# 851. Loud and Rich

## Description

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.

## Solutions

• class Solution {
private List<Integer>[] g;
private int n;
private int[] quiet;
private int[] ans;

public int[] loudAndRich(int[][] richer, int[] quiet) {
n = quiet.length;
this.quiet = quiet;
g = new List[n];
ans = new int[n];
Arrays.fill(ans, -1);
Arrays.setAll(g, k -> new ArrayList<>());
for (var r : richer) {
}
for (int i = 0; i < n; ++i) {
dfs(i);
}
return ans;
}

private void dfs(int i) {
if (ans[i] != -1) {
return;
}
ans[i] = i;
for (int j : g[i]) {
dfs(j);
if (quiet[ans[j]] < quiet[ans[i]]) {
ans[i] = ans[j];
}
}
}
}

• class Solution {
public:
vector<int> loudAndRich(vector<vector<int>>& richer, vector<int>& quiet) {
int n = quiet.size();
vector<vector<int>> g(n);
for (auto& r : richer) {
g[r[1]].push_back(r[0]);
}
vector<int> ans(n, -1);
function<void(int)> dfs = [&](int i) {
if (ans[i] != -1) {
return;
}
ans[i] = i;
for (int j : g[i]) {
dfs(j);
if (quiet[ans[j]] < quiet[ans[i]]) {
ans[i] = ans[j];
}
}
};
for (int i = 0; i < n; ++i) {
dfs(i);
}
return ans;
}
};

• class Solution:
def loudAndRich(self, richer: List[List[int]], quiet: List[int]) -> List[int]:
def dfs(i: int):
if ans[i] != -1:
return
ans[i] = i
for j in g[i]:
dfs(j)
if quiet[ans[j]] < quiet[ans[i]]:
ans[i] = ans[j]

g = defaultdict(list)
for a, b in richer:
g[b].append(a)
n = len(quiet)
ans = [-1] * n
for i in range(n):
dfs(i)
return ans


• func loudAndRich(richer [][]int, quiet []int) []int {
n := len(quiet)
g := make([][]int, n)
ans := make([]int, n)
for i := range g {
ans[i] = -1
}
for _, r := range richer {
a, b := r[0], r[1]
g[b] = append(g[b], a)
}
var dfs func(int)
dfs = func(i int) {
if ans[i] != -1 {
return
}
ans[i] = i
for _, j := range g[i] {
dfs(j)
if quiet[ans[j]] < quiet[ans[i]] {
ans[i] = ans[j]
}
}
}
for i := range ans {
dfs(i)
}
return ans
}

• function loudAndRich(richer: number[][], quiet: number[]): number[] {
const n = quiet.length;
const g: number[][] = new Array(n).fill(0).map(() => []);
for (const [a, b] of richer) {
g[b].push(a);
}
const ans: number[] = new Array(n).fill(-1);
const dfs = (i: number) => {
if (ans[i] != -1) {
return ans;
}
ans[i] = i;
for (const j of g[i]) {
dfs(j);
if (quiet[ans[j]] < quiet[ans[i]]) {
ans[i] = ans[j];
}
}
};
for (let i = 0; i < n; ++i) {
dfs(i);
}
return ans;
}