README_EN.md

851. Loud and Rich

中文文档

Description

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.

Solutions

Python3

class Solution:
    def loudAndRich(self, richer: List[List[int]], quiet: List[int]) -> List[int]:
        n = len(quiet)
        g = defaultdict(list)
        for a, b in richer:
            g[b].append(a)
        ans = [-1] * n

        def dfs(i):
            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]

        for i in range(n):
            dfs(i)
        return ans

Java

class Solution {
    private Map<Integer, List<Integer>> g;
    private int[] quiet;
    private int[] ans;

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

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

C++

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;
    }
};

Go

func loudAndRich(richer [][]int, quiet []int) []int {
    n := len(quiet)
    ans := make([]int, n)
    g := make([][]int, n)
    for i := 0; i < n; i++ {
        ans[i] = -1
        g[i] = make([]int, 0)
    }
    for _, r := range richer {
        g[r[1]] = append(g[r[1]], r[0])
    }

    var dfs func(i 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 := 0; i < n; i++ {
        dfs(i)
    }
    return ans
}

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