Build confidence with an intuition-first walkthrough focused on stack fundamentals.
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 <= 100s consists of digits 0-9 and characters '+', '-', '*', '/', '(', and ')'.s is a VPS.Problem summary: Given a valid parentheses string s, return the nesting depth of s. The nesting depth is the maximum number of nested parentheses.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Stack
"(1+(2*3)+((8)/4))+1"
"(1)+((2))+(((3)))"
"()(())((()()))"
maximum-nesting-depth-of-two-valid-parentheses-strings)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #1614: Maximum Nesting Depth of the Parentheses
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;
}
}
// Accepted solution for LeetCode #1614: Maximum Nesting Depth of the Parentheses
func maxDepth(s string) (ans int) {
d := 0
for _, c := range s {
if c == '(' {
d++
ans = max(ans, d)
} else if c == ')' {
d--
}
}
return
}
# Accepted solution for LeetCode #1614: Maximum Nesting Depth of the Parentheses
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
// Accepted solution for LeetCode #1614: Maximum Nesting Depth of the Parentheses
struct Solution;
impl Solution {
fn max_depth(s: String) -> i32 {
let mut res = 0;
let mut left = 0;
for c in s.chars() {
if c == '(' {
left += 1;
}
if c == ')' {
left -= 1;
}
res = res.max(left);
}
res
}
}
#[test]
fn test() {
let s = "(1+(2*3)+((8)/4))+1".to_string();
let res = 3;
assert_eq!(Solution::max_depth(s), res);
let s = "(1)+((2))+(((3)))".to_string();
let res = 3;
assert_eq!(Solution::max_depth(s), res);
let s = "1+(2*3)/(2-1)".to_string();
let res = 1;
assert_eq!(Solution::max_depth(s), res);
let s = "1".to_string();
let res = 0;
assert_eq!(Solution::max_depth(s), res);
}
// Accepted solution for LeetCode #1614: Maximum Nesting Depth of the Parentheses
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;
}
Use this to step through a reusable interview workflow for this problem.
For each element, scan left (or right) to find the next greater/smaller element. The inner scan can visit up to n elements per outer iteration, giving O(n²) total comparisons. No extra space needed beyond loop variables.
Each element is pushed onto the stack at most once and popped at most once, giving 2n total operations = O(n). The stack itself holds at most n elements in the worst case. The key insight: amortized O(1) per element despite the inner while-loop.
Review these before coding to avoid predictable interview regressions.
Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.
Usually fails on: Indices point to blocked elements and outputs shift.
Fix: Pop while invariant is violated before pushing current element.