What is the time complexity of Group the People Given the Group Size They Belong To?

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

What data structures are used to solve Group the People Given the Group Size They Belong To?

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

What are the key insights for solving Group the People Given the Group Size They Belong To?

Insight 1: Use a hash map (or dictionary) to store the people IDs based on their group size. This allows efficient grouping. Insight 2: Iterate through the `groupSizes` array, and for each person, add their ID to the corresponding group size's list in the hash map.

Which companies ask LeetCode 1282. Group the People Given the Group Size They Belong To in interviews?

This problem is frequently asked at Roblox.

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Group the People Given the Group Size They Belong To - LeetCode 1282 Solution

Group the People Given the Group Size They Belong To - Complete Solution Guide

Group the People Given the Group Size They Belong To is LeetCode problem 1282, 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 are n people that are split into some unknown number of groups. Each person is labeled with a unique ID from 0 to n - 1 . You are given an integer array groupSizes , where groupSizes[i] is the size of the group that person i is in. For example, if groupSizes[1] = 3 , then person 1 must be in a group of size 3 . Return a list of groups such that each person i is in a group of size groupSizes[i] . Each person should appear in exactly one group , and every person must be in a group. If there

Detailed Explanation

The problem requires us to group people into groups based on their specified group sizes. We're given an array `groupSizes` where `groupSizes[i]` indicates the size of the group that person `i` must belong to. The task is to return a list of groups (lists of person IDs) such that each person is in exactly one group of the size specified in `groupSizes`. It's guaranteed that at least one valid solution exists, and we can return any valid solution if multiple are possible. Essentially, we need to partition the indices (person IDs) of the `groupSizes` array into sublists where the size of each sublist matches the value in `groupSizes` at those indices.

Solution Approach

The solution uses a hash map (or dictionary) to organize the people based on their group size. The algorithm iterates through the `groupSizes` array. For each person `i`, it appends their ID `i` to the list associated with `groupSizes[i]` in the hash map. When the list for a particular group size reaches the required size (i.e., the length of the list equals the group size), the algorithm adds the list to the `result` list and then removes the entry for that size from the hash map (or resets the list to an empty state).

Step-by-Step Algorithm

Step 1: Initialize an empty hash map called `groups` to store lists of person IDs for each group size and an empty list called `result` to store the final list of groups.
Step 2: Iterate through the `groupSizes` array with its index `i`.
Step 3: For each element `groupSizes[i]`, append the index `i` (person ID) to the list associated with `groupSizes[i]` in the `groups` hash map.
Step 4: Check if the length of the list `groups[groupSizes[i]]` is equal to `groupSizes[i]`. If it is, this means we have formed a complete group.
Step 5: If a complete group is formed, append the list to the `result` list and remove the `groupSizes[i]` entry from the `groups` hash map to start forming new groups of that size.
Step 6: After iterating through the entire `groupSizes` array, return the `result` list, which contains lists of person IDs forming the required groups.

Key Insights

Insight 1: Use a hash map (or dictionary) to store the people IDs based on their group size. This allows efficient grouping.
Insight 2: Iterate through the `groupSizes` array, and for each person, add their ID to the corresponding group size's list in the hash map.
Insight 3: When a group size's list reaches the size specified by the group size, add that list as a group to the result and clear the list in the hash map.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table, Greedy.

Companies

Asked at: Roblox.

Group the People Given the Group Size They Belong To | LeetCode 1282

Medium

Array Hash Table Greedy

Time: O(n)Space: O(n)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

class Solution {
    public List<List<Integer>> groupThePeople(int[] groupSizes) {
        Map<Integer, List<Integer>> groups = new HashMap<>();
        List<List<Integer>> result = new ArrayList<>();
        
        for (int i = 0; i < groupSizes.length; i++) {
            int size = groupSizes[i];
            
            if (!groups.containsKey(size)) {
                groups.put(size, new ArrayList<>());
            }
            
            groups.get(size).add(i);
            
            if (groups.get(size).size() == size) {
                result.add(new ArrayList<>(groups.get(size)));
                groups.remove(size);
            }
        }
        
        return result;
    }
}

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Problem Info

Description

Problem 1282: Group the People Given the Group Size They Belong To - Detailed explanation and solution approaches.

Topics

Array Hash Table Greedy

Companies

Roblox

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1282

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: Initialize an empty hash map called `groups` to store lists of person IDs for each group size and an empty list called `result` to store the final list of groups.
2
Step 2: Iterate through the `groupSizes` array with its index `i`.
3
Step 3: For each element `groupSizes[i]`, append the index `i` (person ID) to the list associated with `groupSizes[i]` in the `groups` hash map.
4
Step 4: Check if the length of the list `groups[groupSizes[i]]` is equal to `groupSizes[i]`. If it is, this means we have formed a complete group.
5
Step 5: If a complete group is formed, append the list to the `result` list and remove the `groupSizes[i]` entry from the `groups` hash map to start forming new groups of that size.
6
Step 6: After iterating through the entire `groupSizes` array, return the `result` list, which contains lists of person IDs forming the required groups.

💡

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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is linear because each person is processed exactly once. The space complexity is linear because, in the worst case, we might need to store all the indices in the hash map before forming any complete groups. For instance, if groupSizes = [1,1,1,1,1], then all indices will be stored in map.

📖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

Sort the `groupSizes` along with their indices. Then iterate through the sorted array, creating groups of the corresponding size. This approach avoids using a hash map.

Time: O(n log n) due to the sorting step.Space: O(n) for storing sorted indices.

Trade-offs: Sorting introduces O(n log n) complexity, making it less efficient than the hash map approach. However, it can be space-efficient if the original array can be modified in place to track indices.

✨

Key Insights

1Insight 1: Use a hash map (or dictionary) to store the people IDs based on their group size. This allows efficient grouping.
2Insight 2: Iterate through the `groupSizes` array, and for each person, add their ID to the corresponding group size's list in the hash map.
3Insight 3: When a group size's list reaches the size specified by the group size, add that list as a group to the result and clear the list in the hash map.

📚

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!

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: A group size appears only once in the input. The algorithm correctly handles this by creating a group of size 1 if such input comes.
Edge case 2: All people belong to the same group size. The algorithm correctly forms multiple groups of that size, dividing people in the `groupSizes` into multiple groups having similar sizes as present in the given input.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to remove/reset the group size entry in the hash map after forming a complete group, which can lead to incorrect grouping or infinite loops.

✗Common mistake 2: Not handling the case where a group size appears multiple times in the input, potentially resulting in groups not being formed correctly.

✅

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: Task scheduling where tasks need to be assigned to workers based on their capacity (group size).

🏦Real-world scenario 2: Social network group formation where users are categorized into groups based on interests or roles, each group having a predefined capacity.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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 23, 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 "Group the People Given the Group Size They Belong To"?

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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