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438. Find All Anagrams in a String 👍

  • Time: $O(n)$
  • Space: $O(26)$
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class Solution {
 public:
  vector<int> findAnagrams(string s, string p) {
    vector<int> ans;
    vector<int> count(128);
    int required = p.length();

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

    for (int l = 0, r = 0; r < s.length(); ++r) {
      if (--count[s[r]] >= 0)
        --required;
      while (required == 0) {
        if (r - l + 1 == p.length())
          ans.push_back(l);
        if (++count[s[l++]] > 0)
          ++required;
      }
    }

    return ans;
  }
};

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class Solution {
  public List<Integer> findAnagrams(String s, String p) {
    List<Integer> ans = new ArrayList<>();
    int[] count = new int[128];
    int required = p.length();

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

    for (int l = 0, r = 0; r < s.length(); ++r) {
      if (--count[s.charAt(r)] >= 0)
        --required;
      while (required == 0) {
        if (r - l + 1 == p.length())
          ans.add(l);
        if (++count[s.charAt(l++)] > 0)
          ++required;
      }
    }

    return ans;
  }
}

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class Solution:
  def findAnagrams(self, s: str, p: str) -> list[int]:
    ans = []
    count = collections.Counter(p)
    required = len(p)

    for r, c in enumerate(s):
      count[c] -= 1
      if count[c] >= 0:
        required -= 1
      if r >= len(p):
        count[s[r - len(p)]] += 1
        if count[s[r - len(p)]] > 0:
          required += 1
      if required == 0:
        ans.append(r - len(p) + 1)

    return ans