1111

1111. Maximum Nesting Depth of Two Valid Parentheses Strings (Medium)

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

Related Topics:
Binary Search, Greedy

Solution 1.

// OJ: https://leetcode.com/problems/maximum-nesting-depth-of-two-valid-parentheses-strings/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    vector<int> maxDepthAfterSplit(string A) {
        int N = A.size(), lv = 0;
        vector<int> ans(N);
        for (int i = 0; i < N; ++i) {
            if (A[i] == '(') ans[i] = lv++ % 2;
            else ans[i] = --lv % 2;
        }
        return ans;
    }
};