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Kotlin-based LeetCode algorithm problem solutions, regularly updated
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.