# Loud And Rich Problem

## Description

LeetCode Problem 851.

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] = [a_i, b_i] indicates that a_i has more money than b_i and an integer array quiet where quiet[i] is the quietness of the i^th 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:

``````1
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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:

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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 <= a_i, b_i < n
• a_i != b_i
• All the pairs of richer are unique.
• The observations in richer are all logically consistent.

``````1
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class Solution {
public:
vector<int> loudAndRich(vector<vector<int>>& richer, vector<int>& quiet) {
int n = quiet.size();

vector<int> indegree(n, 0);
for (auto& rich : richer) {
indegree[rich[1]] ++;
}

queue<int> q;

for (int i = 0; i < n; i++) {
if (indegree[i] == 0)
q.push(i);
}

while (q.size()) {
int node = q.front();
q.pop();

}

}
}
}
};
``````