242. Valid Anagram
## description
## 242. Valid Anagram
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 * 104
- –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?
## notes
### Intuition
Two strings are anagrams if they contain the same characters with the same frequencies. A fixed-size frequency array can track character counts efficiently.
### Implementation
Create a 26-element array for lowercase letters. Iterate through string s, incrementing the count at each character's index (ord(char) - ord('a')). Then iterate through t, decrementing the counts. If the strings are anagrams, all counts will be zero at the end.
### Edge-cases
Strings of different lengths cannot be anagrams. This is implicitly checked—the frequency array won't be all zeros if lengths differ.
- –Time: O(n) — two passes through the strings
- –Space: O(1) — frequency array is fixed at 26 elements
### Complexity
## solution
1class Solution:
2 def isAnagram(self, s: str, t: str) -> bool:
3 freq = [0] * 26
4
5 for char in s:
6 freq[ord(char) - ord('a')] += 1
7
8 for char in t:
9 freq[ord(char) - ord('a')] -= 1
10
11 return all(f == 0 for f in freq)
12
13
14
15
16if __name__ == "__main__":
17 # Include one-off tests here or debugging logic that can be run by running this file
18 # e.g. print(solution.two_sum([1, 2, 3, 4], 3))
19 solution = Solution()
20