# 12.01.2025 [2116. Check if a Parentheses String Can Be Valid]

Balance parenthesis with wildcards #medium

12.01.2025

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Problem TLDR

Balance parenthesis with wildcards #medium

Intuition

Didn't solve it without a hint.

Some examples to observe the problem:

    // 100000
    // (((()(
    

    // 000111   
    // ()((()

    // 101111  f b
    // ((()))
    // *       0 1
    //  *      1 2

The corner cases that can't be balanced:

• odd string length

• locked unbalanced open brace (which is why we have to check the reversed order too)

The hint is: greedy solution just works, consider unlocked positions as wildcards, balance otherwise and check corner cases.

Approach

• separate counters wildcards and balance can just be a single balance variable, if wildcards + balance >= 0

Complexity

• Time complexity: $$O(n)$$

• Space complexity: $$O(1)$$

Code

    fun canBeValid(s: String, locked: String, o: Char = '('): Boolean {
        if (s.length % 2 > 0) return false; var b = 0
        for (i in s.indices)
            if (s[i] == o || locked[i] == '0') b++
            else if (--b < 0) return false
        return o == ')' || canBeValid(s.reversed(), locked.reversed(), ')')
    }

    pub fn can_be_valid(s: String, locked: String) -> bool {
        if s.len() % 2 > 0 { return false }
        let (mut b, mut o, s) = ([0, 0], [b'(', b')'], s.as_bytes());
        for i in 0..s.len() { for j in 0..2 {
            let i = if j > 0 { s.len() - 1 - i } else { i };
            if s[i] == o[j] || locked.as_bytes()[i] == b'0' { b[j] += 1 }
            else { b[j] -= 1; if b[j] < 0 { return false }}
        }}; true
    }

    bool canBeValid(string s, string locked) {
        if (s.size() % 2 > 0) return 0;
        int b[2] = {0}, o[2] = {'(', ')'};
        for (int i = 0; i < s.size(); ++i) for (int j = 0; j < 2; ++j) {
            int k = j ? s.size() - 1 - i : i;
            if (s[k] == o[j] || locked[k] == '0') ++b[j];
            else if (--b[j] < 0) return 0;
        } return 1;
    }