Loud and Rich

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:

class Solution:
    def loudAndRich(self, richer: List[List[int]], quiet: List[int]) -> List[int]:
        n = len(quiet)
        graph = {}
        for a, b in richer:
            graph[b] = graph.get(b, []) + [a]
        answer = [0] * n
        for beg in range(n):
            paths = [beg]
            visited = {beg}
            while len(paths) > 0:
                curr = paths.pop()
                for j in graph.get(curr, []):
                    if j not in visited:
                        visited.add(j)
                        paths.append(j)
            answer[beg] = min(visited, key = lambda p: quiet[p])
        return answer