Question

1111. Maximum Nesting Depth of Two Valid Parentheses Strings

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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:
  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

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;
  }
};

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;
}