Valid Parentheses Problem – Solution, Explanation, and Mid‑Career Insights

A former trainee shares his thoughts on why many developers in their mid‑career are pushed into "maintenance" roles and whether maintaining legacy projects is a dead‑end or a hidden opportunity.

1. Why many mid‑career engineers are assigned to "watch the gate"

Companies often place experienced developers in stable positions to guard technical debt. Typical tasks include taking over a CRM system that has been running for years, handling an old system that nobody wants, or merely fixing small bugs and writing Excel report interfaces.

2. Is maintaining old projects really useless?

No. Most enterprises value system stability over flashy features. Developers who maintain legacy code are familiar with business processes, know system boundaries, can fix critical bugs, and act as the "last line of defense" when issues arise.

3. The mindset trap of mid‑career developers

Many senior engineers adopt a passive attitude: they think their project will never break, nobody wants the old system, or new technologies are for younger developers. This mindset ignores the industry shift toward AI, large language models, and RPA, which can render unchanged systems obsolete.

4. How to break through the "mid‑career crisis"

Modernize the legacy system : introduce automated monitoring, optimize data processing, or integrate AI assistants to showcase technical upgrades.

Embrace new technologies and become a technical advisor : leverage your deep business knowledge to guide AI tool adoption, RAG systems, or agent plugins.

Transition from maintenance to reconstruction : treat the legacy system as a core asset, gradually influence architecture decisions, and propose migration plans.

5. Final note

Maintaining old projects is not "backward"; stagnation is. Use the experience to strengthen stability, collaboration, and system thinking, then turn it into a stepping stone for future growth.

Algorithm Problem: Valid Parentheses

Given a string s consisting only of the characters ( , ) , { , } , [ and ] , determine whether the string is valid. A valid string satisfies:

Every opening bracket must be closed by the same type of bracket.

Brackets must be closed in the correct order.

Solution Idea

The problem is a classic "last‑in‑first‑out" scenario, which is naturally solved with a stack. Iterate over the characters; push opening brackets onto the stack. When encountering a closing bracket, pop the stack and check if it matches the corresponding opening bracket. If the stack is empty at the end, the string is valid.

Java Implementation

class Solution {
    public boolean isValid(String s) {
        // Odd length strings cannot be valid
        if (s.length() % 2 == 1) return false;
        Stack
stack = new Stack<>();
        char[] chars = s.toCharArray();
        for (char c : chars) {
            if (c == '(' || c == '[' || c == '{') {
                stack.push(c);
            } else {
                if (stack.isEmpty()) return false;
                char top = stack.peek();
                if ((top == '(' && c == ')') ||
                    (top == '[' && c == ']') ||
                    (top == '{' && c == '}')) {
                    stack.pop();
                } else {
                    return false;
                }
            }
        }
        return stack.isEmpty();
    }
}

C++ Implementation

class Solution {
public:
    bool isValid(string s) {
        if (s.size() % 2 == 1) return false;
        stack
stk;
        for (char c : s) {
            if (c == '(' || c == '[' || c == '{') {
                stk.push(c);
            } else {
                if (stk.empty()) return false;
                char top = stk.top();
                if ((top == '(' && c == ')') ||
                    (top == '[' && c == ']') ||
                    (top == '{' && c == '}')) {
                    stk.pop();
                } else {
                    return false;
                }
            }
        }
        return stk.empty();
    }
};

Python Implementation

class Solution:
    def isValid(self, s: str) -> bool:
        if len(s) % 2 == 1:
            return False
        stack = []
        for c in s:
            if c in '({[':
                stack.append(c)
            else:
                if not stack:
                    return False
                top = stack[-1]
                if (top == '(' and c == ')') or \
                   (top == '[' and c == ']') or \
                   (top == '{' and c == '}'):
                    stack.pop()
                else:
                    return False
        return not stack

Complexity Analysis

Time Complexity: O(n) – each character is pushed and popped at most once.

Space Complexity: O(n) in the worst case when all characters are opening brackets.

Common Pitfalls

Forgetting to check whether the stack is empty before popping.

Mismatching bracket types.

Not verifying that the stack is empty after processing the entire string.

Related Problems

LeetCode 921 – Minimum Add to Make Parentheses Valid

LeetCode 32 – Longest Valid Parentheses

LeetCode 1249 – Remove Invalid Parentheses