Formatted question description: https://leetcode.ca/all/678.html

678. Valid Parenthesis String (Medium)

Given a string containing only three types of characters: '(', ')' and '*', write a function to check whether this string is valid. We define the validity of a string by these rules:

1. Any left parenthesis '(' must have a corresponding right parenthesis ')'.
2. Any right parenthesis ')' must have a corresponding left parenthesis '('.
3. Left parenthesis '(' must go before the corresponding right parenthesis ')'.
4. '*' could be treated as a single right parenthesis ')' or a single left parenthesis '(' or an empty string.
5. An empty string is also valid.

Example 1:

Input: "()"
Output: True

Example 2:

Input: "(*)"
Output: True

Example 3:

Input: "(*))"
Output: True

Note:

1. The string size will be in the range [1, 100].

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Facebook, Amazon, Bloomberg

Related Topics:
String

Similar Questions:

• Special Binary String (Hard)

Solution 1.

This solution is not performant if s contains lots of "*".

// OJ: https://leetcode.com/problems/valid-parenthesis-string/
// Time: O(3^S) where S is the length of string s. In the worst case every character
//       is "*", so every step has 3 choices.
// Space: O(S)
class Solution {
private:
    bool dfs(string &s, int start, int leftParenCnt) {
        if (start == s.size()) return !leftParenCnt;
        if (s[start] == '(') {
            return dfs(s, start + 1, leftParenCnt + 1);
        } else if (s[start] == ')') {
            if (--leftParenCnt < 0) return false;
            return dfs(s, start + 1, leftParenCnt);
        } else {
            if (dfs(s, start + 1, leftParenCnt + 1)) return true;
            if (leftParenCnt >= 1 && dfs(s, start + 1, leftParenCnt - 1)) return true;
            return dfs(s, start + 1, leftParenCnt);
        }
    }
public:
    bool checkValidString(string s) {
        return dfs(s, 0, 0);
    }
};

Solution 2.

Let diff be count of left parenthesis minus count of right parenthesis.

When we meet:

• (, we increment diff.
• ), we decrement diff.
• *, we have three choices which makes the diff become a range – [diff - 1, diff + 1].

So we use maxDiff/minDiff to record the maximum/minimum diff we can get.

When we meet:

• (, ++maxDiff and ++minDiff.
• ), --maxDiff and --minDiff.
• *, ++maxDiff and --minDiff.

If maxDiff become negative, it means it’s already invalid, we should return false.

Whenever minDiff falls below 0, we should force it to be 0 because we never accept negative diff during the process.

After scanning through string s, as long as minDiff is 0, this string can be a valid one.

Whenever minDiff falls below 0, we should force it to be 0 because we never accept negative diff during the process.

minDiff means the diff we got if we always try to replace * with ). If minDiff become -1, it means that this replacement results in more ) than (, so it should be avoided. To avoid it, we simply reset minDiff from -1 to 0 which implies we only replace * with ( or empty string.

Example: (**)

• Seeing (, the range becomes [1, 1].
• Seeing *, the range becomes [0, 2]. 0 correponds to (), 1 to (_, 2 to ((.
• Seeing *, the range becomes [-1,3]. But -1 is invalid because it means ()) and should be avoided. So we correct the range to [0, 3].
• Seeing ), the range becomes [-1, 2]. Again, we correct the range to [0, 2] (because -1 means ()_) or (_)))

The final [0, 2] range means that we can either get a perfect string, or has 1 or 2 more ( available (which are created by *).

// OJ: https://leetcode.com/problems/valid-parenthesis-string/
// Time: O(S)
// Space: O(1)
class Solution {
public:
    bool checkValidString(string s) {
        int maxDiff = 0, minDiff = 0;
        for (char c : s) {
            maxDiff += (c == '(' || c == '*') ? 1 : -1;
            minDiff += (c == ')' || c == '*') ? -1 : 1;
            if (maxDiff < 0) return false;
            minDiff = max(0, minDiff);
        }
        return minDiff == 0;
    }
};

Java

• Java
• C++
• Python

• class Solution {
      public boolean checkValidString(String s) {
          Stack<Integer> parenthesesStack = new Stack<Integer>();
          Queue<Integer> asterisksQueue = new LinkedList<Integer>();
          int length = s.length();
          for (int i = 0; i < length; i++) {
              char c = s.charAt(i);
              if (c == '*')
                  asterisksQueue.offer(i);
              else if (c == '(')
                  parenthesesStack.push(i);
              else if (c == ')') {
                  if (!parenthesesStack.isEmpty())
                      parenthesesStack.pop();
                  else if (!asterisksQueue.isEmpty())
                      asterisksQueue.poll();
                  else
                      return false;
              }
          }
          if (parenthesesStack.isEmpty())
              return true;
          else {
              Stack<Integer> newParenthesesStack = new Stack<Integer>();
          	while (!parenthesesStack.isEmpty())
          		newParenthesesStack.push(parenthesesStack.pop());
              while (!newParenthesesStack.isEmpty() && !asterisksQueue.isEmpty()) {
                  int parenthesisIndex = newParenthesesStack.peek();
                  int asteriskIndex = asterisksQueue.poll();
                  if (parenthesisIndex < asteriskIndex)
                  	newParenthesesStack.pop();
              }
              return newParenthesesStack.isEmpty();
          }
      }
  }
  

• // OJ: https://leetcode.com/problems/valid-parenthesis-string/
  // Time: O(3^S) where S is the length of string s. In the worst case every character
  //       is "*", so every step has 3 choices.
  // Space: O(S)
  class Solution {
  private:
      bool dfs(string &s, int start, int leftParenCnt) {
          if (start == s.size()) return !leftParenCnt;
          if (s[start] == '(') {
              return dfs(s, start + 1, leftParenCnt + 1);
          } else if (s[start] == ')') {
              if (--leftParenCnt < 0) return false;
              return dfs(s, start + 1, leftParenCnt);
          } else {
              if (dfs(s, start + 1, leftParenCnt + 1)) return true;
              if (leftParenCnt >= 1 && dfs(s, start + 1, leftParenCnt - 1)) return true;
              return dfs(s, start + 1, leftParenCnt);
          }
      }
  public:
      bool checkValidString(string s) {
          return dfs(s, 0, 0);
      }
  };
  

• class Solution(object):
      def checkValidString(self, s):
          """
          :type s: str
          :rtype: bool
          """
          old_set = set([0])
          for c in s:
              new_set = set()
              if c == '(':
                  for t in old_set:
                      new_set.add(t + 1)
              elif c == ')':
                  for t in old_set:
                      if t > 0:
                          new_set.add(t - 1)
              elif c == '*':
                  for t in old_set:
                      new_set.add(t + 1)
                      new_set.add(t)
                      if t > 0:
                          new_set.add(t - 1)
              old_set = new_set
          return 0 in old_set
  

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