What is the time complexity of Minimum Number of Groups to Create a Valid Assignment?

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

What data structures are used to solve Minimum Number of Groups to Create a Valid Assignment?

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 Minimum Number of Groups to Create a Valid Assignment?

Insight 1: The key idea is to iterate through all possible box sizes 'k' and check if a valid assignment can be created using boxes of sizes 'k' or 'k+1'. We only need to consider 'k' values from 1 up to the minimum count of any ball type, because if k is larger than the smallest count, it's impossible to form groups of size k for all ball types. Insight 2: For a given count 'c' of a ball type and a potential box size 'k', the number of boxes needed is 'c / (k+1)' if 'c' is perfectly divisible by 'k+1'. Otherwise, if 'k - (c % (k+1)) <= c // (k+1)', it means the balls can be grouped into boxes of sizes 'k' and 'k+1', requiring 'c // (k+1) + 1' boxes. The logic here ensures that balls of the same type are placed together such that the difference in box sizes is at most 1.

Which companies ask LeetCode 2910. Minimum Number of Groups to Create a Valid Assignment in interviews?

This problem is frequently asked at BNY Mellon.

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Minimum Number of Groups to Create a Valid Assignment - LeetCode 2910 Solution

Q: How to solve Minimum Number of Groups to Create a Valid Assignment?

The solution utilizes a greedy approach combined with iteration. First, it counts the occurrences of each ball type using a hash map. Then, it iterates through possible box sizes 'k' from 1 up to the minimum count among all ball types. For each 'k', it checks if a valid assignment can be created, determining how many boxes would be needed for each ball type. If a valid assignment is possible for a given 'k', it updates the minimum number of boxes found so far.

Q: Which companies ask LeetCode 2910. Minimum Number of Groups to Create a Valid Assignment in interviews?

This problem is frequently asked at BNY Mellon.

Minimum Number of Groups to Create a Valid Assignment - Complete Solution Guide

Minimum Number of Groups to Create a Valid Assignment is LeetCode problem 2910, 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

You are given a collection of numbered balls and instructed to sort them into boxes for a nearly balanced distribution. There are two rules you must follow: Balls with the same box must have the same value. But, if you have more than one ball with the same number, you can put them in different boxes. The biggest box can only have one more ball than the smallest box. Return the fewest number of boxes to sort these balls following these rules. Example 1: Input: balls = [3,2,3,2,3] Output: 2 Expla

Detailed Explanation

The problem asks us to divide a collection of balls of various numbered types into boxes such that all balls in a box have the same number and the difference between the largest and smallest box sizes is at most 1. The goal is to minimize the number of boxes used. The input is an array of integers representing the ball types, and the output is the minimum number of boxes required to sort the balls according to the rules.

Solution Approach

The solution utilizes a greedy approach combined with iteration. First, it counts the occurrences of each ball type using a hash map. Then, it iterates through possible box sizes 'k' from 1 up to the minimum count among all ball types. For each 'k', it checks if a valid assignment can be created, determining how many boxes would be needed for each ball type. If a valid assignment is possible for a given 'k', it updates the minimum number of boxes found so far.

Step-by-Step Algorithm

Step 1: Count the frequency of each ball type using a hash map.
Step 2: Extract the frequency counts into a list.
Step 3: Find the minimum frequency count among all ball types. This limits the potential sizes for 'k'.
Step 4: Iterate from k = 1 up to the minimum frequency count. For each 'k', attempt to form a valid assignment.
Step 5: For each frequency count 'c', calculate the number of boxes needed. Calculate q = c // (k + 1) and r = c % (k + 1).
Step 6: If r == 0, then c is divisible by k+1 and q boxes are needed.
Step 7: Otherwise, check if k - r <= q. This condition determines if we can form boxes of size k and k+1. If true, then q + 1 boxes are needed.
Step 8: If k - r > q, then a valid assignment is impossible with the current 'k', so break the inner loop.
Step 9: If a valid assignment is possible for the current 'k', update the minimum number of boxes found so far.
Step 10: After iterating through all possible 'k' values, return the minimum number of boxes.

Key Insights

Insight 1: The key idea is to iterate through all possible box sizes 'k' and check if a valid assignment can be created using boxes of sizes 'k' or 'k+1'. We only need to consider 'k' values from 1 up to the minimum count of any ball type, because if k is larger than the smallest count, it's impossible to form groups of size k for all ball types.
Insight 2: For a given count 'c' of a ball type and a potential box size 'k', the number of boxes needed is 'c / (k+1)' if 'c' is perfectly divisible by 'k+1'. Otherwise, if 'k - (c % (k+1)) <= c // (k+1)', it means the balls can be grouped into boxes of sizes 'k' and 'k+1', requiring 'c // (k+1) + 1' boxes. The logic here ensures that balls of the same type are placed together such that the difference in box sizes is at most 1.
Insight 3: Optimization can be achieved by recognizing that the minimum possible box size 'k' needs to be between 1 and the smallest value in the counts of the numbers of each ball type. Also if no assignment is possible return 0.

Complexity Analysis

Time Complexity: O(n*m)

Space Complexity: O(m)

Topics

This problem involves: Array, Hash Table, Greedy.

Companies

Asked at: BNY Mellon.

