LeetCode 851: Loud and Rich Solution

Master LeetCode problem 851 (Loud and Rich), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

851. Loud and Rich

Medium

Array Depth-First Search Graph Topological Sort

Problem Explanation

Explanation

To solve this problem, we can build a graph where each node represents a person and each directed edge represents the relationship that a person has more money than another person. Then, we can perform a depth-first search (DFS) starting from each person to find the person with the minimum quietness among those who have equal to or more money.

Construct a directed graph based on the richer array.
Initialize an array answer with -1 for all people.
Perform a DFS for each person to find the quietest person with equal to or more money.

Time Complexity: O(n^2)
Space Complexity: O(n)

Solution Code

class Solution {
    public int[] loudAndRich(int[][] richer, int[] quiet) {
        int n = quiet.length;
        List<List<Integer>> graph = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            graph.add(new ArrayList<>());
        }
        
        for (int[] pair : richer) {
            graph.get(pair[1]).add(pair[0]);
        }
        
        int[] answer = new int[n];
        Arrays.fill(answer, -1);
        
        for (int i = 0; i < n; i++) {
            if (answer[i] == -1) {
                dfs(graph, quiet, answer, i);
            }
        }
        
        return answer;
    }
    
    private int dfs(List<List<Integer>> graph, int[] quiet, int[] answer, int person) {
        if (answer[person] != -1) {
            return answer[person];
        }
        
        answer[person] = person;
        for (int richerPerson : graph.get(person)) {
            int candidate = dfs(graph, quiet, answer, richerPerson);
            if (quiet[candidate] < quiet[answer[person]]) {
                answer[person] = candidate;
            }
        }
        
        return answer[person];
    }
}

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