Very interesting Tree problem II
Another interesting Tree problem, this time more of a design problem: design and implement a locking tree structure. I used hashtable to represent the tree, another hashtable to represent the locked elements, and a copy of the parents array. The upgrade method is O(N), and since it may get called N times, the total complexity is N^2 where N=2x10^3 (reasonable). Full implementation down below, cheers, ACC.
You are given a tree with n
nodes numbered from 0
to n - 1
in the form of a parent array parent
where parent[i]
is the parent of the ith
node. The root of the tree is node 0
, so parent[0] = -1
since it has no parent. You want to design a data structure that allows users to lock, unlock, and upgrade nodes in the tree.
The data structure should support the following functions:
- Lock: Locks the given node for the given user and prevents other users from locking the same node. You may only lock a node using this function if the node is unlocked.
- Unlock: Unlocks the given node for the given user. You may only unlock a node using this function if it is currently locked by the same user.
- Upgrade: Locks the given node for the given user and unlocks all of its descendants regardless of who locked it. You may only upgrade a node if all 3 conditions are true:
- The node is unlocked,
- It has at least one locked descendant (by any user), and
- It does not have any locked ancestors.
Implement the LockingTree
class:
LockingTree(int[] parent)
initializes the data structure with the parent array.lock(int num, int user)
returnstrue
if it is possible for the user with iduser
to lock the nodenum
, orfalse
otherwise. If it is possible, the nodenum
will become locked by the user with iduser
.unlock(int num, int user)
returnstrue
if it is possible for the user with iduser
to unlock the nodenum
, orfalse
otherwise. If it is possible, the nodenum
will become unlocked.upgrade(int num, int user)
returnstrue
if it is possible for the user with iduser
to upgrade the nodenum
, orfalse
otherwise. If it is possible, the nodenum
will be upgraded.
Example 1:
Input ["LockingTree", "lock", "unlock", "unlock", "lock", "upgrade", "lock"] [[[-1, 0, 0, 1, 1, 2, 2]], [2, 2], [2, 3], [2, 2], [4, 5], [0, 1], [0, 1]] Output [null, true, false, true, true, true, false] Explanation LockingTree lockingTree = new LockingTree([-1, 0, 0, 1, 1, 2, 2]); lockingTree.lock(2, 2); // return true because node 2 is unlocked. // Node 2 will now be locked by user 2. lockingTree.unlock(2, 3); // return false because user 3 cannot unlock a node locked by user 2. lockingTree.unlock(2, 2); // return true because node 2 was previously locked by user 2. // Node 2 will now be unlocked. lockingTree.lock(4, 5); // return true because node 4 is unlocked. // Node 4 will now be locked by user 5. lockingTree.upgrade(0, 1); // return true because node 0 is unlocked and has at least one locked descendant (node 4). // Node 0 will now be locked by user 1 and node 4 will now be unlocked. lockingTree.lock(0, 1); // return false because node 0 is already locked.
Constraints:
n == parent.length
2 <= n <= 2000
0 <= parent[i] <= n - 1
fori != 0
parent[0] == -1
0 <= num <= n - 1
1 <= user <= 104
parent
represents a valid tree.- At most
2000
calls in total will be made tolock
,unlock
, andupgrade
.
public class LockingTree { private Hashtable tree = null; private int[] parent = null; private Hashtable nodeLock = null; public LockingTree(int[] parent) { tree = new Hashtable(); this.parent = new int[parent.Length]; nodeLock = new Hashtable(); for (int i = 0; i < parent.Length; i++) { this.parent[i] = parent[i]; if (!tree.ContainsKey(parent[i])) tree.Add(parent[i], new Hashtable()); Hashtable children = (Hashtable)tree[parent[i]]; if (!children.ContainsKey(i)) children.Add(i, true); } } public bool Lock(int num, int user) { if (!nodeLock.ContainsKey(num)) { nodeLock.Add(num, user); return true; } return false; } public bool Unlock(int num, int user) { if (!nodeLock.ContainsKey(num) || (int)nodeLock[num] != user) return false; nodeLock.Remove(num); return true; } public bool Upgrade(int num, int user) { if (!nodeLock.ContainsKey(num) && HasAnyLockedDescendant(num) && !HasAnyLockedAnscestor(num)) { nodeLock.Add(num, user); UnlockAllDescendants(num); return true; } return false; } private void UnlockAllDescendants(int node) { if (tree.ContainsKey(node)) { Hashtable children = (Hashtable)tree[node]; if (children != null) { foreach (int child in children.Keys) _UnlockAllDescendants(child); } } } private void _UnlockAllDescendants(int node) { if (nodeLock.ContainsKey(node)) nodeLock.Remove(node); if (tree.ContainsKey(node)) { Hashtable children = (Hashtable)tree[node]; if (children != null) { foreach (int child in children.Keys) _UnlockAllDescendants(child); } } } private bool HasAnyLockedDescendant(int node) { if (tree.ContainsKey(node)) { Hashtable children = (Hashtable)tree[node]; if (children != null) { foreach (int child in children.Keys) { if (nodeLock.ContainsKey(child) || HasAnyLockedDescendant(child)) return true; } } } return false; } private bool HasAnyLockedAnscestor(int node) { int parentNode = parent[node]; while (parentNode != -1) { if (nodeLock.ContainsKey(parentNode)) return true; parentNode = parent[parentNode]; } return false; } }
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