept of anagrams, provided examples, and presented a solution approach using hashing. We analyzed the time and space complexity of the solution and addressed a follow-up question related to handling Unicode characters. By understanding the solution approach and the algorithm's time and space complexity, we can effectively solve the problem of validating anagrams.