LeetCode: Valid Anagram Solution

Approach 1

Sort strings and compare

Time complexity: O(nlog(n))

Implementation

1var isAnagram = function (s, t) {
2  const sortString = str =>
3    str
4      .split("")
5      .sort((a, b) => a.localeCompare(b))
6      .join("")
7  return sortString(s) === sortString(t)
8}

Approach 2

Alphabetically count and compare

Time complexity: O(n)

Implementation

1var isAnagram = function (s, t) {
2  const [countS, countT] = [new Uint16Array(26), new Uint16Array(26)]
3
4  for (let char of s) {
5    countS[char.charCodeAt(0) - "a".charCodeAt(0)] += 1
6  }
7
8  for (let char of t) {
9    countT[char.charCodeAt(0) - "a".charCodeAt(0)] += 1
10  }
11
12  for (let i = 0; i < 26; i++) {
13    if (countS[i] !== countT[i]) {
14      return false
15    }
16  }
17
18  return true
19}

References

Original problem

Similar problems

Group Anagrams

Palindrome Permutation

Find All Anagrams in a String

Find Resultant Array After Removing Anagrams