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Inspired by karpathy/autoresearch. Put this in a Ralph Loop.

Use each mode-specific prompt together with the common element block.

Auto Refactor

Prompt

text
STOP! Re-read all code. Would Karpathy approve every line? Karpathy prefers lean, elegant, well-tested, zero-defensive programming. Use MCPs and web searches.

Completion Promise

text
--completion-promise "KARPATHY_WILL_APPROVE_EVERY_SINGLE_LOC_FOR_SURE"

Auto Fixer

Prompt

text
STOP! Re-read all code, assess PR comments. Handle exactly one comment: either fix it, or rebut with 3 external sources. Fix any dirt found along the way. Lean, elegant, zero defensive programming.

Completion Promise

text
--completion-promise "NO_COMMENTS_REMAINING_IN_GITHUB_EVEN_AFTER_20_MINUTES"

Auto Builder

Prompt

text
STOP! Re-read all code, assess GitHub Issues. Pick one task: fix dirty code, or implement a new feature after MCP research. Lean, elegant, zero defensive programming.

Completion Promise

text
--completion-promise "NO_REMAINING_TASK_AND_KARPATHY_APPROVES_EVERY_SINGLE_LOC_IN_ITS_ENTIRETY"

Common Element

text
Also, I am a fresh agent—free to criticize and radically change previous work. Karpathy's philosophy: delete and simplify. Code is liability; prefer well-maintained libraries over custom code. UI libraries: optimize, don't delete. Re-read all the sources from zero. Use MCPs and web searches—traditional knowledge is stale. Commit and push at the loop end. Any edit means I need a fresh iteration. SWOT analysis first, then work.

Detailed review


<task>You are a ruthless engineering critic applying Andrej Karpathy's design philosophy. Read the architecture plan at PLAN LINK.
Karpathy's core principles:- Code is liability. Every line you write is a line you must maintain.- Delete and simplify. If something can be removed without breaking the system, remove it.- Prefer well-maintained libraries over custom code.- Zero-defensive design. Don't code for hypotheticals that haven't happened yet.- Start with the simplest thing that works. Add complexity only when forced by reality.- "Demo is works.any(), product is works.all()" -- but V1 is closer to demo than product.- Overfit a single batch before scaling up.
Apply these principles to the plan. For each section, ask:1. Is this needed for V1, or is it speculative engineering?2. Can this be deleted or simplified without losing core value?3. Is this solving a problem we actually have, or a problem we might have?4. Would a 10x engineer look at this and say "too much"?
Be brutal. Identify:- **OVER-ENGINEERING**: Things designed for scale/problems that don't exist yet- **UNNECESSARY COMPLEXITY**: Things that add cognitive load without proportional value- **PREMATURE ABSTRACTIONS**: Separations that aren't justified at V1 scale- **DELETE CANDIDATES**: Sections, tables, fields, or features that should be cut from V1
This is a V1 product being built by a small team. The goal is to ship a working product, not to architect for 10M traffic on day one.
Use web search and tools to verify any claims you make about simpler alternatives.</task>
<structured_output_contract>Return findings in these sections:1. VERDICT: Would Karpathy approve? One line.2. DELETE: Things to remove entirely3. SIMPLIFY: Things to keep but make simpler4. KEEP: Things that are correctly lean5. THE LEAN V1: What the plan SHOULD look like if you strip it to essentials</structured_output_contract>
<grounding_rules>- Be specific. Don't say "simplify the schema" -- say which fields to cut.- Every DELETE must justify what you lose and why it's acceptable for V1.- Every KEEP must justify why it's essential, not just nice-to-have.- Think from the perspective of "what do I need to ship in 2 weeks?"</grounding_rules>
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Definition for a binary tree node.
Definition for a binary tree node.

Definition for a binary tree node.

Solved at: 220925. Took 17m 09s

Question

Given a binary search tree (BST), find the lowest common ancestor (LCA) node of two given nodes in the BST.

According to thedefinition of LCA on Wikipedia: "The lowest common ancestor is defined between two nodespandqas the lowest node inTthat has bothpandqas descendants (where we allowa node to be a descendant of itself)."

Solution

python
# Definition for a binary tree node.# class TreeNode:#     def __init__(self, x):#         self.val = x#         self.left = None#         self.right = None
class Solution:    def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
        found_p = False        found_q = False
        queue = [root]
        ancestor = {}
        while((not found_p or not found_q) and queue):            current = queue.pop()            if not current:                continue            left = current.left            right = current.right            if left:                ancestor[left] = current                if p.val == left.val:                    found_p = True                if q.val == left.val:                    found_q = True            if right:                ancestor[right] = current                if p.val == right.val:                    found_p = True                if q.val == right.val:                    found_q = True            queue.append(left)            queue.append(right)
        current = p        p_path = [current]
        while current != root:            current = ancestor[current]            p_path.append(current)
        current = q        q_path = [current]
        while current != root:            current = ancestor[current]            q_path.append(current)
        lca = root
        for i in p_path:            print(i.val, "<-", end=" ")        print()        for i in q_path:            print(i.val, "<-", end=" ")
        for i in range(-1, -1 * min(len(q_path), len(p_path)) - 1, -1):            if q_path[i] == p_path[i]:                lca = p_path[i]            elif q_path[i] != p_path[i]:                break
        return lca

I misunderstood LCA as being the minimum in value.

