# 851. Loud and Rich

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:
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.

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.

``````class Solution {
public:
vector<int> loudAndRich\(vector<vector<int>>& richer, vector<int>& quiet) {
vector<int> res(quiet.size(), -1);
unordered_map<int, vector<int>> findRicher;
for (auto a : richer) findRicher[a[1]].push_back(a[0]);
for (int i = 0; i < quiet.size(); ++i) {
helper(findRicher, quiet, i, res);
}
return res;
}
int helper(unordered_map<int, vector<int>>& findRicher, vector<int>& quiet, int i, vector<int>& res) {
if (res[i] > 0) return res[i];
res[i] = i;
for (int j : findRicher[i]) {
if (quiet[res[i]] > quiet[helper(findRicher, quiet, j, res)]) {
res[i] = res[js];
}
}
return res[i];
}
};
``````

``````class Solution {
public:
vector<int> loudAndRich(vector<vector<int>>& richer, vector<int>& quiet) {
int n = quiet.size();
vector<int> res(n, -1), inDegree(n);
unordered_map<int, vector<int>> findPoorer;
queue<int> q;
for (auto a : richer) {
findPoorer[a[0]].push_back(a[1]);
++inDegree[a[1]];
}
for (int i = 0; i < quiet.size(); ++i) {
if (inDegree[i] == 0) q.push(i);
res[i] = i;
}
while (!q.empty()) {
int cur = q.front(); q.pop();
for (int next : findPoorer[cur]) {
if (quiet[res[next]] > quiet[res[cur]]) res[next] = res[cur];
if (--inDegree[next] == 0) q.push(next);
}
}
return res;
}
};
``````

Github 同步地址:

https://github.com/grandyang/leetcode/issues/851

Course Schedule

Course Schedule II

Minimum Height Trees

Reconstruct Itinerary

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

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

https://leetcode.com/problems/loud-and-rich/discuss/138088/C%2B%2B-with-topological-sorting

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