Given two strings `s` and `t`, return `true` if `t` is an anagram of `s` — that is, `t` uses exactly the same letters as `s`, the same number of times, just reordered. Otherwise return `false`.
Example: s = "anagram", t = "nagaram" → true
Sort & compare: Sort both strings; they're anagrams iff the sorted forms are equal. (time O(n log n), space O(n))
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Why the best approach wins
Sorting both strings works but costs O(n log n). Counting letters is O(n): walk s adding to a tally and t subtracting, then check every tally is zero. With a fixed alphabet the count table is constant size, so space is O(1).
Sort & compare: O(n log n) time / O(n) spaceLetter counts: O(n) time / O(1) space
Your turn — implement isAnagram
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