Results

Runtime

  • 89 ms, faster than 86.27% of Python3 online submissions for the Lowest Common Ancestor of a Binary Search Tree.

Memory Usage

  • 18.9 MB, less than 23.33% of Python3 online submissions for the Lowest Common Ancestor of a Binary Search Tree.

Improvements

I ignored that this is a Binary Search Tree, not just Binary Tree. Therefore one can binary search to reduce the search scope.

Complexity Analysis

For this, It will be O(N)O(N)O(N) where NNN is the count of the nodes. Space complexity will be O(N)O(N)O(N) for constructing lists and dictionaries.

If we use the above improvement, space complexity will reduce to O(1)O(1)O(1). However, time complexity will not change.

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STOP! Re-read all code. Would Karpathy approve every line? Karpathy prefers lean, elegant, well-tested, zero-defensive programming. Use MCPs and web searches.
--completion-promise "KARPATHY_WILL_APPROVE_EVERY_SINGLE_LOC_FOR_SURE"
STOP! Re-read all code, assess PR comments. Handle exactly one comment: either fix it, or rebut with 3 external sources. Fix any dirt found along the way. Lean, elegant, zero defensive programming.
--completion-promise "NO_COMMENTS_REMAINING_IN_GITHUB_EVEN_AFTER_20_MINUTES"
STOP! Re-read all code, assess GitHub Issues. Pick one task: fix dirty code, or implement a new feature after MCP research. Lean, elegant, zero defensive programming.
--completion-promise "NO_REMAINING_TASK_AND_KARPATHY_APPROVES_EVERY_SINGLE_LOC_IN_ITS_ENTIRETY"
Also, I am a fresh agent—free to criticize and radically change previous work. Karpathy's philosophy: delete and simplify. Code is liability; prefer well-maintained libraries over custom code. UI libraries: optimize, don't delete. Re-read all the sources from zero. Use MCPs and web searches—traditional knowledge is stale. Commit and push at the loop end. Any edit means I need a fresh iteration. SWOT analysis first, then work.

<task>You are a ruthless engineering critic applying Andrej Karpathy's design philosophy. Read the architecture plan at PLAN LINK.
Karpathy's core principles:- Code is liability. Every line you write is a line you must maintain.- Delete and simplify. If something can be removed without breaking the system, remove it.- Prefer well-maintained libraries over custom code.- Zero-defensive design. Don't code for hypotheticals that haven't happened yet.- Start with the simplest thing that works. Add complexity only when forced by reality.- "Demo is works.any(), product is works.all()" -- but V1 is closer to demo than product.- Overfit a single batch before scaling up.
Apply these principles to the plan. For each section, ask:1. Is this needed for V1, or is it speculative engineering?2. Can this be deleted or simplified without losing core value?3. Is this solving a problem we actually have, or a problem we might have?4. Would a 10x engineer look at this and say "too much"?
Be brutal. Identify:- **OVER-ENGINEERING**: Things designed for scale/problems that don't exist yet- **UNNECESSARY COMPLEXITY**: Things that add cognitive load without proportional value- **PREMATURE ABSTRACTIONS**: Separations that aren't justified at V1 scale- **DELETE CANDIDATES**: Sections, tables, fields, or features that should be cut from V1
This is a V1 product being built by a small team. The goal is to ship a working product, not to architect for 10M traffic on day one.
Use web search and tools to verify any claims you make about simpler alternatives.</task>
<structured_output_contract>Return findings in these sections:1. VERDICT: Would Karpathy approve? One line.2. DELETE: Things to remove entirely3. SIMPLIFY: Things to keep but make simpler4. KEEP: Things that are correctly lean5. THE LEAN V1: What the plan SHOULD look like if you strip it to essentials</structured_output_contract>
<grounding_rules>- Be specific. Don't say "simplify the schema" -- say which fields to cut.- Every DELETE must justify what you lose and why it's acceptable for V1.- Every KEEP must justify why it's essential, not just nice-to-have.- Think from the perspective of "what do I need to ship in 2 weeks?"</grounding_rules>
Warning
This post is more than a year old. Information may be outdated.
# Definition for a binary tree node.# class TreeNode:#     def __init__(self, x):#         self.val = x#         self.left = None#         self.right = None
class Solution:    def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
        found_p = False        found_q = False
        queue = [root]
        ancestor = {}
        while((not found_p or not found_q) and queue):            current = queue.pop()            if not current:                continue            left = current.left            right = current.right            if left:                ancestor[left] = current                if p.val == left.val:                    found_p = True                if q.val == left.val:                    found_q = True            if right:                ancestor[right] = current                if p.val == right.val:                    found_p = True                if q.val == right.val:                    found_q = True            queue.append(left)            queue.append(right)
        current = p        p_path = [current]
        while current != root:            current = ancestor[current]            p_path.append(current)
        current = q        q_path = [current]
        while current != root:            current = ancestor[current]            q_path.append(current)
        lca = root
        for i in p_path:            print(i.val, "<-", end=" ")        print()        for i in q_path:            print(i.val, "<-", end=" ")
        for i in range(-1, -1 * min(len(q_path), len(p_path)) - 1, -1):            if q_path[i] == p_path[i]:                lca = p_path[i]            elif q_path[i] != p_path[i]:                break
        return lca