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‎Problemset/0851-Loud and Rich/README.md

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# 851. Loud and Rich
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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.
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You are given an array `richer` 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 `quiet` where `quiet[i]` is the quietness of the <code>i<sup>th</sup></code> 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).
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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`.
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#### Example 1:
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<pre>
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<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]
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<strong>Output:</strong> [5,5,2,5,4,5,6,7]
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<strong>Explanation:</strong>
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answer[0] = 5.
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Person 5 has more money than 3, which has more money than 1, which has more money than 0.
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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.
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answer[7] = 7.
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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.
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The other answers can be filled out with similar reasoning.
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</pre>
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#### Example 2:
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<pre>
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<strong>Input:</strong> richer = [], quiet = [0]
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<strong>Output:</strong> [0]
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</pre>
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#### Constraints:
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* `n == quiet.length`
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* `1 <= n <= 500`
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* `0 <= quiet[i] < n`
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* All the values of `quiet` are **unique**.
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* `0 <= richer.length <= n * (n - 1) / 2`
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* <code>0 <= a<sub>i</sub>, b<sub>i</sub> < n</code>
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* <code>a<sub>i</sub> != b<sub>i</sub></code>
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* All the pairs of `richer` are **unique**.
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* The observations in `richer` are all logically consistent.
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## Solutions (Rust)
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### 1. Solution
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```Rust
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impl Solution {
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pub fn loud_and_rich(richer: Vec<Vec<i32>>, quiet: Vec<i32>) -> Vec<i32> {
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let n = quiet.len();
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let mut richer_count = vec![0; n];
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let mut poorer_people = vec![vec![]; n];
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let mut people = vec![];
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let mut answer = (0..n as i32).collect::<Vec<_>>();
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for pair in &richer {
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richer_count[pair[1] as usize] += 1;
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poorer_people[pair[0] as usize].push(pair[1] as usize);
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}
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for i in 0..n {
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if richer_count[i] == 0 {
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people.push(i);
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}
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}
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while let Some(x) = people.pop() {
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for &y in &poorer_people[x] {
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richer_count[y] -= 1;
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if richer_count[y] == 0 {
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people.push(y);
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}
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if quiet[answer[x] as usize] < quiet[answer[y] as usize] {
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answer[y] = answer[x];
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}
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}
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}
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answer
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}
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}
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```

‎Problemset/0851-Loud and Rich/README_CN.md

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# 851. 喧闹和富有
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有一组 `n` 个人作为实验对象，从 `0` 到 `n - 1` 编号，其中每个人都有不同数目的钱，以及不同程度的安静值（quietness）。为了方便起见，我们将编号为 `x` 的人简称为 "person `x` "。
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给你一个数组 `richer` ，其中 <code>richer[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> 表示 person <code>a<sub>i</sub></code> 比 person <code>b<sub>i</sub></code> 更有钱。另给你一个整数数组 `quiet` ，其中 `quiet[i]` 是 person `i` 的安静值。`richer` 中所给出的数据 逻辑自洽（也就是说，在 person `x` 比 person `y` 更有钱的同时，不会出现 person `y` 比 person `x` 更有钱的情况 ）。
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现在，返回一个整数数组 `answer` 作为答案，其中 `answer[x] = y` 的前提是，在所有拥有的钱肯定不少于 person `x` 的人中，person `y` 是最不安静的人（也就是安静值 `quiet[y]` 最小的人）。
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#### 示例 1:
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<pre>
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<strong>输入:</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]
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<strong>输出:</strong> [5,5,2,5,4,5,6,7]
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<strong>解释:</strong>
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answer[0] = 5，
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person 5 比 person 3 有更多的钱，person 3 比 person 1 有更多的钱，person 1 比 person 0 有更多的钱。
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唯一较为安静（有较低的安静值 quiet[x]）的人是 person 7，
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但是目前还不清楚他是否比 person 0 更有钱。
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answer[7] = 7，
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在所有拥有的钱肯定不少于 person 7 的人中（这可能包括 person 3，4，5，6 以及 7），
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最安静（有较低安静值 quiet[x]）的人是 person 7。
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其他的答案也可以用类似的推理来解释。
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</pre>
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#### 示例 2:
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<pre>
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<strong>输入:</strong> richer = [], quiet = [0]
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<strong>输出:</strong> [0]
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</pre>
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#### 提示:
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* `n == quiet.length`
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* `1 <= n <= 500`
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* `0 <= quiet[i] < n`
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* `quiet` 的所有值 **互不相同**
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* `0 <= richer.length <= n * (n - 1) / 2`
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* <code>0 <= a<sub>i</sub>, b<sub>i</sub> < n</code>
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* <code>a<sub>i</sub> != b<sub>i</sub></code>
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* `richer` 中的所有数对 **互不相同**
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* 对 `richer` 的观察在逻辑上是一致的
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## 题解 (Rust)
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### 1. 题解
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```Rust
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impl Solution {
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pub fn loud_and_rich(richer: Vec<Vec<i32>>, quiet: Vec<i32>) -> Vec<i32> {
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let n = quiet.len();
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let mut richer_count = vec![0; n];
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let mut poorer_people = vec![vec![]; n];
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let mut people = vec![];
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let mut answer = (0..n as i32).collect::<Vec<_>>();
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for pair in &richer {
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richer_count[pair[1] as usize] += 1;
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poorer_people[pair[0] as usize].push(pair[1] as usize);
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}
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for i in 0..n {
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if richer_count[i] == 0 {
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people.push(i);
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}
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}
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while let Some(x) = people.pop() {
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for &y in &poorer_people[x] {
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richer_count[y] -= 1;
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if richer_count[y] == 0 {
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people.push(y);
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}
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if quiet[answer[x] as usize] < quiet[answer[y] as usize] {
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answer[y] = answer[x];
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}
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}
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}
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answer
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}
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}
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```

