README.md

851. 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.

Solutions (Rust)

1. Solution

impl Solution {
    pub fn loud_and_rich(richer: Vec<Vec<i32>>, quiet: Vec<i32>) -> Vec<i32> {
        let n = quiet.len();
        let mut richer_count = vec![0; n];
        let mut poorer_people = vec![vec![]; n];
        let mut people = vec![];
        let mut answer = (0..n as i32).collect::<Vec<_>>();

        for pair in &richer {
            richer_count[pair[1] as usize] += 1;
            poorer_people[pair[0] as usize].push(pair[1] as usize);
        }

        for i in 0..n {
            if richer_count[i] == 0 {
                people.push(i);
            }
        }

        while let Some(x) = people.pop() {
            for &y in &poorer_people[x] {
                richer_count[y] -= 1;
                if richer_count[y] == 0 {
                    people.push(y);
                }
                if quiet[answer[x] as usize] < quiet[answer[y] as usize] {
                    answer[y] = answer[x];
                }
            }
        }

        answer
    }
}