1614. Maximum Nesting Depth of the Parentheses.md

1614. Maximum Nesting Depth of the Parentheses

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the first one in Weekly Contest 210.

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Difficulty : Easy

Related Topics : String

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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.

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Solution

• mine
  • Java
    • Runtime: 0 ms, faster than 100.00%, Memory Usage: 36.8 MB, less than 100.00% of Java online submissions

      // O(N)time
      // O(N)space
      public int maxDepth(String s) {
          char[] arr = s.toCharArray();
          int res = 0;
          int l = 0;
          for(char c : arr){
              if(c == '(') l++;
              else if(c == ')')l--;
              res = Math.max(l, res);
          }
          return res;
      }
      

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