Formatted question description: https://leetcode.ca/all/2116.html

2116. Check if a Parentheses String Can Be Valid

Difficulty: Medium.
Related Topics: String, Stack, Greedy.
Similar Questions: Valid Parentheses, Generate Parentheses, Valid Parenthesis String, Minimum Remove to Make Valid Parentheses, Check if There Is a Valid Parentheses String Path.

Problem

A parentheses string is a non-empty string consisting only of '(' and ')'. It is valid if any of the following conditions is true:

It is ().
It can be written as AB (A concatenated with B), where A and B are valid parentheses strings.
It can be written as (A), where A is a valid parentheses string.

You are given a parentheses string s and a string locked, both of length n. locked is a binary string consisting only of '0's and '1's. For each index i of locked,

If locked[i] is '1', you cannot change s[i].
But if locked[i] is '0', you can change s[i] to either '(' or ')'.

Return true if you can make s a valid parentheses string. Otherwise, return false.

Example 1:

Input: s = "))()))", locked = "010100"
Output: true
Explanation: locked[1] == '1' and locked[3] == '1', so we cannot change s[1] or s[3].
We change s[0] and s[4] to '(' while leaving s[2] and s[5] unchanged to make s valid.

Example 2:

Input: s = "()()", locked = "0000"
Output: true
Explanation: We do not need to make any changes because s is already valid.

Example 3:

Input: s = ")", locked = "0"
Output: false
Explanation: locked permits us to change s[0]. 
Changing s[0] to either '(' or ')' will not make s valid.

Constraints:

n == s.length == locked.length
1 <= n <= 105
s[i] is either '(' or ')'.
locked[i] is either '0' or '1'.

Solution (Java, C++, Python)

class Solution {
    public boolean canBeValid(String s, String locked) {
        if (s == null || s.isEmpty()) {
            return true;
        }
        if ((s.length() & 1) > 0) {
            return false;
        }
        if (locked == null || locked.isEmpty()) {
            return true;
        }
        int numOfLockedClose = 0;
        int numOfLockedOpen = 0;
        for (int i = 0; i < s.length(); i++) {
            int countOfChars = i + 1;
            if (s.charAt(i) == ')' && locked.charAt(i) == '1') {
                numOfLockedClose++;
                if (numOfLockedClose * 2 > countOfChars) {
                    return false;
                }
            }
            int j = s.length() - 1 - i;
            if (s.charAt(j) == '(' && locked.charAt(j) == '1') {
                numOfLockedOpen++;

                if (numOfLockedOpen * 2 > countOfChars) {
                    return false;
                }
            }
        }
        return true;
    }
}

############

class Solution {
    public boolean canBeValid(String s, String locked) {
        int n = s.length();
        if (n % 2 == 1) {
            return false;
        }
        int x = 0;
        for (int i = 0; i < n; ++i) {
            if (s.charAt(i) == '(' || locked.charAt(i) == '0') {
                ++x;
            } else if (x > 0) {
                --x;
            } else {
                return false;
            }
        }
        x = 0;
        for (int i = n - 1; i >= 0; --i) {
            if (s.charAt(i) == ')' || locked.charAt(i) == '0') {
                ++x;
            } else if (x > 0) {
                --x;
            } else {
                return false;
            }
        }
        return true;
    }
}

class Solution:
    def canBeValid(self, s: str, locked: str) -> bool:
        n = len(s)
        if n & 1:
            return False
        x = 0
        for i in range(n):
            if s[i] == '(' or locked[i] == '0':
                x += 1
            elif x:
                x -= 1
            else:
                return False
        x = 0
        for i in range(n - 1, -1, -1):
            if s[i] == ')' or locked[i] == '0':
                x += 1
            elif x:
                x -= 1
            else:
                return False
        return True

############

# 2116. Check if a Parentheses String Can Be Valid
# https://leetcode.com/problems/check-if-a-parentheses-string-can-be-valid/

class Solution:
    def canBeValid(self, s: str, locked: str) -> bool:
        n = len(s)
        s = list(s)
        
        if n & 1: return False
        
        for i in range(n):
            if locked[i] == '0':
                s[i] = '#'
        
        opened = closed = 0
        for x in s:
            if x == '#' or x == '(':
                opened += 1
            else:
                closed += 1
            
            if closed > opened:
                return False
        
        opened = closed = 0
        for i in range(n - 1, -1, -1):
            if s[i] == '#' or s[i] == ')':
                closed += 1
            else:
                opened += 1
            
            if opened > closed:
                return False
        
        return True

class Solution {
public:
    bool canBeValid(string s, string locked) {
        int n = s.size();
        if (n & 1) {
            return false;
        }
        int x = 0;
        for (int i = 0; i < n; ++i) {
            if (s[i] == '(' || locked[i] == '0') {
                ++x;
            } else if (x) {
                --x;
            } else {
                return false;
            }
        }
        x = 0;
        for (int i = n - 1; i >= 0; --i) {
            if (s[i] == ')' || locked[i] == '0') {
                ++x;
            } else if (x) {
                --x;
            } else {
                return false;
            }
        }
        return true;
    }
};

func canBeValid(s string, locked string) bool {
	n := len(s)
	if n%2 == 1 {
		return false
	}
	x := 0
	for i := range s {
		if s[i] == '(' || locked[i] == '0' {
			x++
		} else if x > 0 {
			x--
		} else {
			return false
		}
	}
	x = 0
	for i := n - 1; i >= 0; i-- {
		if s[i] == ')' || locked[i] == '0' {
			x++
		} else if x > 0 {
			x--
		} else {
			return false
		}
	}
	return true
}

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).

2116 - Check if a Parentheses String Can Be Valid

2116. Check if a Parentheses String Can Be Valid

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Solution (Java, C++, Python)

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2116. Check if a Parentheses String Can Be Valid

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Solution (Java, C++, Python)

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