the problem statemnt is 
You are given a complete, balanced N-Ary Tree and must support Q queries. There are 3 kinds of queries. Return true or false depending on whether the query was successful.

Lock(v, id) — Lock vertex v for user — id Unlock(v, id) — If vertex v is locked by the same id, unlock it. Upgrade(v, id) — If v is unlocked and has at least one locked vertex in it's subtree and every locked vertex in the subtree of v is locked by id, unlock them and lock v instead. Further, here are some additional constraints

A vertex cannot be locked if it has any locked ancestors or descendants, by any ID. When a vertex is upgraded, it's locked descendants are automatically unlocked. An upgrade operation is not possible if the vertex is already locked or has any locked ancestors An unlock operation is only possible if the vertex is already locked and locked by the same id

PS — the lock and unlock function is already done in Log n, Please try to give an approach on how to store information inNode so that the upgrade function takes Log n time.

Locking The Tree of Space
You have a world map represented as an M-Ary tree (sample tree below)

asterisk_green.svg In graph theory, an m-ary tree (for nonnegative integers m) (also known as n-ary, k-ary or k-way tree) is an arborescence (or an ordered tree) in which each node has no more than m children. A binary tree is the special case where m = 2, and a ternary tree is another case with m = 3 that limits its children to three.

asterisk_green.svg In graph theory, an arborescence is a directed graph in which, for a vertex u (called the root) and any other vertex v, there is exactly one directed path from u to v. A graph is a collection of vertices (also known as nodes) and edges that connect these vertices. A tree is a special type of graph that is connected and acyclic, meaning that there are no cycles in the graph. In a tree, there is a unique path between any two vertices, and there is a single vertex called the root that is used as the starting point for traversing the tree.

Sample Tree

You need to define three operations on it:

1. lock(X, uid)
2. unlock(X, uid)
3. upgradeLock(X, uid)
where X the name of a node in the tree (that would be unique) and uid is the user who is performing the operation.
Input format: x is country name (string) and uid is user number (number).

Here are the definitions for the operations:

lock(x, uid)
Lock takes an exclusive access on the subtree rooted at X. It is formally defined like this: Once lock(x, uid) succeeds, then:
lock(A, anyUserid) should fail (returns false), where A is a descendent of X,
lock(B, anyUserld) should fail (returns false), where X is a descendent of B,
Lock operation cannot be performed on a node which is already locked i.e. lock(x, anyUserld) should fail (returns false).

asterisk_green.svg Exclusive access means that only one user can lock a node and its descendants at a time, and no other user can lock any node that is related to the locked node by ancestry or descent. This ensures that there is no conflict or interference between different users who want to perform operations on the tree.
unlock(X, uid)
Unlock reverts what was done by the lock operation. It can only be called on same node on which user uid had called a Lock on before. Returns true if it is successful.

upgradeLock(x, uid)
It helps the user uid upgrade their lock to an ancestor node. It is only possible if the node X already has locked descendants and all of them are only locked by the same user uid. Upgrade should fail if there is any node which is descendant of X that is locked by a different user. Successful Upgrade will 'lock' the node. upgradeLock call shouldn't violate the consistency model that Lock/Unlock function requires.

Notes

The number of nodes in the tree N is very large. So, optimize the time complexity for the above algorithms.
The below section contains the input format.
The first line contains the number of Nodes in the tree (N).

The second line contains number of children per node (value m in m-ary Tree).

The third line contains number of queries (Q).

Next N lines contains the NodeName (string) in the m-Ary Tree.

Next Q lines contains queries which are in format:  Operation Type NodeName Userld

Operation Type

1 for lock
2 for unlock
3 for upgradeLock

NodeName - Name of any node (unique) in m-Ary Tree.

Userld - Integer value representing a unique user.

