README_EN.md

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

Python3

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

Java

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

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

C++

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

Go

func maxDepthAfterSplit(seq string) []int {
	ans := make([]int, len(seq))
	a, b := 0, 0
	for i, c := range seq {
		if c == '(' {
			if a < b {
				a++
			} else {
				b++
				ans[i] = 1
			}
		} else {
			if a > b {
				a--
			} else {
				b--
				ans[i] = 1
			}
		}
	}
	return ans
}

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