Loud and rich

Time: O(Q+R); Space: O(Q+R); medium

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.

Constraints:

  • 1 <= quiet.length = N <= 500

  • 0 <= quiet[i] < N, all quiet[i] are different.

  • 0 <= richer.length <= N * (N-1) / 2

  • 0 <= richer[i][j] < N

  • richer[i][0] != richer[i][1]

  • richer[i]’s are all different.

  • The observations in richer are all logically consistent.

[1]:

class Solution1(object):
    """
    Time: O(Q+R)
    Space: O(Q+R)
    """
    def loudAndRich(self, richer, quiet):
        """
        :type richer: List[List[int]]
        :type quiet: List[int]
        :rtype: List[int]
        """
        def dfs(graph, quiet, node, result):
            if result[node] is None:
                result[node] = node
                for nei in graph[node]:
                    smallest_person = dfs(graph, quiet, nei, result)
                    if quiet[smallest_person] < quiet[result[node]]:
                        result[node] = smallest_person
            return result[node]

        graph = [[] for _ in range(len(quiet))]
        for u, v in richer:
            graph[v].append(u)
        result = [None]*len(quiet)

        return list(map(lambda x: dfs(graph, quiet, x, result), range(len(quiet))))

[2]:

s = Solution1()

richer = [[1,0],[2,1],[3,1],[3,7],[4,3],[5,3],[6,3]]
quiet = [3,2,5,4,6,1,7,0]
assert s.loudAndRich(richer, quiet) == [5,5,2,5,4,5,6,7]