What is the time complexity of Loud and Rich?

The optimal solution has a time complexity of O(n + e).

What is the space complexity of Loud and Rich?

The space complexity is O(n + e).

What data structures are used to solve Loud and Rich?

This problem uses Array, Depth-First Search, Graph, Topological Sort. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Loud and Rich?

Insight 1: The problem can be modeled as a directed graph where an edge from `a` to `b` means `a` is richer than `b`. This allows us to use graph traversal techniques to find all people richer than a given person. Insight 2: Dynamic Programming is useful here by caching the results of the least quiet person at least as rich as person i to avoid redundant computations. DFS helps achieve this.

Which companies ask LeetCode 851. Loud and Rich in interviews?

This problem is frequently asked at PayPal.

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Loud and Rich - LeetCode 851 Solution

Loud and Rich - Complete Solution Guide

Loud and Rich is LeetCode problem 851, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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 i

Detailed Explanation

The problem asks us to determine, for each person in a group, who is the least quiet person that is at least as rich as them. We are given two arrays: `richer` and `quiet`. The `richer` array is a list of pairs `[a, b]`, indicating that person `a` is richer than person `b`. The `quiet` array contains the quietness level of each person. The goal is to return an array `answer` where `answer[x] = y` means person `y` is the least quiet among all people who are at least as rich as person `x`. The quietness levels are unique, and the richness data is consistent (no cycles).

Solution Approach

The solution uses Depth-First Search (DFS) to traverse the implied directed graph. For each person, it explores all people richer than them and identifies the person with the minimum quietness. The DFS employs memoization to avoid recomputing the least quiet person for individuals whose results have already been calculated. This approach efficiently determines the required answer for each person in the group.

Step-by-Step Algorithm

Step 1: Build the adjacency list representation of the 'richer' relationships. The list `adj[i]` will contain the people who are richer than person `i`.
Step 2: Initialize an `answer` array of the same size as `quiet`, with each element initialized to -1. This array will store the least quiet person for each index.
Step 3: Define a recursive DFS function. The DFS function takes a person as input and returns the index of the least quiet person at least as rich as the input.
Step 4: Inside the DFS function, check if the result for the current person is already computed and stored in the `answer` array. If so, return the stored value.
Step 5: If not, initialize the answer for the current person to themselves. This is the base case: the quietest person at least as rich as themselves could be themselves.
Step 6: Iterate through all the richer people. For each richer person, call the DFS function recursively to get the least quiet person at least as rich as that richer person.
Step 7: Update the answer for the current person if the quietness of the result for the richer person is less than the quietness of the current answer.
Step 8: Store the result in the answer array to avoid recomputation. Return the result.
Step 9: Iterate through each person in `quiet` and invoke the DFS function to populate `answer`.
Step 10: Return the `answer` array.

Key Insights

Insight 1: The problem can be modeled as a directed graph where an edge from `a` to `b` means `a` is richer than `b`. This allows us to use graph traversal techniques to find all people richer than a given person.
Insight 2: Dynamic Programming is useful here by caching the results of the least quiet person at least as rich as person i to avoid redundant computations. DFS helps achieve this.
Insight 3: We need to account for transitivity of wealth. If A is richer than B, and B is richer than C, then A is also richer than C. The DFS takes care of this.
Insight 4: Initialize answer array with -1. This array acts as memoization storage. If answer[i] is not -1, it means we have already calculated and cached our answer.

Complexity Analysis

Time Complexity: O(n + e)

Space Complexity: O(n + e)

Topics

This problem involves: Array, Depth-First Search, Graph, Topological Sort.

Companies

Asked at: PayPal.

Loud and Rich | LeetCode 851

Medium

Array Depth-First Search Graph Topological Sort

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

Solution Code

Available in:

import java.util.ArrayList;
import java.util.List;

class Solution {
    public int[] loudAndRich(int[][] richer, int[] quiet) {
        int n = quiet.length;
        List<List<Integer>> adj = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            adj.add(new ArrayList<>());
        }
        for (int[] edge : richer) {
            int r = edge[0];
            int p = edge[1];
            adj.get(p).add(r);
        }

        int[] answer = new int[n];
        for (int i = 0; i < n; i++) {
            answer[i] = -1;
        }

        function dfs = new function(adj, quiet, answer);

        for (int i = 0; i < n; i++) {
            dfs.apply(i);
        }

        return answer;
    }

    class function {
        private List<List<Integer>> adj;
        private int[] quiet;
        private int[] answer;

        public function(List<List<Integer>> adj, int[] quiet, int[] answer) {
            this.adj = adj;
            this.quiet = quiet;
            this.answer = answer;
        }

        public void apply(int node) {
            if (answer[node] != -1) {
                return;
            }

            answer[node] = node;

            for (int neighbor : adj.get(node)) {
                apply(neighbor);
                if (quiet[answer[neighbor]] < quiet[answer[node]]) {
                    answer[node] = answer[neighbor];
                }
            }
        }
    }
}

