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Problem solution · C++

Collect Coins in a Tree

Collect Coins in a Tree: a C++ solution using breadth-first search. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Breadth-first search
Source
walkccc LeetCode Solutions
Length
51 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

Breadth-first search

For Collect Coins in a Tree, the implementation explores reachable states in layers, which is the standard shape for unweighted shortest paths and minimum-step transitions.

  1. Model each valid configuration as a state and each legal move as an edge.
  2. Seed the queue with the starting state and mark it immediately.
  3. Expand each state once, recording distance or reachability for unseen neighbours.

Code notes

  • 51 lines of C++ from the credited upstream file 2603.cpp.
  • The implementation visibly relies on sequence storage, hash lookup, work queue.
  • 5 loop blocks detected.

Complexity

Verify that each state and transition is processed only a bounded number of times; that determines the traversal cost.

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 codeCollect Coins in a Tree · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  int collectTheCoins(vector<int>& coins, vector<vector<int>>& edges) {    const int n = coins.size();    vector<unordered_set<int>> tree(n);    queue<int> leavesToBeRemoved;     for (const vector<int>& edge : edges) {      const int u = edge[0];      const int v = edge[1];      tree[u].insert(v);      tree[v].insert(u);    }     for (int i = 0; i < n; ++i) {      int u = i;      // Remove the leaves that don't have coins.      while (tree[u].size() == 1 && coins[u] == 0) {        const int v = *tree[u].begin();        tree[u].clear();        tree[v].erase(u);        u = v;  // Walk up to its parent.      }      // After trimming leaves without coins, leaves with coins may satisfy      // `leavesToBeRemoved`.      if (tree[u].size() == 1)        leavesToBeRemoved.push(u);    }     // Remove each remaining leaf node and its parent. The remaining nodes are    // the ones that must be visited.    for (int i = 0; i < 2; ++i)      for (int sz = leavesToBeRemoved.size(); sz > 0; --sz) {        const int u = leavesToBeRemoved.front();        leavesToBeRemoved.pop();        if (!tree[u].empty()) {          const int v = *tree[u].begin();          tree[u].clear();          tree[v].erase(u);          if (tree[v].size() == 1)            leavesToBeRemoved.push(v);        }      }     return accumulate(tree.begin(), tree.end(), 0,                      [](int acc, const unordered_set<int>& children) {      return acc + children.size();    });  }};

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