‎Problemset/0851-Loud and Rich/Solution.rs

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impl Solution {
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pub fn loud_and_rich(richer: Vec<Vec<i32>>, quiet: Vec<i32>) -> Vec<i32> {
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let n = quiet.len();
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let mut richer_count = vec![0; n];
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let mut poorer_people = vec![vec![]; n];
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let mut people = vec![];
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let mut answer = (0..n as i32).collect::<Vec<_>>();
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for pair in &richer {
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richer_count[pair[1] as usize] += 1;
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poorer_people[pair[0] as usize].push(pair[1] as usize);
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}
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for i in 0..n {
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if richer_count[i] == 0 {
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people.push(i);
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}
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}
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while let Some(x) = people.pop() {
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for &y in &poorer_people[x] {
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richer_count[y] -= 1;
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if richer_count[y] == 0 {
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people.push(y);
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}
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if quiet[answer[x] as usize] < quiet[answer[y] as usize] {
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answer[y] = answer[x];
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}
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}
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}
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answer
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}
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}

‎README.md

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[846][846l] |[Hand of Straights][846] |![rs]
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[848][848l] |[Shifting Letters][848] |![rs]
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[849][849l] |[Maximize Distance to Closest Person][849] |![rs]
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[851][851l] |[Loud and Rich][851] |![rs]
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[852][852l] |[Peak Index in a Mountain Array][852] |![rs]
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[853][853l] |[Car Fleet][853] |![rs]
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[856][856l] |[Score of Parentheses][856] |![rs]
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[846]:Problemset/0846-Hand%20of%20Straights/README.md#846-hand-of-straights
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[848]:Problemset/0848-Shifting%20Letters/README.md#848-shifting-letters
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[849]:Problemset/0849-Maximize%20Distance%20to%20Closest%20Person/README.md#849-maximize-distance-to-closest-person
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[851]:Problemset/0851-Loud%20and%20Rich/README.md#851-loud-and-rich
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[852]:Problemset/0852-Peak%20Index%20in%20a%20Mountain%20Array/README.md#852-peak-index-in-a-mountain-array
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[853]:Problemset/0853-Car%20Fleet/README.md#853-car-fleet
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[856]:Problemset/0856-Score%20of%20Parentheses/README.md#856-score-of-parentheses
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[846l]:https://leetcode.com/problems/hand-of-straights/
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[848l]:https://leetcode.com/problems/shifting-letters/
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[849l]:https://leetcode.com/problems/maximize-distance-to-closest-person/
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[851l]:https://leetcode.com/problems/loud-and-rich/
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[852l]:https://leetcode.com/problems/peak-index-in-a-mountain-array/
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[853l]:https://leetcode.com/problems/car-fleet/
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[856l]:https://leetcode.com/problems/score-of-parentheses/

‎README_CN.md

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[846][846l] |[一手顺子][846] |![rs]
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[848][848l] |[字母移位][848] |![rs]
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[849][849l] |[到最近的人的最大距离][849] |![rs]
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[851][851l] |[喧闹和富有][851] |![rs]
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[852][852l] |[山脉数组的峰顶索引][852] |![rs]
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[853][853l] |[车队][853] |![rs]
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[856][856l] |[括号的分数][856] |![rs]
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[846]:Problemset/0846-Hand%20of%20Straights/README_CN.md#846-一手顺子
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[848]:Problemset/0848-Shifting%20Letters/README_CN.md#848-字母移位
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[849]:Problemset/0849-Maximize%20Distance%20to%20Closest%20Person/README_CN.md#849-到最近的人的最大距离
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[851]:Problemset/0851-Loud%20and%20Rich/README_CN.md#851-喧闹和富有
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[852]:Problemset/0852-Peak%20Index%20in%20a%20Mountain%20Array/README_CN.md#852-山脉数组的峰顶索引
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[853]:Problemset/0853-Car%20Fleet/README_CN.md#853-车队
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[856]:Problemset/0856-Score%20of%20Parentheses/README_CN.md#856-括号的分数
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[846l]:https://leetcode.cn/problems/hand-of-straights/
33373339
[848l]:https://leetcode.cn/problems/shifting-letters/
33383340
[849l]:https://leetcode.cn/problems/maximize-distance-to-closest-person/
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[851l]:https://leetcode.cn/problems/loud-and-rich/
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[852l]:https://leetcode.cn/problems/peak-index-in-a-mountain-array/
33403343
[853l]:https://leetcode.cn/problems/car-fleet/
33413344
[856l]:https://leetcode.cn/problems/score-of-parentheses/

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