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Problem solution · Java

Number of Good Paths

Number of Good Paths: a Java solution using disjoint set union. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Disjoint set union
Source
walkccc LeetCode Solutions
Length
74 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

Disjoint set union

For Number of Good Paths, the implementation maintains connected components and merges them as relationships are processed.

  1. Give each element a component representative.
  2. Merge representatives when a connection is accepted.
  3. Answer connectivity or component queries from the compressed representatives.

Code notes

  • 74 lines of Java from the credited upstream file 2421.java.
  • The implementation visibly relies on sequence storage, hash lookup, ordered lookup.
  • 9 loop blocks detected.

Complexity

Account for every find and union operation; with path compression and ranked merging, the amortized cost is nearly constant per operation.

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 codeNumber of Good Paths · JavaJava
Use this to learn the idea, then write your own version.
class UnionFind {  public UnionFind(int n) {    id = new int[n];    rank = new int[n];    for (int i = 0; i < n; ++i)      id[i] = i;  }   public void unionByRank(int u, int v) {    final int i = find(u);    final int j = find(v);    if (i == j)      return;    if (rank[i] < rank[j]) {      id[i] = j;    } else if (rank[i] > rank[j]) {      id[j] = i;    } else {      id[i] = j;      ++rank[j];    }  }   public int find(int u) {    return id[u] == u ? u : (id[u] = find(id[u]));  }   private int[] id;  private int[] rank;} class Solution {  public int numberOfGoodPaths(int[] vals, int[][] edges) {    final int n = vals.length;    int ans = n;    UnionFind uf = new UnionFind(n);    List<Integer>[] tree = new List[n];    Map<Integer, List<Integer>> valToNodes = new TreeMap<>();     for (int i = 0; i < n; ++i)      tree[i] = new ArrayList<>();     for (int[] edge : edges) {      final int u = edge[0];      final int v = edge[1];      if (vals[v] <= vals[u])        tree[u].add(v);      if (vals[u] <= vals[v])        tree[v].add(u);    }     for (int i = 0; i < vals.length; ++i) {      valToNodes.putIfAbsent(vals[i], new ArrayList<>());      valToNodes.get(vals[i]).add(i);    }     for (Map.Entry<Integer, List<Integer>> entry : valToNodes.entrySet()) {      final int val = entry.getKey();      List<Integer> nodes = entry.getValue();      for (final int u : nodes)        for (final int v : tree[u])          uf.unionByRank(u, v);      Map<Integer, Integer> rootCount = new HashMap<>();      for (final int u : nodes)        rootCount.merge(uf.find(u), 1, Integer::sum);      // For each group, C(count, 2) := count * (count - 1) / 2      for (final int count : rootCount.values())        ans += count * (count - 1) / 2;    }     return ans;  }}

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