Valid Anagram

Valid Anagram

Problem

Given two strings s and t, check if t is an anagram of s.

An anagram means both strings contain the same characters with the same frequency, just arranged differently.

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Strategy

The idea is simple:

• Count how many times each character appears in s
• Then try to “cancel out” those counts using t

If everything balances to zero, the strings are anagrams.

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Code

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

        count = {}

        for ch in s:
            count[ch] = count.get(ch, 0) + 1

        for ch in t:
            if ch not in count or count[ch] == 0:
                return False
            count[ch] -= 1

        return True

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Key Lines Explained

• if len(s) != len(t):
  If lengths differ, they can’t be anagrams.

• count[ch] = count.get(ch, 0) + 1
  Count frequency of each character in s.

• if ch not in count or count[ch] == 0:
  If a character in t doesn’t match, return False.

• count[ch] -= 1
  Reduce count as we match characters.

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Why This Works

Anagrams have the same characters in the same quantities.

By counting and cancelling, we’re checking if both strings are perfectly balanced.

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Complexity

• Time: O(n)
• Space: O(1) (since only lowercase letters)

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Final Note

This problem is less about strings and more about counting.

Once you think in terms of frequency instead of order, it becomes straightforward.