Description
https://leetcode.com/problems/maximum-nesting-depth-of-the-parentheses/
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
Example 3:
Input: s = "1+(2*3)/(2-1)" Output: 1
Example 4:
Input: s = "1" Output: 0
Constraints:
1 <= s.length <= 100
s
consists of digits0-9
and characters'+'
,'-'
,'*'
,'/'
,'('
, and')'
.- It is guaranteed that parentheses expression
s
is a VPS.
Explanation
Use a stack to track the left parenthesis encountered, if encountering a right parenthesis, just pop the stack. Return the max depth of the stack.
Python Solution
class Solution:
def maxDepth(self, s: str) -> int:
stack = []
max_depth = 0
for c in s:
if c == '(':
stack.append(c)
max_depth = max(max_depth, len(stack))
elif c == ')':
stack.pop()
return max_depth
- Time Complexity: O(N).
- Space Complexity: O(N).