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Problem solution · Java

Maximize the Number of Target Nodes After Connecting Trees II

Maximize the Number of Target Nodes After Connecting Trees II: a Java solution using depth-first search. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Depth-first search
Source
walkccc LeetCode Solutions
Length
49 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

Depth-first search

For Maximize the Number of Target Nodes After Connecting Trees II, the implementation follows one branch at a time, making it suitable for components, trees, backtracking, or dependency exploration.

  1. Define the state carried into one recursive or stack frame.
  2. Mark or choose the current state before exploring children.
  3. Combine child results or undo the choice when the branch finishes.

Code notes

  • 49 lines of Java from the credited upstream file 3373.java.
  • The implementation visibly relies on sequence storage.
  • 4 loop blocks detected, together with recursive traversal.

Complexity

Count unique states for graph traversal; for backtracking, count the branching factor and maximum depth.

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 codeMaximize the Number of Target Nodes After Connecting Trees II · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public int[] maxTargetNodes(int[][] edges1, int[][] edges2) {    final int n = edges1.length + 1;    final int m = edges2.length + 1;    List<Integer>[] graph1 = buildGraph(edges1, n);    List<Integer>[] graph2 = buildGraph(edges2, m);    boolean[] parity1 = new boolean[n];    boolean[] parity2 = new boolean[m]; // Placeholder (not used)    final int even1 = dfs(graph1, 0, -1, parity1, true);    final int even2 = dfs(graph2, 0, -1, parity2, true);    final int odd1 = n - even1;    final int odd2 = m - even2;    int[] ans = new int[n];     for (int i = 0; i < n; i++) {      final int tree1 = parity1[i] ? even1 : odd1;      // Can connect the current node in tree1 to either an even or an odd node      // in tree2.      final int tree2 = Math.max(even2, odd2);      ans[i] = tree1 + tree2;    }     return ans;  }   // Returns the number of nodes that can be reached from u with even steps.  private int dfs(List<Integer>[] graph, int u, int prev, boolean[] parity, boolean isEven) {    int res = isEven ? 1 : 0;    parity[u] = isEven;    for (final int v : graph[u])      if (v != prev)        res += dfs(graph, v, u, parity, !isEven);    return res;  }   private List<Integer>[] buildGraph(int[][] edges, int n) {    List<Integer>[] graph = new ArrayList[n];    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 graph;  }}

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