Example input

7
2
3
World
Asia
Africa
China
India
South Africa
Egypt
1 China 9
2 India 9
3 Asia 9
With the above input you represent a 2-ary tree with 7 nodes as follows:

                 World
            /             \
        Asia               Africa
      /      \          /         \
China    India    South Africa   Egypt
Additional Notes

Here 1 China 3 indicates the following 'Operation Type NodeName Userld'.
The tree is always fully balanced.
Constraints on the inputs are as follows:
1 < N < 5 * 10^5
1 < m < 30
1 < Q <5 * 10^5
1 < length of NodeName < 20
Optimize the time complexity:
Lock - O(logmN)
Unlock - O(logmN)
UpgradeLock - O(numberOfLockedNodes * logmN)
Lock operation on already locked node should fail.
Once Upgrade lock(X, uid) succeeds on X. It is equivalent to X being locked by uid. So, Lock(A/B, anyuser) should fail as per the definition of Lock and Unlock(X, uid) should also work.
Upgrade lock operation on a node having no locked descendants should fail and upgrade lock on already locked node should also fail.
Example input

7
2
5
World
Asia
Africa
China
India
SouthAfrica
Egypt
1 China 9
1 India 9
3 Asia 9
2 India 9
2 Asia 9
Example input

true
true
true
false
true
Query 1: 1 China 9 ⇒ This operation is success as initially China is unlocked.

Query 2: 1 India 9 ⇒ This should be success as none of ancestors and descendants of India are locked.

Query 3: 3 Asia 9 ⇒ This also should be success as upgrade operation is done by same user who has locked descendants.

Query 4: 2 India 9 ⇒ This should fail as the India is now not locked

Query 5: 2 Asia 9 ⇒ This should be success as Asia was earlier (refer Query 3) locked by user 9.
"""
class Node:
def init(self, val):
self.val = val
self.uid = None
self.parent = None
self.child = []
self.isLocked = False
self.lockedDescendantCount = 0 # Use 'descendant' consistently


class Tree:
def init(self, m, n, name_list, queries):
self.node_list = []
self.node_map = {}

for name in name_list:
node = Node(name)
self.node_list.append(node)
self.node_map[name] = node

# Build the tree
for i in range(n):
for j in range(m):
child_index = m * i + j + 1
if child_index < n:
self.node_list[i].child.append(self.node_list[child_index])
self.node_list[child_index].parent = self.node_list[i]

self.root = self.node_list[0]
self.processed_queries = []
for query in queries:
self.processed_queries.append(query.strip().split())

def lock(self, node, uid):
# If the node is already locked, return False
if node.isLocked:
return False

# If any ancestors are locked, cannot lock this node
if self.isAncestorLocked(node):
return False

# If any descendants are locked, cannot lock this node
if node.lockedDescendantCount > 0:
return False

# Lock the node
node.isLocked = True
node.uid = uid

# Update the lockedDescendantCount for ancestors
parent = node.parent
while parent:
parent.lockedDescendantCount += 1
parent = parent.parent

return True

def unlock(self, node, uid):
# The node must be locked and the uid must match
if not node.isLocked or node.uid != uid:
return False

# Unlock the node
node.isLocked = False
node.uid = None

# Update the lockedDescendantCount for ancestors
parent = node.parent
while parent:
parent.lockedDescendantCount -= 1
parent = parent.parent

return True

def upgrade(self, node, uid):
# The node must not be locked
if node.isLocked:
return False

# If any ancestors are locked, cannot upgrade this node
if self.isAncestorLocked(node):
return False

# The node must have at least one locked descendant
if node.lockedDescendantCount == 0:
return False

locked_descendants = []

# Collect all locked descendants
if not self.collectLockedDescendants(node, uid, locked_descendants):
return False

# Unlock all locked descendants
for descendant in locked_descendants:
descendant.isLocked = False
descendant.uid = None

# Update the lockedDescendantCount for ancestors
parent = descendant.parent
while parent:
parent.lockedDescendantCount -= 1
parent = parent.parent

