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1898. Maximum Number of Removable Characters 👍

  • Time: $O(n\log n)$
  • Space: $O(n)$
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class Solution {
 public:
  int maximumRemovals(string s, string p, vector<int>& removable) {
    int l = 0;
    int r = removable.size() + 1;

    while (l < r) {
      const int m = (l + r) / 2;
      const string removed = remove(s, removable, m);
      if (isSubsequence(p, removed))
        l = m + 1;
      else
        r = m;
    }

    return l - 1;
  }

 private:
  string remove(const string& s, const vector<int>& removable, int k) {
    string removed(s);
    for (int i = 0; i < k; ++i)
      removed[removable[i]] = '*';
    return removed;
  }

  bool isSubsequence(const string& p, const string& s) {
    int i = 0;  // p's index
    for (int j = 0; j < s.length(); ++j)
      if (p[i] == s[j])
        if (++i == p.length())
          return true;
    return false;
  }
};

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class Solution {
  public int maximumRemovals(String s, String p, int[] removable) {
    int l = 0;
    int r = removable.length + 1;

    while (l < r) {
      final int m = (l + r) / 2;
      final String removed = remove(s, removable, m);
      if (isSubsequence(p, removed))
        l = m + 1;
      else
        r = m;
    }

    return l - 1;
  }

  private String remove(final String s, int[] removable, int k) {
    StringBuilder sb = new StringBuilder(s);
    for (int i = 0; i < k; ++i)
      sb.setCharAt(removable[i], '*');
    return sb.toString();
  }

  private boolean isSubsequence(final String p, final String s) {
    int i = 0; // p's index
    for (int j = 0; j < s.length(); ++j)
      if (p.charAt(i) == s.charAt(j))
        if (++i == p.length())
          return true;
    return false;
  }
}

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class Solution:
  def maximumRemovals(self, s: str, p: str, removable: List[int]) -> int:
    l = 0
    r = len(removable) + 1

    def remove(k: int) -> str:
      removed = [c for c in s]
      for i in range(k):
        removed[removable[i]] = '*'
      return ''.join(removed)

    def isSubsequence(p: str, s: str) -> bool:
      i = 0
      for j, c in enumerate(s):
        if p[i] == s[j]:
          i += 1
          if i == len(p):
            return True
      return False

    while l < r:
      m = (l + r) // 2
      removed = remove(m)
      if isSubsequence(p, removed):
        l = m + 1
      else:
        r = m

    return l - 1