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3544. Subtree Inversion Sum 👍

  • Time: $O(nk)$
  • Space: $O(nk)$
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class Solution {
 public:
  long long subtreeInversionSum(vector<vector<int>>& edges, vector<int>& nums,
                                int k) {
    const int n = edges.size() + 1;
    vector<int> parent(n, -1);
    vector<vector<int>> graph(n);

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

    vector<vector<vector<long>>> mem(
        n, vector<vector<long>>(k + 1, vector<long>(2, -1)));
    return dfs(graph, /*u=*/0, /*stepsSinceInversion=*/k,
               /*inverted=*/false, nums, k, parent, mem);
  }

 private:
  // Returns the maximum sum for subtree rooted at u, with `stepsSinceInversion`
  // steps of inversion and `inverted` is true if the subtree is inverted.
  long dfs(const vector<vector<int>>& graph, int u, int stepsSinceInversion,
           bool inverted, const vector<int>& nums, int k, vector<int>& parent,
           vector<vector<vector<long>>>& mem) {
    if (mem[u][stepsSinceInversion][inverted] != -1)
      return mem[u][stepsSinceInversion][inverted];
    long num = inverted ? -nums[u] : nums[u];
    long negNum = -num;
    for (const int v : graph[u]) {
      if (v == parent[u])
        continue;
      parent[v] = u;
      num += dfs(graph, v, min(k, stepsSinceInversion + 1), inverted, nums, k,
                 parent, mem);
      if (stepsSinceInversion == k)
        negNum += dfs(graph, v, /*stepsSinceInversion=*/1, !inverted, nums, k,
                      parent, mem);
    }
    return mem[u][stepsSinceInversion][inverted] =
               (stepsSinceInversion == k) ? max(num, negNum) : num;
  }
};

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class Solution {
  public long subtreeInversionSum(int[][] edges, int[] nums, int k) {
    final int n = edges.length + 1;
    int[] parent = new int[n];
    List<Integer>[] graph = new List[n];
    Arrays.fill(parent, -1);

    for (int i = 0; i < n; ++i)
      graph[i] = new ArrayList<>();

    for (int[] edge : edges) {
      final int u = edge[0];
      final int v = edge[1];
      graph[u].add(v);
      graph[v].add(u);
    }

    return dfs(graph, /*u=*/0, /*stepsSinceInversion=*/k,
               /*inverted=*/false, nums, k, parent, new Long[n][k + 1][2]);
  }

  private long dfs(List<Integer>[] graph, int u, int stepsSinceInversion, boolean inverted,
                   int[] nums, int k, int[] parent, Long[][][] mem) {
    if (mem[u][stepsSinceInversion][inverted ? 1 : 0] != null)
      return mem[u][stepsSinceInversion][inverted ? 1 : 0];
    long num = inverted ? -nums[u] : nums[u];
    long negNum = -num;
    for (final int v : graph[u]) {
      if (v == parent[u])
        continue;
      parent[v] = u;
      num += dfs(graph, v, Math.min(k, stepsSinceInversion + 1), inverted, nums, k, parent, mem);
      if (stepsSinceInversion == k)
        negNum += dfs(graph, v, 1, !inverted, nums, k, parent, mem);
    }
    return mem[u][stepsSinceInversion][inverted ? 1 : 0] =
               (stepsSinceInversion == k) ? Math.max(num, negNum) : num;
  }
}

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class Solution:
  def subtreeInversionSum(
      self,
      edges: list[list[int]],
      nums: list[int],
      k: int
  ) -> int:
    n = len(edges) + 1
    parent = [-1] * n
    graph = [[] for _ in range(n)]

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

    @functools.lru_cache(None)
    def dp(u: int, stepsSinceInversion: int, inverted: bool) -> int:
      """
      Returns the maximum sum for subtree rooted at u, with
      `stepsSinceInversion` steps of inversion and `inverted` is true if the
      subtree is inverted.
      """
      num = -nums[u] if inverted else nums[u]
      negNum = -num
      for v in graph[u]:
        if v == parent[u]:
          continue
        parent[v] = u
        num += dp(v, min(k, stepsSinceInversion + 1), inverted)
        if stepsSinceInversion == k:
          negNum += dp(v, 1, not inverted)
      return max(num, negNum) if stepsSinceInversion == k else num

    return dp(0, k, False)