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Problem Info

Description

Problem 851: Loud and Rich - Detailed explanation and solution approaches.

Topics

Array Depth-First Search Graph Topological Sort

Companies

PayPal

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:851

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

10 steps

1
Step 1: Build the adjacency list representation of the 'richer' relationships. The list `adj[i]` will contain the people who are richer than person `i`.
2
Step 2: Initialize an `answer` array of the same size as `quiet`, with each element initialized to -1. This array will store the least quiet person for each index.
3
Step 3: Define a recursive DFS function. The DFS function takes a person as input and returns the index of the least quiet person at least as rich as the input.
4
Step 4: Inside the DFS function, check if the result for the current person is already computed and stored in the `answer` array. If so, return the stored value.
5
Step 5: If not, initialize the answer for the current person to themselves. This is the base case: the quietest person at least as rich as themselves could be themselves.
6
Step 6: Iterate through all the richer people. For each richer person, call the DFS function recursively to get the least quiet person at least as rich as that richer person.
7
Step 7: Update the answer for the current person if the quietness of the result for the richer person is less than the quietness of the current answer.
8
Step 8: Store the result in the answer array to avoid recomputation. Return the result.
9
Step 9: Iterate through each person in `quiet` and invoke the DFS function to populate `answer`.
10
Step 10: Return the `answer` array.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n + e)

💾

Space Complexity

O(n + e)

📝

Why?

The adjacency list takes O(e) space. The answer array takes O(n) space. The DFS visits each node at most once due to memoization, and for each node, it iterates through its neighbors (richer people). Thus, the time complexity is O(n + e).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Topological Sort

Alternatively, you could use topological sort on the richer relationships. After obtaining a topological order, iterate through the people in reverse topological order. For each person, update the answer based on people who are poorer than them. This ensures that the answer for the richer people are already computed.

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

Trade-offs: Topological sort might be slightly more complex to implement, but it can be useful in cases where the graph needs to be processed in a specific order. The main DFS solution is often easier to understand.

✨

Key Insights

1Insight 1: The problem can be modeled as a directed graph where an edge from `a` to `b` means `a` is richer than `b`. This allows us to use graph traversal techniques to find all people richer than a given person.
2Insight 2: Dynamic Programming is useful here by caching the results of the least quiet person at least as rich as person i to avoid redundant computations. DFS helps achieve this.
3Insight 3: We need to account for transitivity of wealth. If A is richer than B, and B is richer than C, then A is also richer than C. The DFS takes care of this.
4Insight 4: Initialize answer array with -1. This array acts as memoization storage. If answer[i] is not -1, it means we have already calculated and cached our answer.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Graph

Nodes connected by edges. Like a social network or city map!

Edge Cases

Edge case 1: Empty richer array (no rich relationships). In this case, the answer for each person is themselves, because no one is richer than them, so the quietest person who is at least as rich is themselves. The provided solution handles this correctly because the initial answer is themselves before exploring richer people.
Edge case 2: A single person in the input. The richer array would be empty, and the quiet array would contain a single element. The provided solution correctly returns an array containing just that single person's index.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not using memoization. Without memoization, the DFS could revisit the same nodes multiple times, resulting in a time complexity of O(n*e) in the worst case.

✗Common mistake 2: Forgetting to initialize the answer for the current person to themselves before exploring richer people. This can lead to incorrect results when there are no richer people for a given person.

✗Common mistake 3: Not handling the edge case where the input list of richer people is empty.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Social network influence analysis. Identifying the most influential (analogous to quietest) person who has at least the same level of network connections (analogous to richer) as a given individual.

🏦Real-world scenario 2: Supply chain analysis. Determining the most reliable (analogous to quietest) supplier who can meet the same or higher level of demand (analogous to richer) as a given supplier.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 10, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 14

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 7

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Loud and Rich"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether depth-first search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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