Description Link to heading

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.

Solution Link to heading

Update ans when doing topo sort

Code Link to heading 

class Solution {
public:
    vector<int> loudAndRich(vector<vector<int>>& richer, vector<int>& quiet) {
        int n = quiet.size();
        vector<vector<int>> graph(n);
        vector<int> in(n);
        for (auto &vec : richer) {
            graph[vec[0]].push_back(vec[1]);
            ++in[vec[1]];
        }
        std::queue<int> in0;
        vector<int> res(n);
        for (int i = 0; i < n; ++i) {
            if (in[i] == 0) {
                in0.push(i);
            }
            res[i] = i;
        }
        while (!in0.empty()) {
            auto idx = in0.front();
            in0.pop();
            for (auto v : graph[idx]) {
                if (quiet[res[idx]] < quiet[res[v]]) {
                    res[v] = res[idx];
                }
                if (--in[v] == 0) {
                    in0.push(v);
                }
            }
        }
        return res;
    }
};