How do you solve Valid Anagram on LeetCode?

Valid Anagram (LeetCode #242) 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: Anagrams must have the same characters with the same frequency.

What is the time complexity of Valid Anagram?

The optimal solution for Valid Anagram runs in O(n) time and uses O(n) space. Counting characters takes O(n) time, and we use a hash map to store counts, which takes O(n) space in the worst case.

What is the brute force solution for Valid Anagram?

The brute force approach for Valid Anagram is Brute Force, with O(n²) time complexity and O(1) space complexity. The simplest way to check if two strings are anagrams is to sort both strings and compare them. If they are identical after sorting, they are anagrams. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Valid Anagram?

Valid Anagram uses the following concepts: Hash Table, String, Sorting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Valid Anagram?

Always consider edge cases, such as empty strings or strings of different lengths. Explain your thought process clearly while coding; it shows your understanding. Discuss the trade-offs between different approaches before jumping into coding.

What are common mistakes when solving Valid Anagram?

Not checking if the lengths of the strings are equal before proceeding. Confusing character counts or not resetting counts correctly.

#242

Valid Anagram

Easy

Hash TableStringSortingHash 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)

Instead of sorting, we can count the frequency of each character in both strings using a hash map. If the counts match, the strings are anagrams.

⚙️

Algorithm

3 steps

1Step 1: Create a frequency map for characters in string s.
2Step 2: Decrease the count for each character found in string t.
3Step 3: Check if all counts in the frequency map are zero.

solution.py4 lines

1from collections import Counter
2
3def isAnagram(s, t):
4    return Counter(s) == Counter(t)

ℹ

Complexity note: Counting characters takes O(n) time, and we use a hash map to store counts, which takes O(n) space in the worst case.

1Anagrams must have the same characters with the same frequency.
2Sorting is a straightforward approach but not the most efficient.

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