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LeetCode Solutions

567. Permutation in String

Time: $O(|\texttt{s1}| + |\texttt{s2}|)$

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

			


class Solution {
 public:
  bool checkInclusion(string s1, string s2) {
    vector<int> count(128);
    int required = s1.length();

    for (const char c : s1)
      ++count[c];

    for (int l = 0, r = 0; r < s2.length(); ++r) {
      if (--count[s2[r]] >= 0)
        --required;
      while (required == 0) {
        if (r - l + 1 == s1.length())
          return true;
        if (++count[s2[l++]] > 0)
          ++required;
      }
    }

    return false;
  }
};




			


class Solution {
  public boolean checkInclusion(String s1, String s2) {
    int[] count = new int[128];
    int required = s1.length();

    for (final char c : s1.toCharArray())
      ++count[c];

    for (int l = 0, r = 0; r < s2.length(); ++r) {
      if (--count[s2.charAt(r)] >= 0)
        --required;
      while (required == 0) {
        if (r - l + 1 == s1.length())
          return true;
        if (++count[s2.charAt(l++)] > 0)
          ++required;
      }
    }

    return false;
  }
}




			


class Solution {
 public:
  bool checkInclusion(string s1, string s2) {
    vector<int> count(128);
    int required = s1.length();

    for (const char c : s1)
      ++count[c];

    for (int r = 0; r < s2.length(); ++r) {
      if (--count[s2[r]] >= 0)
        --required;
      if (r >= s1.length())  // The window is oversized
        if (++count[s2[r - s1.length()]] > 0)
          ++required;
      if (required == 0)
        return true;
    }

    return false;
  }
};