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LeetCode Solutions

242. Valid Anagram

Time: $O(n)$

Space: $O(128) = O(1)$

			


class Solution {
 public:
  bool isAnagram(string s, string t) {
    if (s.length() != t.length())
      return false;

    vector<int> count(128);

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

    for (const char c : t)
      if (--count[c] < 0)
        return false;

    return true;
  }
};




			


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

    int[] count = new int[128];

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

    for (final char c : t.toCharArray())
      if (--count[c] < 0)
        return false;

    return true;
  }
}




			


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

    dict = Counter(s)

    for c in t:
      dict[c] -= 1
      if dict[c] < 0:
        return False

    return True