Skip to content

3372. Maximize the Number of Target Nodes After Connecting Trees I

  • Time: $O(k(n + m))$
  • Space: $O(n + m)$
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
class Solution {
 public:
  vector<int> maxTargetNodes(vector<vector<int>>& edges1,
                             vector<vector<int>>& edges2, int k) {
    vector<int> ans;
    const vector<vector<int>> graph1 = buildGraph(edges1);
    const vector<vector<int>> graph2 = buildGraph(edges2);
    int maxReachableInGraph2 = 0;

    if (k > 0)
      for (int i = 0; i < edges2.size() + 1; ++i)
        maxReachableInGraph2 =
            max(maxReachableInGraph2, dfs(graph2, i, -1, k - 1));

    for (int i = 0; i < edges1.size() + 1; ++i)
      ans.push_back(maxReachableInGraph2 + dfs(graph1, i, -1, k));

    return ans;
  }

 private:
  // Returns the number of nodes that can be reached from u with k steps.
  int dfs(const vector<vector<int>>& graph, int u, int prev, int k) {
    if (k == 0)
      return 1;
    int res = 0;
    for (const int v : graph[u])
      if (v != prev)
        res += dfs(graph, v, u, k - 1);
    return 1 + res;
  }

  vector<vector<int>> buildGraph(const vector<vector<int>>& edges) {
    vector<vector<int>> graph(edges.size() + 1);
    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);
    }
    return graph;
  }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
class Solution {
  public int[] maxTargetNodes(int[][] edges1, int[][] edges2, int k) {
    int[] ans = new int[edges1.length + 1];
    List<Integer>[] graph1 = buildGraph(edges1);
    List<Integer>[] graph2 = buildGraph(edges2);
    int maxReachableInGraph2 = 0;

    if (k > 0)
      for (int i = 0; i < edges2.length + 1; ++i)
        maxReachableInGraph2 = Math.max(maxReachableInGraph2, dfs(graph2, i, -1, k - 1));

    for (int i = 0; i < edges1.length + 1; ++i)
      ans[i] = maxReachableInGraph2 + dfs(graph1, i, -1, k);

    return ans;
  }

  // Returns the number of nodes that can be reached from u with k steps.
  private int dfs(List<Integer>[] graph, int u, int prev, int k) {
    if (k == 0)
      return 1;
    int res = 0;
    for (final int v : graph[u])
      if (v != prev)
        res += dfs(graph, v, u, k - 1);
    return 1 + res;
  }

  private List<Integer>[] buildGraph(int[][] edges) {
    List<Integer>[] graph = new ArrayList[edges.length + 1];
    for (int i = 0; i < edges.length + 1; ++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 graph;
  }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
class Solution:
  def maxTargetNodes(
      self,
      edges1: list[list[int]],
      edges2: list[list[int]],
      k: int
  ) -> list[int]:
    graph1 = self._buildGraph(edges1)
    graph2 = self._buildGraph(edges2)
    maxReachableInGraph2 = 0

    if k > 0:
      for i in range(len(edges2) + 1):
        maxReachableInGraph2 = max(maxReachableInGraph2,
                                   self._dfs(graph2, i, -1, k - 1))

    return [maxReachableInGraph2 + self._dfs(graph1, i, -1, k)
            for i in range(len(edges1) + 1)]

  def _dfs(self, graph: list[list[int]], u: int, prev: int, k: int) -> int:
    """Returns the number of nodes that can be reached from u with k steps."""
    if k == 0:
      return 1
    res = 0
    for v in graph[u]:
      if v != prev:
        res += self._dfs(graph, v, u, k - 1)
    return 1 + res

  def _buildGraph(self, edges: list[list[int]]) -> list[list[int]]:
    graph = [[] for _ in range(len(edges) + 1)]
    for u, v in edges:
      graph[u].append(v)
      graph[v].append(u)
    return graph