Valid Anagram in Python Using Two Methods (Sorting & Counting)

Problem Explanation

You are given two strings s and t.
Your task is to return True if t is an anagram of s, otherwise return False.

An anagram means both strings contain the same characters with the same frequency, just in a different order.

Example:

• Input: s = "anagram", t = "nagaram"
  Output: True

• Input: s = "rat", t = "car"
  Output: False

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Method 1: Sorting (Simplest Approach)

Idea

If two strings are anagrams, their sorted forms will be equal.

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Code

class Solution:
    def isAnagram(self, s, t):
        return sorted(s) == sorted(t)

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Explanation

• sorted(s) → sorts characters of string s
• sorted(t) → sorts characters of string t
• If both sorted results are equal → strings are anagrams

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Why Use This?

• Very easy to write and understand
• Best for beginners
• Clean one-line solution

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Method 2: Counting Characters (Optimal Approach)

Idea

Count how many times each character appears in both strings and compare.

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Code

class Solution:
    def isAnagram(self, s, t):
        if len(s) != len(t):
            return False

        count = [0] * 26

        for i in range(len(s)):
            count[ord(s[i]) - ord('a')] += 1
            count[ord(t[i]) - ord('a')] -= 1

        for num in count:
            if num != 0:
                return False

        return True

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Explanation (Step-by-Step)

• if len(s) != len(t):
  If lengths differ → cannot be anagrams

• count = [0] * 26
  Create a list to store frequency of each character

• for i in range(len(s)):
  Loop through both strings

• count[...] += 1
  Increase count for character in s

• count[...] -= 1
  Decrease count for character in t

• if num != 0:
  If any count is not zero → not anagram

• return True
  All counts match → valid anagram

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Key Takeaway

• Sorting → simplest and clean
• Counting → more efficient and optimal

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