Algorithm
Problem Name: 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(Aconcatenated withB), whereAandBare VPS's, or - It can be written as
(A), whereAis a VPS.
We can similarly define the nesting depth depth(S) of any VPS S as follows:
depth("") = 0depth(A + B) = max(depth(A), depth(B)), whereAandBare VPS'sdepth("(" + A + ")") = 1 + depth(A), whereAis 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
Code Examples
#1 Code Example with Java Programming
Code -
Java Programming
start coding.
class Solution {
public int[] maxDepthAfterSplit(String seq) {
int[] result = new int[seq.length()];
int idx = 0;
int depth = 0;
for (char c : seq.toCharArray()) {
result[idx++] = c == '(' ? (depth++ % 2) : (--depth % 2);
}
return result;
}
}
..Input
Output
#2 Code Example with Javascript Programming
Code -
Javascript Programming
const maxDepthAfterSplit = function(seq) {
const n = seq.length
const res = Array(n).fill(0)
let depth = 0
for(let i = 0; i < n; i++) {
const ch = seq[i]
if(ch === '(') {
depth++
}
res[i] = depth % 2
if(ch === ')') depth--
}
return res
};
Input
Output
#3 Code Example with Python Programming
Code -
Python Programming
class Solution:
def maxDepthAfterSplit(self, seq: str) -> List[int]:
stack = []
res = [0] * len(seq)
for i, c in enumerate(seq):
if c == '(':
stack.append(i if not stack or stack[-1] < 0 else -i)
else:
ind = stack.pop()
if ind >= 0:
res[i] = res[ind] = 1
return res
Input
Output
#4 Code Example with C# Programming
Code -
C# Programming
namespace LeetCode
{
public class _1111_MaximumNestingDepthOfTwoValidParenthesesStrings
{
public int[] MaxDepthAfterSplit(string seq)
{
var result = new int[seq.Length];
var currDepth = -1;
for (var i = 0; i < seq.Length; i++)
{
if (seq[i] == '(')
{
currDepth++;
result[i] = currDepth % 2;
}
else
{
result[i] = currDepth % 2;
currDepth--;
}
}
return result;
}
}
}
Input
Output