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Problem solution · Java

Count Number of Possible Root Nodes

Count Number of Possible Root Nodes: 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
67 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 Count Number of Possible Root Nodes, 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

  • 67 lines of Java from the credited upstream file 2581.java.
  • The implementation visibly relies on sequence storage, hash lookup, ordered lookup.
  • 6 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 codeCount Number of Possible Root Nodes · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public int rootCount(int[][] edges, int[][] guesses, int k) {    final int n = edges.length + 1;    List<Integer>[] graph = new List[n];    Set<Integer>[] guessGraph = new Set[n];    int[] parent = new int[n];     for (int i = 0; i < n; ++i) {      graph[i] = new ArrayList<>();      guessGraph[i] = new HashSet<>();    }     for (int[] edge : edges) {      final int u = edge[0];      final int v = edge[1];      graph[u].add(v);      graph[v].add(u);    }     for (int[] guess : guesses) {      final int u = guess[0];      final int v = guess[1];      guessGraph[u].add(v);    }     // Precalculate `parent`.    dfs(graph, 0, /*prev=*/-1, parent);     // Calculate `correctGuess` for tree rooted at 0.    int correctGuess = 0;    for (int i = 1; i < n; ++i)      if (guessGraph[parent[i]].contains(i))        ++correctGuess;     reroot(graph, 0, /*prev=*/-1, correctGuess, guessGraph, k);    return ans;  }   private int ans = 0;   private void dfs(List<Integer>[] graph, int u, int prev, int[] parent) {    parent[u] = prev;    for (final int v : graph[u])      if (v != prev)        dfs(graph, v, u, parent);  }   private void reroot(List<Integer>[] graph, int u, int prev, int correctGuess,                      Set<Integer>[] guessGraph, int k) {    if (u != 0) {      // The tree is rooted at u, so a guess edge (u, prev) will match the new      // `parent` relationship.      if (guessGraph[u].contains(prev))        ++correctGuess;      // A guess edge (prev, u) matching the old `parent` relationship will no      // longer be true.      if (guessGraph[prev].contains(u))        --correctGuess;    }    if (correctGuess >= k)      ++ans;    for (final int v : graph[u])      if (v != prev)        reroot(graph, v, u, correctGuess, guessGraph, k);  }}

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