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2065. Maximum Path Quality of a Graph 👍

Approach 1: DFS

  • Time: $O(|V| + |E|)$, where $|V| = |\texttt{values}|$ and $|E| = |\texttt{edges}|$
  • Space: $O(|V| + |E|)$
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class Solution {
 public:
  int maximalPathQuality(vector<int>& values, vector<vector<int>>& edges,
                         int maxTime) {
    const int n = values.size();
    int ans = 0;
    vector<vector<pair<int, int>>> graph(n);
    vector<int> seen(n);
    seen[0] = 1;

    for (const vector<int>& edge : edges) {
      const int u = edge[0];
      const int v = edge[1];
      const int time = edge[2];
      graph[u].emplace_back(v, time);
      graph[v].emplace_back(u, time);
    }

    dfs(graph, values, 0, values[0], maxTime, seen, ans);
    return ans;
  }

 private:
  void dfs(const vector<vector<pair<int, int>>>& graph,
           const vector<int>& values, int u, int quality, int remainingTime,
           vector<int>& seen, int& ans) {
    if (u == 0)
      ans = max(ans, quality);
    for (const auto& [v, time] : graph[u]) {
      if (time > remainingTime)
        continue;
      const int newQuality = quality + values[v] * (seen[v] == 0);
      ++seen[v];
      dfs(graph, values, v, newQuality, remainingTime - time, seen, ans);
      --seen[v];
    }
  }
};

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class Solution {
  public int maximalPathQuality(int[] values, int[][] edges, int maxTime) {
    final int n = values.length;
    List<Pair<Integer, Integer>>[] graph = new List[n];
    int[] seen = new int[n];
    seen[0] = 1;
    Arrays.setAll(graph, i -> new ArrayList<>());

    for (int[] edge : edges) {
      final int u = edge[0];
      final int v = edge[1];
      final int time = edge[2];
      graph[u].add(new Pair<>(v, time));
      graph[v].add(new Pair<>(u, time));
    }

    dfs(graph, values, 0, values[0], maxTime, seen);
    return ans;
  }

  private int ans = 0;

  private void dfs(List<Pair<Integer, Integer>>[] graph, int[] values, int u, int quality,
                   int remainingTime, int[] seen) {
    if (u == 0)
      ans = Math.max(ans, quality);
    for (Pair<Integer, Integer> pair : graph[u]) {
      final int v = pair.getKey();
      final int time = pair.getValue();
      if (time > remainingTime)
        continue;
      final int newQuality = quality + values[v] * (seen[v] == 0 ? 1 : 0);
      ++seen[v];
      dfs(graph, values, v, newQuality, remainingTime - time, seen);
      --seen[v];
    }
  }
}

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class Solution:
  def maximalPathQuality(
      self,
      values: list[int],
      edges: list[list[int]],
      maxTime: int,
  ) -> int:
    n = len(values)
    ans = 0
    graph = [[] for _ in range(n)]
    seen = [0] * n
    seen[0] = 1

    for u, v, time in edges:
      graph[u].append((v, time))
      graph[v].append((u, time))

    def dfs(u: int, quality: int, remainingTime: int):
      nonlocal ans
      if u == 0:
        ans = max(ans, quality)
      for v, time in graph[u]:
        if time > remainingTime:
          continue
        newQuality = quality + values[v] * (seen[v] == 0)
        seen[v] += 1
        dfs(v, newQuality, remainingTime - time)
        seen[v] -= 1

    dfs(0, values[0], maxTime)
    return ans

Approach 2: BFS

  • Time: $O(|V| + |E|)$
  • Space: $O(|V| + |E|)$
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class Solution:
  def maximalPathQuality(
      self,
      values: list[int],
      edges: list[list[int]],
      maxTime: int,
  ) -> int:
    ans = 0
    graph = [[] for _ in range(len(values))]
    # (node, quality, remainingTime, seen)
    q = collections.deque([(0, values[0], maxTime, {0})])

    for u, v, time in edges:
      graph[u].append((v, time))
      graph[v].append((u, time))

    while q:
      u, quality, remainingTime, seen = q.popleft()
      if u == 0:
        ans = max(ans, quality)
      for v, time in graph[u]:
        if time <= remainingTime:
          q.append(
              (v, quality + values[v] * (v not in seen),
               remainingTime - time, seen | set([v])))

    return ans