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Problem solution · C++

Operations on Tree

Operations on Tree: a C++ solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Direct simulation
Source
walkccc LeetCode Solutions
Length
58 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Direct simulation

For Operations on Tree, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 58 lines of C++ from the credited upstream file 1993.cpp.
  • The implementation visibly relies on sequence storage.
  • 3 loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeOperations on Tree · C++C++
Use this to learn the idea, then write your own version.
struct Node {  vector<int> children;  int lockedBy = -1;}; class LockingTree { public:  LockingTree(vector<int>& parent) : parent(parent) {    nodes.resize(parent.size());    for (int i = 1; i < parent.size(); ++i)      nodes[parent[i]].children.push_back(i);  }   bool lock(int num, int user) {    if (nodes[num].lockedBy != -1)      return false;    return nodes[num].lockedBy = user;  }   bool unlock(int num, int user) {    if (nodes[num].lockedBy != user)      return false;    return nodes[num].lockedBy = -1;  }   bool upgrade(int num, int user) {    if (nodes[num].lockedBy != -1)      return false;    if (!anyLockedDescendant(num))      return false;     // Walk up the hierarchy to ensure that there are no locked ancestors.    for (int i = num; i != -1; i = parent[i])      if (nodes[i].lockedBy != -1)        return false;     unlockDescendants(num);    return nodes[num].lockedBy = user;  }  private:  const vector<int> parent;  vector<Node> nodes;   bool anyLockedDescendant(int i) {    return nodes[i].lockedBy != -1 ||           ranges::any_of(nodes[i].children, [this](const int child) {      return anyLockedDescendant(child);    });  }   void unlockDescendants(int i) {    nodes[i].lockedBy = -1;    for (const int child : nodes[i].children)      unlockDescendants(child);  }};

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