## DEV Community

Abhishek Chaudhary

Posted on

# 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:
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.
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]
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: