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by AskGif | Oct 18, 2020 | Category :coding | Tags : algorithm data-structure easy leetcode stack

Maximum Nesting Depth of the Parentheses - Stack - Easy - LeetCode

Maximum Nesting Depth of the Parentheses - Stack - Easy - LeetCode

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 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 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 digits 0-9 and characters '+', '-', '*', '/', '(', and ')'.
It is guaranteed that parentheses expression s is a VPS.

public class Solution {
    public int MaxDepth(string s) {
        var stack = new Stack<char>();
        int max = 0;
        for(int i=0; i< s.Length;i++){
            if(s[i]=='('){
                stack.Push('(');
                if(max<stack.Count()){
                    max = stack.Count();
                }
            }
            else if(s[i]==')'){
                if(stack.Count()==0){
                    return 0;
                }
                stack.Pop();
            }
        }
        
        return max;
    }
}

Time Complexity: O(n)

Space Complexity: O(n)