A string is a valid parentheses string (denoted VPS) if it meets one of the following:
- It is an empty string
""
, or a single character not equal to"("
or")"
, - It can be written as
AB
(A
concatenated withB
), whereA
andB
are VPS's, or - It can be written as
(A)
, whereA
is a VPS.
We can similarly define the nesting depth depth(S)
of any VPS S
as follows:
depth("") = 0
depth(C) = 0
, whereC
is a string with a single character not equal to"("
or")"
.depth(A + B) = max(depth(A), depth(B))
, whereA
andB
are VPS's.depth("(" + A + ")") = 1 + depth(A)
, whereA
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 represented as string s
, return the nesting depth of s
.
Example 1:
Input: s = "(1+(2*3)+((8)/4))+1" Output: 3 Explanation: Digit 8 is inside of 3 nested parentheses in the string.
Example 2:
Input: s = "(1)+((2))+(((3)))" Output: 3
Constraints:
1 <= s.length <= 100
s
consists of digits0-9
and characters'+'
,'-'
,'*'
,'/'
,'('
, and')'
.- It is guaranteed that parentheses expression
s
is a VPS.
class Solution:
def maxDepth(self, s: str) -> int:
n = ans = 0
for c in s:
if c == '(':
n += 1
ans = max(ans, n)
elif c == ')':
n -= 1
return ans
class Solution {
public int maxDepth(String s) {
int n = 0, ans = 0;
for (char c : s.toCharArray()) {
if (c == '(') {
ans = Math.max(ans, ++n);
} else if (c == ')') {
--n;
}
}
return ans;
}
}
class Solution {
public:
int maxDepth(string s) {
int n = 0, ans = 0;
for (char c : s) {
if (c == '(')
ans = max(ans, ++n);
else if (c == ')')
--n;
}
return ans;
}
};
func maxDepth(s string) int {
n, ans := 0, 0
for _, c := range s {
if c == '(' {
n++
if ans < n {
ans = n
}
} else if c == ')' {
n--
}
}
return ans
}
/**
* @param {string} s
* @return {number}
*/
var maxDepth = function (s) {
let n = 0,
ans = 0;
for (let c of s) {
if (c == '(') ans = Math.max(ans, ++n);
else if (c == ')') --n;
}
return ans;
};
public class Solution {
public int MaxDepth(string s) {
int n = 0, ans = 0;
foreach (char c in s)
{
if (c == '(')
{
ans = Math.Max(ans, ++n);
}
else if (c == ')')
{
--n;
}
}
return ans;
}
}