• AutoBuilder

Cursor is incredibly good harness.

Opus is also not Oputhetic anymore

Why? Multi-tenant environments. First, we need to understand a few differences between environments:

• End-user UI
• Agent Runtime Environment
• LLM Server

So

• When you run Claude Code on your local MacBook, the first two are always local. The third is usually the Claude.ai server.
• When you ssh to a virtual private server (VPS) and install Claude Code there, the first two are your remote server. The third is still the Claude.ai server.
• When you run Claude RC on your virtual private server and code from your iPad using the Claude app, the end-user UI is on your iPad, the agent runtime environment is on your VPS, and the server is still Claude.ai.

Most people physically separate their tenancy, such as Claude Code, from their personal vs. work laptops. So in most cases, it's not a big deal.

But when you need multi-tenancy, it becomes super stressful. For example, say you have two different toolkits:

• personal toolkits (personal Notion, personal Sentry, personal Linear)
• workplace toolkits (company Notion, company Sentry, company Linear)

Most MCP auth states or code harnesses don't support profiles, so you can only log in to one.

So therefore... a natural evolution was to have both:

• a personal VPS with all personal toolkits set up
• a workplace VPS with all workspace toolkits set up

to physically isolate tenancies.

Now we've solved the multiple-profile issue, but the client's problems persist. Now let's get back to the environments:

• End-user UI
• Agent Runtime Environment
• LLM Server

All MCP auth or toolkit auth info should always be saved in the Agent Runtime Environment IMHO. However, a surprising number of harnesses tie them to the LLM server (such as Codex Apps or Claude.ai Plugins) or put them in the end-user UI (Claude Desktop or Codex Desktop).

Now the problem is:

• If the auth data is put on the LLM server, you cannot reuse LLM accounts across tenants
• If the auth data is put on the end-user UI, you cannot use the same app to access multi-tenants.

The only way to reliably isolate different auth information is thus:

• You ssh to a virtual private server (VPS) and run Claude Code there. Never use LLM server plugins.

Then

• End-user UI
• Agent Runtime Environment

are both isolated VPS, and

• LLM Server holds no information on the tenancy

This way, you can provide different toolkits, creating multiple dev environments.

Solved at: 220710

Question

• Two Sum - LeetCode

Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target. You may assume that each input would have exactly one solution, and you may not use the same element twice. You can return the answer in any order.

Solution

So the first obvious answer is to iterate twice. This finishes calculations in $O(n^2)$ time.

class Solution:    def twoSum(self, nums: List[int], target: int) -> List[int]:        for idx1, val1 in enumerate(nums):            for idx2, val2 in enumerate(nums):                if idx1 == idx2:                    continue                if val1 + val2 == target:                    return [idx1, idx2]

However, this gives a timeout.

Improved

I used Python Dictionary to store complementing values. Python Dictionary will have $O(1)$ access time for most cases. This solution will run in $O(n)$ time.

• One caveat: depending on the hash function, it can go as bad as $O(n^2)$.

class Solution:    def twoSum(self, nums: List[int], target: int) -> List[int]:
        # map for complementing elements: complementary-idx        complementing_map = {}
        for idx, val in enumerate(nums):            if val in complementing_map:                return [complementing_map[val], idx]            complementing_map[target - val] = idx

Results

Runtime

• 60 ms, faster than 97.16% of Python3 online submissions for Two Sum.

Memory Usage

• 15.4 MB, less than 14.24% of Python3 online submissions for Two Sum.

Other Answers Online

• Sort first, $O(n \log n)$
• For all elements, $O(n)$
  • Perform binary search $O(\log n)$
• In total: $O(n \log n)$