LeetCode #851 — MEDIUM

Loud and Rich

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

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] = [a_i, b_i] indicates that a_i has more money than b_i and an integer array quiet where quiet[i] is the quietness of the i^th person. All the given data in richer are logically correct (i.e., the data will not lead you to a situation where x is richer than y and y is richer than x at the same time).

Return an integer array 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 the 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 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.

Example 2:

Input: richer = [], quiet = [0]
Output: [0]

Constraints:

n == quiet.length
1 <= n <= 500
0 <= quiet[i] < n
All the values of quiet are unique.
0 <= richer.length <= n * (n - 1) / 2
0 <= a_i, b_i < n
a_i!= b_i
All the pairs of richer are unique.
The observations in richer are all logically consistent.

Step 01

Brute Force Baseline

Problem summary: 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. All the given data in richer are logically correct (i.e., the data will not lead you to a situation where x is richer than y and y is richer than x at the same time). Return an integer array 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 the person x.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Topological Sort

Example 1

[[1,0],[2,1],[3,1],[3,7],[4,3],[5,3],[6,3]]
[3,2,5,4,6,1,7,0]

Example 2

[]
[0]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #851: Loud and Rich
class Solution {
    private List<Integer>[] g;
    private int n;
    private int[] quiet;
    private int[] ans;

    public int[] loudAndRich(int[][] richer, int[] quiet) {
        n = quiet.length;
        this.quiet = quiet;
        g = new List[n];
        ans = new int[n];
        Arrays.fill(ans, -1);
        Arrays.setAll(g, k -> new ArrayList<>());
        for (var r : richer) {
            g[r[1]].add(r[0]);
        }
        for (int i = 0; i < n; ++i) {
            dfs(i);
        }
        return ans;
    }

    private void dfs(int i) {
        if (ans[i] != -1) {
            return;
        }
        ans[i] = i;
        for (int j : g[i]) {
            dfs(j);
            if (quiet[ans[j]] < quiet[ans[i]]) {
                ans[i] = ans[j];
            }
        }
    }
}

// Accepted solution for LeetCode #851: Loud and Rich
func loudAndRich(richer [][]int, quiet []int) []int {
	n := len(quiet)
	g := make([][]int, n)
	ans := make([]int, n)
	for i := range g {
		ans[i] = -1
	}
	for _, r := range richer {
		a, b := r[0], r[1]
		g[b] = append(g[b], a)
	}
	var dfs func(int)
	dfs = func(i int) {
		if ans[i] != -1 {
			return
		}
		ans[i] = i
		for _, j := range g[i] {
			dfs(j)
			if quiet[ans[j]] < quiet[ans[i]] {
				ans[i] = ans[j]
			}
		}
	}
	for i := range ans {
		dfs(i)
	}
	return ans
}

# Accepted solution for LeetCode #851: Loud and Rich
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

// Accepted solution for LeetCode #851: Loud and Rich
struct Solution;
use std::collections::HashSet;

impl Solution {
    fn loud_and_rich(richer: Vec<Vec<i32>>, quiet: Vec<i32>) -> Vec<i32> {
        let n = quiet.len();
        let mut graph: Vec<HashSet<usize>> = vec![HashSet::new(); n];
        for e in richer {
            let u = e[0] as usize;
            let v = e[1] as usize;
            graph[v].insert(u);
        }
        let mut res = vec![n; n];
        for i in 0..n {
            Self::dfs(i, &mut res, &graph, &quiet, n);
        }
        res.into_iter().map(|x| x as i32).collect()
    }
    fn dfs(
        u: usize,
        stack: &mut Vec<usize>,
        graph: &[HashSet<usize>],
        quiet: &[i32],
        n: usize,
    ) -> usize {
        if stack[u] == n {
            stack[u] = u;
            for &v in &graph[u] {
                let w = Self::dfs(v, stack, graph, quiet, n);
                if quiet[w] < quiet[stack[u]] {
                    stack[u] = w;
                }
            }
        }
        stack[u]
    }
}

#[test]
fn test() {
    let richer = vec_vec_i32![[1, 0], [2, 1], [3, 1], [3, 7], [4, 3], [5, 3], [6, 3]];
    let quiet = vec![3, 2, 5, 4, 6, 1, 7, 0];
    let res = vec![5, 5, 2, 5, 4, 5, 6, 7];
    assert_eq!(Solution::loud_and_rich(richer, quiet), res);
}

// Accepted solution for LeetCode #851: Loud and Rich
function loudAndRich(richer: number[][], quiet: number[]): number[] {
    const n = quiet.length;
    const g: number[][] = new Array(n).fill(0).map(() => []);
    for (const [a, b] of richer) {
        g[b].push(a);
    }
    const ans: number[] = new Array(n).fill(-1);
    const dfs = (i: number) => {
        if (ans[i] != -1) {
            return ans;
        }
        ans[i] = i;
        for (const j of g[i]) {
            dfs(j);
            if (quiet[ans[j]] < quiet[ans[i]]) {
                ans[i] = ans[j];
            }
        }
    };
    for (let i = 0; i < n; ++i) {
        dfs(i);
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(V + E)

Space

O(V + E)

Approach Breakdown

REPEATED DFS

O(V × E) time

O(V) space

Repeatedly find a vertex with no incoming edges, remove it and its outgoing edges, and repeat. Finding the zero-in-degree vertex scans all V vertices, and we do this V times. Removing edges touches E edges total. Without an in-degree array, this gives O(V × E).

TOPOLOGICAL SORT

O(V + E) time

O(V + E) space

Build an adjacency list (O(V + E)), then either do Kahn's BFS (process each vertex once + each edge once) or DFS (visit each vertex once + each edge once). Both are O(V + E). Space includes the adjacency list (O(V + E)) plus the in-degree array or visited set (O(V)).

Shortcut: Process each vertex once + each edge once → O(V + E). Same as BFS/DFS on a graph.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.