LeetCode in Kotlin

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

Solution

class Solution {
    fun maxDepthAfterSplit(seq: String): IntArray {
        val n = seq.length
        val ans = IntArray(n)
        val chars = seq.toCharArray()
        var depth = 0
        for (i in 0 until n) {
            if (chars[i] == '(') {
                depth++
                if (depth % 2 == 0) {
                    ans[i] = 0
                } else {
                    ans[i] = 1
                }
            } else {
                if (depth % 2 == 0) {
                    ans[i] = 0
                } else {
                    ans[i] = 1
                }
                depth--
            }
        }
        return ans
    }
}

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