CPMagical CP Library

Problem solution · Java

Maximum Employees to Be Invited to a Meeting

Maximum Employees to Be Invited to a Meeting: a Java solution using breadth-first search. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Breadth-first search
Source
walkccc LeetCode Solutions
Length
77 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

Breadth-first search

For Maximum Employees to Be Invited to a Meeting, the implementation explores reachable states in layers, which is the standard shape for unweighted shortest paths and minimum-step transitions.

  1. Model each valid configuration as a state and each legal move as an edge.
  2. Seed the queue with the starting state and mark it immediately.
  3. Expand each state once, recording distance or reachability for unseen neighbours.

Code notes

  • 77 lines of Java from the credited upstream file 2127.java.
  • The implementation visibly relies on sequence storage, work queue.
  • 7 loop blocks detected.

Complexity

Verify that each state and transition is processed only a bounded number of times; that determines the traversal cost.

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 codeMaximum Employees to Be Invited to a Meeting · JavaJava
Use this to learn the idea, then write your own version.
enum State { INIT, VISITING, VISITED } class Solution {  public int maximumInvitations(int[] favorite) {    final int n = favorite.length;    int sumComponentsLength = 0; // the component: a -> b -> c <-> x <- y    List<Integer>[] graph = new List[n];    int[] inDegrees = new int[n];    int[] maxChainLength = new int[n];    Arrays.fill(maxChainLength, 1);    Arrays.setAll(graph, i -> new ArrayList<>());     // Build the graph.    for (int i = 0; i < n; ++i) {      graph[i].add(favorite[i]);      ++inDegrees[favorite[i]];    }     // Perform topological sorting.    Queue<Integer> q = IntStream.range(0, n)                           .filter(i -> inDegrees[i] == 0)                           .boxed()                           .collect(Collectors.toCollection(ArrayDeque::new));     while (!q.isEmpty()) {      final int u = q.poll();      for (final int v : graph[u]) {        if (--inDegrees[v] == 0)          q.offer(v);        maxChainLength[v] = Math.max(maxChainLength[v], 1 + maxChainLength[u]);      }    }     for (int i = 0; i < n; ++i)      if (favorite[favorite[i]] == i)        // i <-> favorite[i] (the cycle's length = 2)        sumComponentsLength += maxChainLength[i] + maxChainLength[favorite[i]];     int[] parent = new int[n];    Arrays.fill(parent, -1);    boolean[] seen = new boolean[n];    State[] states = new State[n];     for (int i = 0; i < n; ++i)      if (!seen[i])        findCycle(graph, i, parent, seen, states);     return Math.max(sumComponentsLength / 2, maxCycleLength);  }   private int maxCycleLength = 0; // the cycle : a -> b -> c -> a   private void findCycle(List<Integer>[] graph, int u, int[] parent, boolean[] seen,                         State[] states) {    seen[u] = true;    states[u] = State.VISITING;     for (final int v : graph[u]) {      if (!seen[v]) {        parent[v] = u;        findCycle(graph, v, parent, seen, states);      } else if (states[v] == State.VISITING) {        // Find the cycle's length.        int curr = u;        int cycleLength = 1;        while (curr != v) {          curr = parent[curr];          ++cycleLength;        }        maxCycleLength = Math.max(maxCycleLength, cycleLength);      }    }     states[u] = State.VISITED;  }}

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