LeetCode Q 1111 - Maximum Nesting Depth of Two Valid Parentheses Strings

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

Solution:

Intuition: Seperate the adjacent "("s to different groups.

Code:

public int[] maxDepthAfterSplit(String seq) {
  int[] res = new int[seq.length()]; 
  int index = 0, depth = 1;
  
  for (char ch: seq.toCharArray()) {
    if (ch == '(') depth++;
    res[index++] = depth % 2;
    if (ch == ')') depth--;
  }
  
  return res;
}

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