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Loud and Rich⚓︎

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Description⚓︎

There is a group of n people labeled from 0 to n - 1 where each person has a different amount of money and a different level of quietness.

You are given an array richer where richer[i] = [ai, bi] indicates that ai has more money than bi and an integer array quiet where quiet[i] is the quietness of the i-th person. All the given data in richer are logically correct (i.e., the data will not lead you to a situation where x is richer than y and y is richer than x at the same time).

Return an integer array 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 the 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:
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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 is not 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.

Example 2:

  • Input: richer = [], quiet = [0]
  • Output: [0]

Constraints:

  • n == quiet.length
  • 1 <= n <= 500
  • 0 <= quiet[i] < n
  • All the values of quiet are unique.
  • 0 <= richer.length <= n * (n - 1) / 2
  • 0 <= ai, bi < n
  • ai != bi
  • All the pairs of richer are unique.
  • The observations in richer are all logically consistent.

Solution⚓︎

Topological Sort⚓︎

See reference (Chinese).

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class Solution {
public:
    vector<int> loudAndRich(vector<vector<int>>& richer, vector<int>& quiet) {
        int n = quiet.size();
        vector<vector<int>> edges(n);
        vector<int> inDegree(n, 0);

        for (const auto& edge : richer) {
            edges[edge[0]].push_back(edge[1]);
            inDegree[edge[1]]++;
        }

        queue<int> que;
        for (int i = 0; i < n; i++) {
            if (inDegree[i] == 0)
                que.push(i);
        }

        vector<int> res(n);
        for (int i = 0; i < n; i++)
            res[i] = i;

        while (!que.empty()) {
            int current = que.front();
            que.pop();

            for (auto v : edges[current]) {
                if (quiet[res[current]] < quiet[res[v]])
                    res[v] = res[current];
                if (--inDegree[v] == 0)
                    que.push(v);
            }
        }

        return res;
    }
};

DFS⚓︎

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class Solution {
private:
    void dfs(int u, vector<int>& res, const vector<vector<int>>& edges, const vector<int>& quiet) {
        if (res[u] != -1) return;

        res[u] = u;
        for (int vertex : edges[u]) {
            dfs(vertex, res, edges, quiet);
            if (quiet[res[u]] > quiet[res[vertex]])
                res[u] = res[vertex];
        }
    }

public:
    vector<int> loudAndRich(vector<vector<int>>& richer, vector<int>& quiet) {
        int n = quiet.size();
        vector<vector<int>> edges(n);

        for (const auto& edge : richer)
            edges[edge[1]].push_back(edge[0]);

        vector<int> res(n, -1);
        for (int i = 0; i < n; i++)
            dfs(i, res, edges, quiet);

        return res;
    }
};