Valid Parenthesis String

Jun 24, 2026Madhav

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Description:

Given a string s containing only three types of characters: '(', ')' and '*', return true if s is valid.

The following rules define a valid string:

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 "".

Example 1:

Input: s = "()" Output: true

Example 2:

Input: s = "(*)" Output: true Explanation: The '*' can be treated as an empty string.

Example 3:

Input: s = "(*))" Output: true Explanation: The '*' can be treated as a left parenthesis '('.

---

Approaches:

1. Greedy Approach (Optimal)

Instead of using a stack or dynamic programming, we can solve this problem optimally in a single pass using a greedy approach. Since the '*' character introduces branching possibilities, we can maintain the range of possible open left parentheses at any given step.

We maintain two variables:

• minOpen: The minimum possible number of open left parentheses.
• maxOpen: The maximum possible number of open left parentheses.

1. Iterate through the string:
   • If we see a '(': Both minOpen and maxOpen increment by 1.
   • If we see a ')': Both minOpen and maxOpen decrement by 1.
   • If we see a '*': It could be an open parenthesis (maxOpen++), a close parenthesis (minOpen--), or empty (leaves limits unchanged, but effectively expands our range of possibilities).
2. Validation Checks:
   • If at any point maxOpen < 0, it means even if we converted every '*' to a '(', we still have too many ')'. Thus, the string is invalid, and we immediately return false.
   • If minOpen < 0, it means we might have counted too many '*' as ')'. A negative open count is impossible in reality, so we just reset minOpen to 0 (essentially treating those extra '*' as empty strings instead).
3. Final Check: At the end of the string, if minOpen == 0, it means we successfully matched all parentheses.

Complexity

• Time Complexity: $O(N)$ where $N$ is the length of the string s. We only traverse the string exactly once.
• Space Complexity: $O(1)$ as we only use two integer variables (minOpen and maxOpen) regardless of the size of the input.

---

Solutions:

C++

class Solution {
public:
    bool checkValidString(const string& s) {
        int minOpen = 0;
        int maxOpen = 0;
        
        for (char c : s) {
            if (c == '(') {
                minOpen++;
                maxOpen++;
            } else if (c == ')') {
                minOpen--;
                maxOpen--;
            } else { // c == '*'
                minOpen--;
                maxOpen++;
            }
            
            // If maxOpen is negative, we have too many ')'
            if (maxOpen < 0) return false;
            
            // minOpen can't be negative, reset to 0
            if (minOpen < 0) minOpen = 0;
        }
        
        return minOpen == 0;
    }
};

Java

class Solution {
    public boolean checkValidString(String s) {
        int minOpen = 0;
        int maxOpen = 0;
        
        for (char c : s.toCharArray()) {
        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            if (c == '(') {
                minOpen++;
                maxOpen++;
            } else if (c == ')') {
                minOpen--;
                maxOpen--;
            } else { // c == '*'
                minOpen--;
                maxOpen++;
            }
            // If maxOpen is negative, we have too many ')'
            if (maxOpen < 0) return false;
            
            // minOpen can't be negative, reset to 0
            if (minOpen < 0) minOpen = 0;
        }
        
        return minOpen == 0;
    }
}

Python

class Solution:
    def checkValidString(self, s: str) -> bool:
        min_open = 0
        max_open = 0
        
        for char in s:
            if char == '(':
                min_open += 1
                max_open += 1
            elif char == ')':
                min_open -= 1
                max_open -= 1
            else: # char == '*'
                min_open -= 1
                max_open += 1
            
            # If max_open is negative, we have too many ')'
            if max_open < 0:
                return False
                
            # min_open can't be negative, reset to 0
            if min_open < 0:
                min_open = 0
                
        return min_open == 0

JavaScript

/**
 * @param {string} s
 * @return {boolean}
 */
var checkValidString = function(s) {
    let minOpen = 0;
    let maxOpen = 0;
    
    for (let i = 0; i < s.length; i++) {
        let char = s[i];
        
        if (char === '(') {
            minOpen++;
            maxOpen++;
        } else if (char === ')') {
            minOpen--;
            maxOpen--;
        } else { // char === '*'
            minOpen--;
            maxOpen++;
        }
        
        // If maxOpen is negative, we have too many ')'
        if (maxOpen < 0) return false;
        
        // minOpen can't be negative, reset to 0
        if (minOpen < 0) minOpen = 0;
    }
    
    return minOpen === 0;
};