# Lock the current node
node.isLocked = True
node.uid = uid

# Update the lockedDescendantCount for ancestors
parent = node.parent
while parent:
parent.lockedDescendantCount += 1
parent = parent.parent

return True

def isAncestorLocked(self, node):
curr = node.parent
while curr:
if curr.isLocked:
return True
curr = curr.parent
return False

def collectLockedDescendants(self, node, uid, locked_nodes):
# If the node is locked
if node.isLocked:
if node.uid == uid:
locked_nodes.append(node)
else:
return False # Locked by different uid

# Recursively check all children
for child_node in node.child:
if not self.collectLockedDescendants(child_node, uid, locked_nodes):
return False
return True

def process_queries(self):
results = []
for query in self.processed_queries:
operation, node_name, uid = query
node = self.node_map[node_name]
if operation == "1":
result = self.lock(node, uid)
elif operation == "2":
result = self.unlock(node, uid)
elif operation == "3":
result = self.upgrade(node, uid)
else:
result = False # Invalid operation
results.append('true' if result else 'false')
return results


def main():
# Test case 1: Basic lock and unlock
n = 3
m = 2
name_list = ["A", "B", "C"]
queries = [
"1 A 1", # Lock A with uid 1
"2 A 1", # Unlock A with uid 1
]
tree = Tree(m, n, name_list, queries)
print(tree.process_queries()) # Expected: ['true', 'true']

# Test case 2: Locking a node with a locked ancestor
queries = [
"1 A 1", # Lock A with uid 1
"1 B 2", # Try to lock B with uid 2 (should fail because A is locked)
]
tree = Tree(m, n, name_list, queries)
print(tree.process_queries()) # Expected: ['true', 'false']

# Test case 3: Unlocking a node with wrong uid
queries = [
"1 A 1", # Lock A with uid 1
"2 A 2", # Try to unlock A with uid 2 (should fail because uid is wrong)
]
tree = Tree(m, n, name_list, queries)
print(tree.process_queries()) # Expected: ['true', 'false']

# Test case 4: Upgrade a node with locked descendants
queries = [
"1 B 1", # Lock B with uid 1
"1 C 1", # Lock C with uid 1
"3 A 1", # Upgrade A with uid 1 (should succeed and unlock B and C)
]
tree = Tree(m, n, name_list, queries)
print(tree.process_queries()) # Expected: ['true', 'true', 'true']

# Test case 5: Upgrade a node with no locked descendants
queries = [
"3 A 1", # Try to upgrade A with uid 1 (should fail because no locked descendants)
]
tree = Tree(m, n, name_list, queries)
print(tree.process_queries()) # Expected: ['false']

# Test case 6: Locking a node with locked descendants
queries = [
"1 B 1", # Lock B with uid 1
"1 A 2", # Try to lock A with uid 2 (should fail because B is locked)
]
tree = Tree(m, n, name_list, queries)
print(tree.process_queries()) # Expected: ['true', 'false']

# Test case 7: Complex scenario
n = 7
m = 2
name_list = ["A", "B", "C", "D", "E", "F", "G"]
queries = [
"1 B 1", # Lock B with uid 1
"1 D 1", # Lock D with uid 1
"1 E 1", # Lock E with uid 1
"3 A 1", # Upgrade A with uid 1 (should succeed and unlock B, D, E)
"1 A 2", # Lock A with uid 2
"2 A 1", # Try to unlock A with uid 1 (should fail because uid is wrong)
"2 A 2", # Unlock A with uid 2
]
tree = Tree(m, n, name_list, queries)
print(tree.process_queries()) # Expected: ['true', 'true', 'true', 'true', 'true', 'false', 'true']


if name == "main":
main()

""""
"""
in terminal
"""
(base) asvinir@Asvinis-MacBook-Air learnDsa % /opt/homebrew/bin/python3 "/Users/asvinir/Library/Mobile Documents/comappleCloudDocs/Downloads/learnDsa/tree_of_spa
ceAtest.py"
['true', 'true']
['true', 'false']
['true', 'false']
['true', 'true', 'true']
['false']
['true', 'false']
['true', 'false', 'false', 'true', 'false', 'true', 'false']
(base) asvinir@Asvinis-MacBook-Air learnDsa %