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1111. Maximum Nesting Depth of Two Valid Parentheses Strings

Description

A string is a valid parentheses string (denoted VPS) if and only if it consists of "(" and ")" characters only, and:

  • It is the empty string, or
  • It can be written as AB (A concatenated with B), where A and B are VPS's, or
  • It can be written as (A), where A is a VPS.

We can similarly define the nesting depth depth(S) of any VPS S as follows:

  • depth("") = 0
  • depth(A + B) = max(depth(A), depth(B)), where A and B are VPS's
  • depth("(" + A + ")") = 1 + depth(A), where A is a VPS.

For example,  "", "()()", and "()(()())" are VPS's (with nesting depths 0, 1, and 2), and ")(" and "(()" are not VPS's.

 

Given a VPS seq, split it into two disjoint subsequences A and B, such that A and B are VPS's (and A.length + B.length = seq.length).

Now choose any such A and B such that max(depth(A), depth(B)) is the minimum possible value.

Return an answer array (of length seq.length) that encodes such a choice of A and B:  answer[i] = 0 if seq[i] is part of A, else answer[i] = 1.  Note that even though multiple answers may exist, you may return any of them.

 

Example 1:

Input: seq = "(()())"
Output: [0,1,1,1,1,0]

Example 2:

Input: seq = "()(())()"
Output: [0,0,0,1,1,0,1,1]

 

Constraints:

  • 1 <= seq.size <= 10000

Solutions

Solution 1: Greedy

We use a variable $x$ to maintain the current balance of parentheses, which is the number of left parentheses minus the number of right parentheses.

We traverse the string $seq$, updating the value of $x$. If $x$ is odd, we assign the current left parenthesis to $A$, otherwise we assign it to $B$.

The time complexity is $O(n)$, where $n$ is the length of the string $seq$. Ignoring the space consumption of the answer, the space complexity is $O(1)$.

  • class Solution {
    public int[] maxDepthAfterSplit(String seq) {
        int n = seq.length();
        int[] ans = new int[n];
        for (int i = 0, x = 0; i < n; ++i) {
            if (seq.charAt(i) == '(') {
                ans[i] = x++ & 1;
            } else {
                ans[i] = --x & 1;
            }
        }
        return ans;
    }
}

  • class Solution {
public:
    vector<int> maxDepthAfterSplit(string seq) {
        int n = seq.size();
        vector<int> ans(n);
        for (int i = 0, x = 0; i < n; ++i) {
            if (seq[i] == '(') {
                ans[i] = x++ & 1;
            } else {
                ans[i] = --x & 1;
            }
        }
        return ans;
    }
};

  • class Solution:
    def maxDepthAfterSplit(self, seq: str) -> List[int]:
        ans = [0] * len(seq)
        x = 0
        for i, c in enumerate(seq):
            if c == "(":
                ans[i] = x & 1
                x += 1
            else:
                x -= 1
                ans[i] = x & 1
        return ans


  • func maxDepthAfterSplit(seq string) []int {
	n := len(seq)
	ans := make([]int, n)
	for i, x := 0, 0; i < n; i++ {
		if seq[i] == '(' {
			ans[i] = x & 1
			x++
		} else {
			x--
			ans[i] = x & 1
		}
	}
	return ans
}

  • function maxDepthAfterSplit(seq: string): number[] {
    const n = seq.length;
    const ans: number[] = new Array(n);
    for (let i = 0, x = 0; i < n; ++i) {
        if (seq[i] === '(') {
            ans[i] = x++ & 1;
        } else {
            ans[i] = --x & 1;
        }
    }
    return ans;
}

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