What is the time complexity of People Whose List of Favorite Companies Is Not a Subset of Another List?

The optimal solution has a time complexity of O(n^2 * m).

What is the space complexity of People Whose List of Favorite Companies Is Not a Subset of Another List?

The space complexity is O(n * m).

What data structures are used to solve People Whose List of Favorite Companies Is Not a Subset of Another List?

This problem uses Array, Hash Table, String. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving People Whose List of Favorite Companies Is Not a Subset of Another List?

Insight 1: The core operation is checking if one list is a subset of another. Converting lists to sets allows for efficient subset checking. Insight 2: A naive approach involves comparing each list against every other list. It's important to ensure each list is compared with every other list *excluding itself*.

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People Whose List of Favorite Companies Is Not a Subset of Another List - LeetCode 1452 Solution

Q: Which companies ask LeetCode 1452. People Whose List of Favorite Companies Is Not a Subset of Another List in interviews?

This problem is frequently asked at Datadog.

People Whose List of Favorite Companies Is Not a Subset of Another List - Complete Solution Guide

People Whose List of Favorite Companies Is Not a Subset of Another List is LeetCode problem 1452, 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

Given the array favoriteCompanies where favoriteCompanies[i] is the list of favorites companies for the ith person ( indexed from 0 ). Return the indices of people whose list of favorite companies is not a subset of any other list of favorites companies . You must return the indices in increasing order. Example 1: Input: favoriteCompanies = [["leetcode","google","facebook"],["google","microsoft"],["google","facebook"],["google"],["amazon"]] Output: [0,1,4] Explanation: Person with index=2 has fa

Detailed Explanation

The problem requires us to identify people whose lists of favorite companies are not subsets of any other person's list of favorite companies. We are given an array `favoriteCompanies` where each element `favoriteCompanies[i]` is a list of strings representing the favorite companies of the i-th person. The goal is to return a list of indices (in increasing order) of those people whose company lists are not subsets of any other lists. The constraints specify the size of the input array and the length of the strings involved.

Solution Approach

The provided solutions utilize a straightforward approach: iterate through each person's list of favorite companies and, for each person, check if their list is a subset of any other person's list. If a person's list is not a subset of any other list, their index is added to the result list. To efficiently check for subsets, the lists of companies are converted into sets. The `issubset()` method (or equivalent in other languages) is used for the subset check.

Step-by-Step Algorithm

Step 1: Convert each person's list of favorite companies into a set. This allows for efficient subset checking using the `issubset()` method or `containsAll()` in Java or simulating it in C and C++.
Step 2: Iterate through each person's set (outer loop).
Step 3: For each person, iterate through all other people's sets (inner loop).
Step 4: Inside the inner loop, check if the current person's set is a subset of the other person's set. If it is, mark the current person as being a subset.
Step 5: If, after iterating through all other people, the current person is not marked as a subset, add their index to the result list.
Step 6: Sort the result list and return it.

Key Insights

Insight 1: The core operation is checking if one list is a subset of another. Converting lists to sets allows for efficient subset checking.
Insight 2: A naive approach involves comparing each list against every other list. It's important to ensure each list is compared with every other list *excluding itself*.
Insight 3: The problem explicitly mentions that the output should be a sorted list of indices.

Complexity Analysis

Time Complexity: O(n^2 * m)

Space Complexity: O(n * m)

Topics

This problem involves: Array, Hash Table, String.

Companies

Asked at: Datadog.

People Whose List of Favorite Companies Is Not a Subset of Another List | LeetCode 1452

Medium

Array Hash Table String

Time: O(n^2 * m)Space: O(n * m)

Solution Code

Available in:

import java.util.*;

class Solution {
    public List<Integer> peopleIndexes(List<List<String>> favoriteCompanies) {
        int n = favoriteCompanies.size();
        List<Set<String>> companySets = new ArrayList<>();
        for (List<String> companies : favoriteCompanies) {
            companySets.add(new HashSet<>(companies));
        }

        List<Integer> result = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            boolean isSubset = false;
            for (int j = 0; j < n; j++) {
                if (i == j) {
                    continue;
                }

                if (companySets.get(j).containsAll(companySets.get(i))) {
                    isSubset = true;
                    break;
                }
            }

            if (!isSubset) {
                result.add(i);
            }
        }

        return result;
    }
}

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Problem Info

Description

Problem 1452: People Whose List of Favorite Companies Is Not a Subset of Another List - Detailed explanation and solution approaches.

