Leetcode #851: Loud and Rich
In this guide, we solve Leetcode #851 Loud and Rich in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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 quietness. You are given an array richer where richer[i] = [ai, bi] indicates that ai has more money than bi and an integer array quiet where quiet[i] is the quietness of the ith person.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Depth-First Search, Graph, Topological Sort, Array
Intuition
The data forms a graph, so we should explore nodes and edges systematically.
A traversal ensures we visit each node once while maintaining the needed state.
Approach
Build an adjacency list and traverse with BFS or DFS.
Aggregate results as you visit nodes.
Steps:
- Build the graph.
- Traverse with BFS/DFS.
- Accumulate the required output.
Example
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 is not 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.
Python Solution
class Solution:
def loudAndRich(self, richer: List[List[int]], quiet: List[int]) -> List[int]:
def dfs(i: int):
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]
g = defaultdict(list)
for a, b in richer:
g[b].append(a)
n = len(quiet)
ans = [-1] * n
for i in range(n):
dfs(i)
return ans
Complexity
The time complexity is O(V+E). The space complexity is O(V).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.