851. Loud and Rich

In a group of N people (labelled 0, 1, 2, ..., N-1), each person has different amounts of money, and different levels of quietness.

For convenience, we'll call the person with label x, simply "person x".

We'll say that richer[i] = [x, y] if person x definitely has more money than person y.  Note that richer may only be a subset of valid observations.

Also, we'll say quiet[x] = q if person x has quietness q.

Now, return 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 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 isn't 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.

Note:

  1. 1 <= quiet.length = N <= 500
  2. 0 <= quiet[i] < N, all quiet[i] are different.
  3. 0 <= richer.length <= N * (N-1) / 2
  4. 0 <= richer[i][j] < N
  5. richer[i][0] != richer[i][1]
  6. richer[i]'s are all different.
  7. The observations in richer are all logically consistent.

Rust Solution

struct Solution;
use std::collections::HashSet;

impl Solution {
    fn loud_and_rich(richer: Vec<Vec<i32>>, quiet: Vec<i32>) -> Vec<i32> {
        let n = quiet.len();
        let mut graph: Vec<HashSet<usize>> = vec![HashSet::new(); n];
        for e in richer {
            let u = e[0] as usize;
            let v = e[1] as usize;
            graph[v].insert(u);
        }
        let mut res = vec![n; n];
        for i in 0..n {
            Self::dfs(i, &mut res, &graph, &quiet, n);
        }
        res.into_iter().map(|x| x as i32).collect()
    }
    fn dfs(
        u: usize,
        stack: &mut Vec<usize>,
        graph: &[HashSet<usize>],
        quiet: &[i32],
        n: usize,
    ) -> usize {
        if stack[u] == n {
            stack[u] = u;
            for &v in &graph[u] {
                let w = Self::dfs(v, stack, graph, quiet, n);
                if quiet[w] < quiet[stack[u]] {
                    stack[u] = w;
                }
            }
        }
        stack[u]
    }
}

#[test]
fn test() {
    let richer = vec_vec_i32![[1, 0], [2, 1], [3, 1], [3, 7], [4, 3], [5, 3], [6, 3]];
    let quiet = vec![3, 2, 5, 4, 6, 1, 7, 0];
    let res = vec![5, 5, 2, 5, 4, 5, 6, 7];
    assert_eq!(Solution::loud_and_rich(richer, quiet), res);
}

Having problems with this solution? Click here to submit an issue on github.