851

851. Loud and Rich (Medium)

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 ith 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: 
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.

Companies:
Amazon

Related Topics:
Array, Depth-First Search, Graph, Topological Sort

Solution 1. Topological Sort (BFS)

// OJ: https://leetcode.com/problems/loud-and-rich/
// Author: github.com/lzl124631x
// Time: O(N + E)
// Space: O(N + E)
class Solution {
public:
    vector<int> loudAndRich(vector<vector<int>>& R, vector<int>& Q) {
        int N = Q.size();
        vector<vector<int>> G(N);
        vector<int> indegree(N);
        for (auto &r : R) {
            int u = r[0], v = r[1];
            G[u].push_back(v);
            indegree[v]++;
        }
        queue<int> q;
        vector<int> ans(N), minQuite(N, INT_MAX);
        for (int i = 0; i < N; ++i) {
            minQuite[i] = Q[i];
            ans[i] = i;
            if (indegree[i] == 0) q.push(i);
        }
        while (q.size()) {
            auto u = q.front();
            q.pop();
            for (int v : G[u]) {
                if (minQuite[u] < minQuite[v]) {
                    ans[v] = ans[u];
                    minQuite[v] = minQuite[u];
                }
                if (--indegree[v]) continue;
                q.push(v);
            }
        }
        return ans;
    }
};

Solution 2. Topological Sort (Post-order DFS)

// OJ: https://leetcode.com/problems/loud-and-rich/
// Author: github.com/lzl124631x
// Time: O(N + E)
// Space: O(N + E)
class Solution {
public:
    vector<int> loudAndRich(vector<vector<int>>& R, vector<int>& Q) {
        int N = Q.size();
        vector<vector<int>> G(N);
        vector<int> ans(N, -1);
        for (auto &r : R) G[r[1]].push_back(r[0]);
        function<int(int)> dfs = [&](int u) {
            if (ans[u] != -1) return ans[u];
            ans[u] = u;
            for (int v : G[u]) {
                int minNode = dfs(v);
                if (Q[minNode] < Q[ans[u]]) ans[u] = minNode;
            }
            return ans[u];
        };
        for (int i = 0; i < N; ++i) dfs(i);
        return ans;
    }
};