README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/1600-1699/1614.Maximum%20Nesting%20Depth%20of%20the%20Parentheses/README_EN.md	1322	Weekly Contest 210 Q1	Stack String

1614. Maximum Nesting Depth of the Parentheses

中文文档

Description

Given a valid parentheses string s, return the nesting depth of s. The nesting depth is the maximum number of nested parentheses.

 

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

Explanation:

Digit 3 is inside of 3 nested parentheses in the string.

Example 3:

Input: s = "()(())((()()))"

Output: 3

 

Constraints:

• 1 <= s.length <= 100
• s consists of digits 0-9 and characters '+', '-', '*', '/', '(', and ')'.
• It is guaranteed that parentheses expression s is a VPS.

Solutions

Solution 1: Traversal

We use a variable $d$ to record the current depth, initially $d = 0$.

Traverse the string $s$. When encountering a left parenthesis, increment the depth $d$ by one and update the answer to be the maximum of the current depth $d$ and the answer. When encountering a right parenthesis, decrement the depth $d$ by one.

Finally, return the answer.

The time complexity is $O(n)$, where $n$ is the length of the string $s$. The space complexity is $O(1)$.

Python3

class Solution:
    def maxDepth(self, s: str) -> int:
        ans = d = 0
        for c in s:
            if c == '(':
                d += 1
                ans = max(ans, d)
            elif c == ')':
                d -= 1
        return ans

Java

class Solution {
    public int maxDepth(String s) {
        int ans = 0, d = 0;
        for (int i = 0; i < s.length(); ++i) {
            char c = s.charAt(i);
            if (c == '(') {
                ans = Math.max(ans, ++d);
            } else if (c == ')') {
                --d;
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int maxDepth(string s) {
        int ans = 0, d = 0;
        for (char& c : s) {
            if (c == '(') {
                ans = max(ans, ++d);
            } else if (c == ')') {
                --d;
            }
        }
        return ans;
    }
};

Go

func maxDepth(s string) (ans int) {
	d := 0
	for _, c := range s {
		if c == '(' {
			d++
			ans = max(ans, d)
		} else if c == ')' {
			d--
		}
	}
	return
}

TypeScript

function maxDepth(s: string): number {
    let ans = 0;
    let d = 0;
    for (const c of s) {
        if (c === '(') {
            ans = Math.max(ans, ++d);
        } else if (c === ')') {
            --d;
        }
    }
    return ans;
}

JavaScript

/**
 * @param {string} s
 * @return {number}
 */
var maxDepth = function (s) {
    let ans = 0;
    let d = 0;
    for (const c of s) {
        if (c === '(') {
            ans = Math.max(ans, ++d);
        } else if (c === ')') {
            --d;
        }
    }
    return ans;
};

C#

public class Solution {
    public int MaxDepth(string s) {
        int ans = 0, d = 0;
        foreach(char c in s) {
            if (c == '(') {
                ans = Math.Max(ans, ++d);
            } else if (c == ')') {
                --d;
            }
        }
        return ans;
    }
}