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Problem solution · Java

Critical Connections in a Network

Critical Connections in a Network: a Java solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Direct simulation
Source
walkccc LeetCode Solutions
Length
48 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

Direct simulation

For Critical Connections in a Network, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 48 lines of Java from the credited upstream file 1192.java.
  • The implementation visibly relies on sequence storage.
  • 2 loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

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 codeCritical Connections in a Network · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public List<List<Integer>> criticalConnections(int n, List<List<Integer>> connections) {    List<List<Integer>> ans = new ArrayList<>();    List<Integer>[] graph = new List[n];    Arrays.setAll(graph, i -> new ArrayList<>());     for (List<Integer> connection : connections) {      final int u = connection.get(0);      final int v = connection.get(1);      graph[u].add(v);      graph[v].add(u);    }     // rank[i] := the minimum node that node i can reach with forward edges    // Initialize with NO_RANK = -2 to indicate not visited.    int[] rank = new int[n];    Arrays.fill(rank, NO_RANK);    getRank(graph, 0, 0, rank, ans);    return ans;  }   private static final int NO_RANK = -2;   // Gets the minimum rank that u can reach with forward edges.  private int getRank(List<Integer>[] graph, int u, int myRank, int[] rank,                      List<List<Integer>> ans) {    if (rank[u] != NO_RANK) // The rank is already been determined.      return rank[u];     rank[u] = myRank;    int minRank = myRank;     for (final int v : graph[u]) {      // visited || parent (that's why NO_RANK = -2 instead of -1)      if (rank[u] == rank.length || rank[v] == myRank - 1)        continue;      final int nextRank = getRank(graph, v, myRank + 1, rank, ans);      // (u, v) is the only way for u go to v.      if (nextRank == myRank + 1)        ans.add(Arrays.asList(u, v));      minRank = Math.min(minRank, nextRank);    }     rank[u] = rank.length; // Mark as visited.    return minRank;  }}

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