Stella

OpenClaw, Hermes, Poke를 쓰며 느낀 단점들

내가 만약 워크스페이스 에이전트를 만든다면 어떻게 만들까?

메모리

육하원칙에 맞게 기억해야하는데, 이걸 잘 못 함
- 맥락을 종종 헷갈려함
메모리 툴을 명시적으로 불러야 함
- 이는 Hermes에서 많이 완화된 부분

Slack

답장을 매번 하려 함. 쓰레드에 달리는 모든 글에 답을 함.
- 매번 답할 필요는 없음. 필요할 때만 하면 됨.
언급하지 않은 메시지는 읽지 않음.
- 모든 워크스페이스의 변화를 계속 보고 있어야 함, 그리고 기억해야 함

Backlinks (2)

260413

Ramp의 AX (회사를 AI로 물들이는 법)

Backlinks (0)

No backlinks found.

Debian Setup

sudo apt update && sudo apt install git && /bin/bash -c "$(curl -fsSL https://raw.githubusercontent.com/Homebrew/install/HEAD/install.sh)" && echo >> ~/.bashrc && echo 'eval "$(/home/linuxbrew/.linuxbrew/bin/brew shellenv bash)"' >> ~/.bashrc && eval "$(/home/linuxbrew/.linuxbrew/bin/brew shellenv bash)" && sudo apt-get install build-essential && brew install gcc btop

Backlinks (1)

260415

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 $O(N)$ where $N$ is the count of the nodes. Space complexity will be $O(N)$ for constructing lists and dictionaries.

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

Backlinks (2)

Stella

OpenClaw, Hermes, Poke를 쓰며 느낀 단점들

내가 만약 워크스페이스 에이전트를 만든다면 어떻게 만들까?

메모리

육하원칙에 맞게 기억해야하는데, 이걸 잘 못 함
- 맥락을 종종 헷갈려함
메모리 툴을 명시적으로 불러야 함
- 이는 Hermes에서 많이 완화된 부분

Slack

답장을 매번 하려 함. 쓰레드에 달리는 모든 글에 답을 함.
- 매번 답할 필요는 없음. 필요할 때만 하면 됨.
언급하지 않은 메시지는 읽지 않음.
- 모든 워크스페이스의 변화를 계속 보고 있어야 함, 그리고 기억해야 함

Backlinks (2)

260413

Ramp의 AX (회사를 AI로 물들이는 법)

Backlinks (0)

No backlinks found.

Debian Setup

sudo apt update && sudo apt install git && /bin/bash -c "$(curl -fsSL https://raw.githubusercontent.com/Homebrew/install/HEAD/install.sh)" && echo >> ~/.bashrc && echo 'eval "$(/home/linuxbrew/.linuxbrew/bin/brew shellenv bash)"' >> ~/.bashrc && eval "$(/home/linuxbrew/.linuxbrew/bin/brew shellenv bash)" && sudo apt-get install build-essential && brew install gcc btop

sudo apt update && sudo apt install git && /bin/bash -c "$(curl -fsSL https://raw.githubusercontent.com/Homebrew/install/HEAD/install.sh)" && echo >> ~/.bashrc && echo 'eval "$(/home/linuxbrew/.linuxbrew/bin/brew shellenv bash)"' >> ~/.bashrc && eval "$(/home/linuxbrew/.linuxbrew/bin/brew shellenv bash)" && sudo apt-get install build-essential && brew install gcc btop

Backlinks (1)

260415

Definition for a binary tree node.

Warning

This post is more than a year old. Information may be outdated.

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.

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

# 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 $O(N)$ where $N$ is the count of the nodes. Space complexity will be $O(N)$ for constructing lists and dictionaries.

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

Backlinks (2)