Minimum Number of Groups to Create a Valid Assignment | LeetCode 2910

Medium

Array Hash Table Greedy

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

Solution Code

Available in:

import java.util.HashMap;
import java.util.List;
import java.util.ArrayList;
import java.util.Collections;
import java.lang.Math;

class Solution {
    public int minGroupsForValidAssignment(int[] balls) {
        HashMap<Integer, Integer> counts = new HashMap<>();
        for (int ball : balls) {
            counts.put(ball, counts.getOrDefault(ball, 0) + 1);
        }
        
        List<Integer> counts_values = new ArrayList<>(counts.values());
        
        int min_c = Integer.MAX_VALUE;
        for (int c : counts_values) {
            min_c = Math.min(min_c, c);
        }
        
        int min_total_groups = Integer.MAX_VALUE;
        
        for (int k = 1; k <= min_c; k++) {
            int current_total_groups = 0;
            boolean possible = true;
            for (int c : counts_values) {
                int q = c / (k + 1);
                int r = c % (k + 1);
                
                if (r == 0) {
                    current_total_groups += q;
                } else {
                    if (k - r <= q) {
                        current_total_groups += q + 1;
                    } else {
                        possible = false;
                        break;
                    }
                }
            }
            
            if (possible) {
                min_total_groups = Math.min(min_total_groups, current_total_groups);
            }
        }
        
        return min_total_groups == Integer.MAX_VALUE ? 0 : min_total_groups;
    }
}

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Problem Info

Description

Problem 2910: Minimum Number of Groups to Create a Valid Assignment - Detailed explanation and solution approaches.

Topics

Array Hash Table Greedy

Companies

BNY Mellon

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2910

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: Count the frequency of each ball type using a hash map.
2
Step 2: Extract the frequency counts into a list.
3
Step 3: Find the minimum frequency count among all ball types. This limits the potential sizes for 'k'.
4
Step 4: Iterate from k = 1 up to the minimum frequency count. For each 'k', attempt to form a valid assignment.
5
Step 5: For each frequency count 'c', calculate the number of boxes needed. Calculate q = c // (k + 1) and r = c % (k + 1).
6
Step 6: If r == 0, then c is divisible by k+1 and q boxes are needed.
7
Step 7: Otherwise, check if k - r <= q. This condition determines if we can form boxes of size k and k+1. If true, then q + 1 boxes are needed.
8
Step 8: If k - r > q, then a valid assignment is impossible with the current 'k', so break the inner loop.
9
Step 9: If a valid assignment is possible for the current 'k', update the minimum number of boxes found so far.
10
Step 10: After iterating through all possible 'k' values, return the minimum number of boxes.

💡

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*m)

💾

Space Complexity

O(m)

📝

Why?

The time complexity is determined by the nested loops. The outer loop runs up to the minimum count, and the inner loop iterates through each distinct ball type to determine the minimum groups needed. The space complexity is determined primarily by the space used to store the counts of each ball type, which is proportional to the number of distinct ball types.

📖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

Binary Search

Instead of linearly iterating through possible box sizes 'k', we could use binary search to find the optimal 'k'. This involves defining a predicate function that checks if a given 'k' results in a valid assignment. The binary search would be performed on the range [1, min_c].

Time: O(n * log m), where n is the number of distinct ball types and m is the minimum count of any ball type.Space: O(n)

Trade-offs: Binary search could potentially offer a slight improvement in time complexity, especially when the minimum count is large. However, the added complexity of implementing binary search and the predicate function might not always be worth the effort for smaller input sizes.

✨

Key Insights

1Insight 1: The key idea is to iterate through all possible box sizes 'k' and check if a valid assignment can be created using boxes of sizes 'k' or 'k+1'. We only need to consider 'k' values from 1 up to the minimum count of any ball type, because if k is larger than the smallest count, it's impossible to form groups of size k for all ball types.
2Insight 2: For a given count 'c' of a ball type and a potential box size 'k', the number of boxes needed is 'c / (k+1)' if 'c' is perfectly divisible by 'k+1'. Otherwise, if 'k - (c % (k+1)) <= c // (k+1)', it means the balls can be grouped into boxes of sizes 'k' and 'k+1', requiring 'c // (k+1) + 1' boxes. The logic here ensures that balls of the same type are placed together such that the difference in box sizes is at most 1.
3Insight 3: Optimization can be achieved by recognizing that the minimum possible box size 'k' needs to be between 1 and the smallest value in the counts of the numbers of each ball type. Also if no assignment is possible return 0.

📚

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: Empty input array. While the code doesn't explicitly handle this, it should return a correct 0 since the loop will not execute and min_total_groups remains INT_MAX and the condition at the end is triggered.
Edge case 2: All balls are of the same type. The algorithm will still correctly compute the minimum number of boxes by iterating through all possible 'k' values.
Edge case 3: Very large counts. The integer division and modulo operations should handle large counts correctly as long as the intermediate calculations don't overflow.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the case when 'r' is not zero and incorrectly calculating the number of boxes. The condition 'k - r <= q' is crucial for determining if a valid assignment is possible.

✗Common mistake 2: Not considering that the smallest possible 'k' value is 1 and starting the loop from a different value. Starting from 1 ensures all possible box sizes are checked.

✗Common mistake 3: Integer overflow can potentially be a problem if the counts of the balls can be extremely large. Consider using long integers.

✅

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: Resource allocation where resources need to be grouped into sets such that the sets are as balanced as possible and the items in each set are of the same type.

🏦Real-world scenario 2: Manufacturing, where items of the same type need to be packaged into boxes, with constraints on box size variations.

💡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.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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: July 1, 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

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

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

@AlgoMasterJul 6

❓

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 "Minimum Number of Groups to Create a Valid Assignment"?

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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