Topics

Array Hash Table String

Companies

Datadog

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1452

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

6 steps

1
Step 1: Convert each person's list of favorite companies into a set. This allows for efficient subset checking using the `issubset()` method or `containsAll()` in Java or simulating it in C and C++.
2
Step 2: Iterate through each person's set (outer loop).
3
Step 3: For each person, iterate through all other people's sets (inner loop).
4
Step 4: Inside the inner loop, check if the current person's set is a subset of the other person's set. If it is, mark the current person as being a subset.
5
Step 5: If, after iterating through all other people, the current person is not marked as a subset, add their index to the result list.
6
Step 6: Sort the result list and return it.

💡

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^2 * m)

💾

Space Complexity

O(n * m)

📝

Why?

The time complexity is dominated by the nested loops and the subset check. The outer loop runs `n` times, and for each iteration, the inner loop runs `n` times. The `issubset` operation in Python, `containsAll` in Java and the equivalent implementations in C++ and C, on average, takes O(m) to complete, since it needs to check the presence of each element of set A within Set B. Therefore, the time complexity for subset checking is O(n * n * m) which simplifies to O(n^2 * m). The space complexity comes from creating the sets. We are creating n sets, each with at most m elements giving space complexity of O(n * m)

📖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

Sorting and String Comparison

Sort each company list alphabetically. Then, for each pair of lists, use string comparison to check if one is a subset of the other. Requires custom subset checking logic.

Time: O(n^2 * m * log(m)) where n is number of people and m is average number of companies per person (due to sorting each list and comparing lists)Space: O(1) if sorting in place, O(m) if creating sorted copies

Trade-offs: Avoids creating sets, but relies on string comparisons, which can be less efficient than set operations for larger lists. This approach is less readable compared to using sets.

Trie Data Structure

A trie can be built where each path from root to leaf represents a company. Then, checking for subset relationship involves traversing the trie. This may offer benefits when there are many common prefixes among company names.

Time: Depends on Trie construction and lookup, potentially O(n*m*k) where k is avg string lengthSpace: O(total_length_of_all_strings)

Trade-offs: More complex to implement and potentially less efficient than sets for typical cases. Most beneficial when company names share significant prefixes.

✨

Key Insights

1Insight 1: The core operation is checking if one list is a subset of another. Converting lists to sets allows for efficient subset checking.
2Insight 2: A naive approach involves comparing each list against every other list. It's important to ensure each list is compared with every other list *excluding itself*.
3Insight 3: The problem explicitly mentions that the output should be a sorted list of indices.

📚

Key Terms Explained

For Students

Array

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

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Empty list of favorite companies for a person. The code correctly handles this because an empty set is a subset of any other set, so such a person will only be added to the results if all other people's favorite company lists are non-empty.
Edge case 2: A person's favorite companies are exactly the same as another person's favorite companies. The check `if i == j: continue` prevents a person from being considered a subset of themselves. However, if two lists are identical the algorithm considers the first list to be a subset of the second list and will only add the second list's index to the result.
Edge case 3: One large list and many small lists. The algorithm will correctly identify the person with the large list as being a non-subset, assuming that no other list contains all their companies.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to exclude the case where a person is being compared to themselves. This leads to incorrect results because every list is a subset of itself.

✗Common mistake 2: Not converting the lists to sets for efficient subset checking. This leads to a much higher time complexity.

✗Common mistake 3: Returning the indices in the wrong order. The problem explicitly states that the indices must be returned in increasing order. This is handled correctly in the provided solution since indices are added in increasing order.

✅

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: Recommendation systems. If we have users' lists of preferred items (e.g., movies, products), we can identify users whose preferences are not encompassed by other users' preferences. This could be used to identify niche users with unique interests for targeted marketing or product development.

🏦Real-world scenario 2: Network security. In a network, access control lists (ACLs) define which users/devices have access to which resources. If an ACL is a subset of another, it might indicate redundancy or a potential simplification opportunity. Identifying non-subset ACLs can highlight users with unique permissions.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ 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: June 13, 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

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

@InterviewProJul 14

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

@CodeNinjaJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "People Whose List of Favorite Companies Is Not a Subset of Another List"?

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 hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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