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242. Valid Anagram 👍

  • Time: $O(n)$
  • Space: $O(26) = O(1)$
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class Solution {
 public:
  bool isAnagram(string s, string t) {
    if (s.length() != t.length())
      return false;

    vector<int> count(26);

    for (const char c : s)
      ++count[c - 'a'];

    for (const char c : t) {
      if (count[c - 'a'] == 0)
        return false;
      --count[c - 'a'];
    }

    return true;
  }
};

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class Solution {
  public boolean isAnagram(String s, String t) {
    if (s.length() != t.length())
      return false;

    int[] count = new int[26];

    for (final char c : s.toCharArray())
      ++count[c - 'a'];

    for (final char c : t.toCharArray()) {
      if (count[c - 'a'] == 0)
        return false;
      --count[c - 'a'];
    }

    return true;
  }
}

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

    count = collections.Counter(s)
    count.subtract(collections.Counter(t))
    return all(freq == 0 for freq in count.values())

var isAnagram = function(s, t) {
  if (s.length !== t.length) return false;
  const cnt = {};
  for (const c of s) cnt[c] = (cnt[c] || 0) + 1;
  for (const c of t) {
    if (!cnt[c]) return false;
    cnt[c]--;
  }
  return true;
};
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