Leetcode 3 Longest Substring Without Repeating Characters
// 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, find the length of the longest substring without repeating characters.
Example 1:
Input: "abcabcbb"
Output: 3
Explanation: The answer is "abc", with the length of 3.
Example 2:
Input: "bbbbb"
Output: 1
Explanation: The answer is "b", with the length of 1.
Example 3:
Input: "pwwkew" Output: 3 Explanation: The answer is"wke", with the length of 3. Note that the answer must be a substring,"pwke"is a subsequence and not a substring.
Solution 1:
class Solution {
public int lengthOfLongestSubstring(String s) {
char[] stringChar = s.toCharArray();
if(stringChar.length < 2) return stringChar.length;
Stack<String> tempStack = new Stack<String>();
Queue<String> q = new LinkedList<>();
int max = 0;
int i = 0;
for(char c: stringChar){
if(!q.contains(Character.toString(c))){
q.add(Character.toString(c));
i++;
}else{
if(i>=max) max = i;
while(q.size()!=0 && (!q.peek().equals(Character.toString(c)))){
q.remove();
}
if(q.size()!=0 && (q.peek().equals(Character.toString(c)))) {
q.remove();}
q.add(Character.toString(c));
i = q.size();
}
}
if(i>=max) max=i;
return max;For the AI-native engineering side of HackerHeap — building MCP servers, comparing agents (Claude Code, Cursor, Windsurf, Codex, Gemini, Copilot), and weekly working code — see the Friday Build newsletter and the MCP archive.