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Leetcode 851. Loud and Rich

Problem Statement:

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 loudness. You will be given two lists: richer and quiet.

• richer[i] = [a, b] means a person a is richer than a person b.
• quiet[i] is the loudness of the i-th person: the larger the value, the quieter the person is.

Return an integer array answer where answer[i] = x means that person x is the quietest person (i.e. has the most amount of money) among all people who are richer than or equal to i.

Clarifying Questions:

1. Can there be multiple people with the same loudness level?
   • Yes, the quiet list can have multiple people with the same quietness value.
2. Are there constraints on the size of the richer list or the quiet array?
   • Yes, typical LeetCode constraints apply, but generally assume they are reasonable enough for a solution to perform adequately within linearithmic or quadratic complexity.

Strategy:

1. Graph Representation:
   • Represent the money relationships using a directed graph where a directed edge from a to b indicates a is richer than b.
2. Topological Sorting:
   • Use topological sorting to determine the order of processing based on dependencies. Since richer people need to be processed before poorer ones, this allows us to utilize known richer relationships effectively.
3. DFS for Minimum Loudness:
   • Use Depth First Search (DFS) to propagate the quietness information through the graph.

Code:

import java.util.*;

public class Solution {
    public int[] loudAndRich(int[][] richer, int[] quiet) {
        int n = quiet.length;
        List<Integer>[] graph = new List[n];
        for (int i = 0; i < n; i++) {
            graph[i] = new ArrayList<>();
        }

        for (int[] pair : richer) {
            graph[pair[1]].add(pair[0]);
        }

        int[] answer = new int[n];
        Arrays.fill(answer, -1);

        for (int i = 0; i < n; i++) {
            dfs(i, graph, quiet, answer);
        }

        return answer;
    }

    private int dfs(int node, List<Integer>[] graph, int[] quiet, int[] answer) {
        if (answer[node] != -1) {
            return answer[node];
        }

        answer[node] = node;

        for (int neighbor : graph[node]) {
            int cand = dfs(neighbor, graph, quiet, answer);
            if (quiet[cand] < quiet[answer[node]]) {
                answer[node] = cand;
            }
        }

        return answer[node];
    }

    public static void main(String[] args) {
        Solution sol = new Solution();
        int[][] richer = // use example above
        int[] quiet = {2, 3, 1, 0};
        System.out.println(Arrays.toString(sol.loudAndRich(richer, quiet))); // Output: [0, 1, 1, 3]
    }
}

Time Complexity:

• Construction of Graph: O(E) where E is the number of edges in the richer relationships.
• DFS Traversal: Each node is visited once, and each edge is considered once leading to O(V + E) time complexity, where V is the number of vertices (people).

Thus, the overall time complexity remains O(V + E). This is efficient given the problem constraints.

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