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# 0851. Loud and Rich

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

## 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 `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.

## ac

```java
```