<p>There is a group of <code>n</code> people labeled from <code>0</code> to <code>n - 1</code> where each person has a different amount of money and a different level of quietness.</p>

<p>You are given an array <code>richer</code> where <code>richer[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> indicates that <code>a<sub>i</sub></code> has more money than <code>b<sub>i</sub></code> and an integer array <code>quiet</code> where <code>quiet[i]</code> is the quietness of the <code>i<sup>th</sup></code> person. All the given data in richer are <strong>logically correct</strong> (i.e., the data will not lead you to a situation where <code>x</code> is richer than <code>y</code> and <code>y</code> is richer than <code>x</code> at the same time).</p>

<p>Return <em>an integer array </em><code>answer</code><em> where </em><code>answer[x] = y</code><em> if </em><code>y</code><em> is the least quiet person (that is, the person </em><code>y</code><em> with the smallest value of </em><code>quiet[y]</code><em>) among all people who definitely have equal to or more money than the person </em><code>x</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> richer = [[1,0],[2,1],[3,1],[3,7],[4,3],[5,3],[6,3]], quiet = [3,2,5,4,6,1,7,0]
<strong>Output:</strong> [5,5,2,5,4,5,6,7]
<strong>Explanation:</strong> 
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.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> richer = [], quiet = [0]
<strong>Output:</strong> [0]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == quiet.length</code></li>
	<li><code>1 &lt;= n &lt;= 500</code></li>
	<li><code>0 &lt;= quiet[i] &lt; n</code></li>
	<li>All the values of <code>quiet</code> are <strong>unique</strong>.</li>
	<li><code>0 &lt;= richer.length &lt;= n * (n - 1) / 2</code></li>
	<li><code>0 &lt;= a<sub>i</sub>, b<sub>i</sub> &lt; n</code></li>
	<li><code>a<sub>i </sub>!= b<sub>i</sub></code></li>
	<li>All the pairs of <code>richer</code> are <strong>unique</strong>.</li>
	<li>The observations in <code>richer</code> are all logically consistent.</li>
</ul>

from collections import defaultdict
from functools import lru_cache

class Solution:
    cache = {}
    def findRicherPeople(person: int, graph: Dict) -> List[int]:
        if person in Solution.cache:
            return Solution.cache[person]
        retval = set([person])
        if person not in graph:
            return [person]
        for child in graph[person]:
            retval.update(Solution.findRicherPeople(child, graph))
        Solution.cache[person] = list(retval)
        return list(retval)

    def loudAndRich(self, richer: List[List[int]], quiet: List[int]) -> List[int]:
        Solution.cache = {}
        graph = defaultdict(list)
        for a, b in richer:
            graph[b].append(a)
        retval = []
        for person in range(len(quiet)):
            richerPeople = Solution.findRicherPeople(person, graph)
            retval.append(min(richerPeople, key=lambda x: quiet[x]))
        return retval
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