LeetCode 851 - Loud and Rich

https://leetcode.com/problems/loud-and-rich/description/

In a group of N people (labelled 0, 1, 2, ..., N-1), each person has different amounts of money, and different levels of quietness.

For convenience, we'll call the person with label x, simply "person x".

We'll say that richer[i] = [x, y] if person x definitely has more money than person y.  Note that richer may only be a subset of valid observations.

Also, we'll say quiet[x] = q if person x has quietness q.

Now, return answer, where answer[x] = y if y is the least quiet person (that is, the person y with the smallest value of quiet[y]), among all people who definitely have equal to or more money than person x.

Example 1:

Input: richer = [[1,0],[2,1],[3,1],[3,7],[4,3],[5,3],[6,3]], quiet = [3,2,5,4,6,1,7,0]
Output: [5,5,2,5,4,5,6,7]
Explanation: 
answer[0] = 5.
Person 5 has more money than 3, which has more money than 1, which has more money than 0.
The only person who is quieter (has lower quiet[x]) is person 7, but
it isn't clear if they have more money than person 0.

answer[7] = 7.
Among all people that definitely have equal to or more money than person 7
(which could be persons 3, 4, 5, 6, or 7), the person who is the quietest (has lower quiet[x])
is person 7.

The other answers can be filled out with similar reasoning.

Note:

1. 1 <= quiet.length = N <= 500
2. 0 <= quiet[i] < N, all quiet[i] are different.
3. 0 <= richer.length <= N * (N-1) / 2
4. 0 <= richer[i][j] < N
5. richer[i][0] != richer[i][1]
6. richer[i]'s are all different.
7. The observations in richer are all logically consistent.

X. DFS
https://leetcode.com/problems/loud-and-rich/discuss/137918/C%2B%2BJavaPython-Concise-DFS

The description is not easy to understand.
In fact it's a basic dfs traversal problem.
For every people, call a sub function dfs to compare the quiet with others, who is richer than him.
Also we will note this answer to avoid repeated calculation.

Time Complexity:
O(richer.length),
Sub function dfs traverse every people only once, and every richer is traversed only one once.

    HashMap<Integer, List<Integer>> richer2 = new HashMap<>();
    int res[];
    public int[] loudAndRich(int[][] richer, int[] quiet) {
        int n = quiet.length;
        for (int i = 0; i < n; ++i) richer2.put(i, new ArrayList<Integer>());
        for (int[] v : richer) richer2.get(v[1]).add(v[0]);
        res = new int[n]; Arrays.fill(res, -1);
        for (int i = 0; i < n; i++) dfs(i, quiet);
        return res;
    }

    int dfs(int i, int[] quiet) {
        if (res[i] >= 0) return res[i];
        res[i] = i;
        for (int j : richer2.get(i)) if (quiet[res[i]] > quiet[dfs(j, quiet)]) res[i] = res[j];
        return res[i];
    }

Consider the directed graph with edge x -> y if y is richer than x.

For each person x, we want the quietest person in the subtree at x.

Algorithm

Construct the graph described above, and say dfs(person) is the quietest person in the subtree at person. Notice because the statements are logically consistent, the graph must be a DAG - a directed graph with no cycles.

Now dfs(person) is either person, or min(dfs(child) for child in person). That is to say, the quietest person in the subtree is either the person itself, or the quietest person in some subtree of a child of person.

We can cache values of dfs(person) as answer[person], when performing our post-order traversal of the graph. That way, we don't repeat work. This technique reduces a quadratic time algorithm down to linear time.

• Time Complexity: $O(N)$, where $N$ is the number of people.
• Space Complexity: $O(N)$, the space used by the answer, and the implicit call stack of dfs.

  ArrayList<Integer>[] graph;

  int[] answer;

  int[] quiet;

  public int[] loudAndRich(int[][] richer, int[] quiet) {

    int N = quiet.length;

    graph = new ArrayList[N];

    answer = new int[N];

    this.quiet = quiet;

    for (int node = 0; node < N; ++node)

      graph[node] = new ArrayList<Integer>();

    for (int[] edge : richer)

      graph[edge[1]].add(edge[0]);

    Arrays.fill(answer, -1);

    for (int node = 0; node < N; ++node)

      dfs(node);

    return answer;

  }

  public int dfs(int node) {

    if (answer[node] == -1) {

      answer[node] = node;

      for (int child : graph[node]) {

        int cand = dfs(child);

        if (quiet[cand] < quiet[answer[node]])

          answer[node] = cand;

      }

    }

    return answer[node];

  }

https://leetcode.com/problems/loud-and-rich/discuss/137987/Java-BFS

    public int[] loudAndRich(int[][] richer, int[] quiet) {
        int n = quiet.length;
        // construct "adjacent list" , record richer people 
        List<List<Integer>> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(new ArrayList<>());
        }
        for (int i = 0; i < richer.length; i++) {
            list.get(richer[i][1]).add(richer[i][0]);
        }
        
        int[] result = new int[n];
        for (int i = 0; i < n; i++) {
            // no one is richer than i
            if (list.get(i).size() == 0) {
                result[i] = i;
                continue;
            }
            // otherwise,do BFS
            result[i] = bfs(list, quiet, i);
        }
        
        return result;
    }
    private int bfs(List<List<Integer>> list, int[] quiet, int index) {
        int result = index;
        int q = Integer.MAX_VALUE; // least quiet
        boolean[] visited = new boolean[quiet.length];
        Queue<Integer> queue = new LinkedList<>();
        visited[index] = true;
        queue.offer(index);
        
        while (!queue.isEmpty()) {
            int curr = queue.poll();
            if (quiet[curr] < q) {
                q = quiet[curr];
                result = curr;
            }
            if (list.get(curr).size() != 0) {
                for (int next : list.get(curr)) {
                    if (visited[next]) {
                        continue;
                    }
                    queue.offer(next);
                    visited[next] = true;
                }
            }
        }
        return result;
    }