Leetcode 20 Valid Parentheses Java Solution
// SOLVING THIS WITH AN AI ASSISTANT (2026)
If you are working through this problem with an AI coding assistant — Claude, ChatGPT, Cursor chat, Gemini, GitHub Copilot, Aider, or any agent — the goal isn’t to ask for the answer. It is to use the tool to understand the pattern. The prompt sequence I’d run:
- Spec it back to me first. “In your own words, what is this problem actually testing? What’s the smallest example that fails the naive approach?”
- Brute-force first, optimize after. “Write the simplest correct solution, even if it’s O(n²). Don’t optimize. Just make it correct, with comments explaining each step.”
- Ask for the upgrade. “Now show me the optimal solution. What insight makes it possible? What pattern is this an instance of?”
- Stress-test it. “Generate 10 edge cases — empty input, single element, duplicates, max size, sorted, reverse-sorted. Run my solution against each.”
The pattern matters more than the answer. If the agent just hands you optimized code, you’ve trained yourself to lose interviews.
Given a string containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid.
An input string is valid if:
- Open brackets must be closed by the same type of brackets.
- Open brackets must be closed in the correct order.
Note that an empty string is also considered valid.
Example 1:
Input: "()" Output: true
Example 2:
Input: "()[]{}"
Output: true
Example 3:
Input: "(]" Output: false
Example 4:
Input: "([)]" Output: false
Example 5:
Input: "{[]}"
Output: true
Solution 1:
class Solution {
public boolean isValid(String s) {
if(s.length()<=1) return false;
Stack<Character> paraS = new Stack<Character>();
for(int i=0;i<s.length();i++){
if(s.charAt(i)=='('||s.charAt(i)=='{'||s.charAt(i)=='['){
paraS.push(s.charAt(i));
}else if(s.charAt(i)==')'){
if(paraS.size()==0 || paraS.pop()!= '(') return false;
}else if(s.charAt(i)==']'){
if(paraS.size()==0 || paraS.pop()!= '[') return false;
}else if(s.charAt(i)=='}'){
if(paraS.size()==0 || paraS.pop()!= '{') return false;
}
}
return paraS.isEmpty();
}
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