How do you solve Loud and Rich on LeetCode?

Loud and Rich (LeetCode #851) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n + e) time complexity and O(n) space complexity. Key insight: Understanding the relationship between quietness and wealth is crucial.

What is the brute force solution for Loud and Rich?

The brute force approach for Loud and Rich is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we check each person against every other person to determine who is richer and quieter. This is straightforward but inefficient, as it requires multiple comparisons for each person. The optimal Optimal Solution approach improves this to O(n + e).

What algorithm or data structure is used to solve Loud and Rich?

Loud and Rich uses the following concepts: Array, Depth-First Search, Graph Theory, Topological Sort. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Loud and Rich?

Clearly explain your thought process while coding. Consider edge cases, such as when the richer array is empty. Practice graph traversal techniques to become comfortable with them.

What are common mistakes when solving Loud and Rich?

Not considering all richer relationships for each person. Failing to use memoization in DFS, leading to redundant calculations.

#851

Loud and Rich

Medium

ArrayDepth-First SearchGraph TheoryTopological SortHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n + e)
Space	O(1)	O(n)

💡

Intuition

Time O(n + e)Space O(n)

The optimal approach uses a graph representation of the relationships and a depth-first search (DFS) to propagate the quietness information. This allows us to efficiently find the quietest person among those who are richer.

⚙️

Algorithm

3 steps

1Step 1: Build a graph where each person points to those they are richer than.
2Step 2: Use DFS to explore each person, keeping track of the quietest person found in the path.
3Step 3: Store results in the answer array, ensuring that we only update if we find a quieter person.

solution.py19 lines

1def loudAndRich(richer, quiet):
2    from collections import defaultdict
3    n = len(quiet)
4    graph = defaultdict(list)
5    for a, b in richer:
6        graph[a].append(b)
7    answer = [-1] * n
8    def dfs(person):
9        if answer[person] != -1:
10            return answer[person]
11        answer[person] = person
12        for richer_person in graph[person]:
13            candidate = dfs(richer_person)
14            if quiet[candidate] < quiet[answer[person]]:
15                answer[person] = candidate
16        return answer[person]
17    for i in range(n):
18        dfs(i)
19    return answer

ℹ

Complexity note: The time complexity is O(n + e), where n is the number of people and e is the number of richer relationships. This is efficient because we only visit each person and each relationship once. The space complexity is O(n) for storing the graph and the answer array.

1Understanding the relationship between quietness and wealth is crucial.
2Graph traversal techniques like DFS can simplify complex relationships.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.