What is the time complexity of Removing Minimum Number of Magic Beans?

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

What data structures are used to solve Removing Minimum Number of Magic Beans?

This problem uses Array, Greedy, Sorting, Enumeration, Prefix Sum. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Removing Minimum Number of Magic Beans?

Insight 1: The key idea is to realize that if we decide to keep a certain number of beans 'x' in the remaining bags, we will need to remove beans from all other bags that have more than 'x' beans and remove the entire bag if it has less than 'x' beans to make them zero. Insight 2: Sorting the array allows us to efficiently iterate through all possible values of 'x' (the number of beans to keep in the remaining bags).

Which companies ask LeetCode 2171. Removing Minimum Number of Magic Beans in interviews?

This problem is frequently asked at DE Shaw.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Removing Minimum Number of Magic Beans - LeetCode 2171 Solution

Removing Minimum Number of Magic Beans - Complete Solution Guide

Removing Minimum Number of Magic Beans is LeetCode problem 2171, 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 an array of positive integers beans , where each integer represents the number of magic beans found in a particular magic bag. Remove any number of beans ( possibly none ) from each bag such that the number of beans in each remaining non-empty bag (still containing at least one bean) is equal . Once a bean has been removed from a bag, you are not allowed to return it to any of the bags. Return the minimum number of magic beans that you have to remove . Example 1: Input: beans = [4,

Detailed Explanation

The problem requires us to minimize the number of magic beans removed from a given array of bags, such that all remaining non-empty bags have the same number of beans. We're given an array 'beans' where each element represents the number of beans in a bag. We can remove any number of beans from any bag, but we cannot add beans back. The goal is to find the minimum total number of beans we need to remove to achieve the condition where all non-empty bags contain the same number of beans.

Solution Approach

The solution involves sorting the 'beans' array and then iterating through each bean value in the sorted array. For each bean value, we treat it as the target number of beans that all remaining bags should have. We calculate the number of beans to remove to achieve this state, keeping track of the minimum number of beans removed so far. By iterating through each bean value and considering it as the target, we find the optimal solution.

Step-by-Step Algorithm

Step 1: Sort the 'beans' array in ascending order.
Step 2: Calculate the total sum of beans in all bags.
Step 3: Initialize 'min_removed' to infinity (or the maximum possible value for a long long integer).
Step 4: Iterate through the sorted 'beans' array using a 'for' loop.
Step 5: In each iteration, let 'target' be the current bean value (beans[i]).
Step 6: Calculate 'removed_count'. This is the total sum of beans minus the number of bags *remaining* after keeping all other values to at least target. This can be expressed as total_sum - (n - i) * target, where 'n' is the length of the array and 'i' is the current index.
Step 7: Update 'min_removed' with the minimum of its current value and 'removed_count'.
Step 8: After the loop finishes, return 'min_removed'. Cast to integer where needed.

Key Insights

Insight 1: The key idea is to realize that if we decide to keep a certain number of beans 'x' in the remaining bags, we will need to remove beans from all other bags that have more than 'x' beans and remove the entire bag if it has less than 'x' beans to make them zero.
Insight 2: Sorting the array allows us to efficiently iterate through all possible values of 'x' (the number of beans to keep in the remaining bags).
Insight 3: Using prefix sums (or in this case, total sum) helps compute the number of beans that need to be removed for each possible 'x' in O(1) time, after the initial sum calculation.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(1)

Topics

This problem involves: Array, Greedy, Sorting, Enumeration, Prefix Sum.

Companies

Asked at: DE Shaw.

Removing Minimum Number of Magic Beans | LeetCode 2171

Medium

Array Greedy Sorting Enumeration Prefix Sum

Time: O(n log n)Space: O(1)

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public long minimumRemoval(int[] beans) {
        int n = beans.length;
        Arrays.sort(beans);
        
        long total_sum = 0;
        for (int bean : beans) {
            total_sum += bean;
        }
        
        long min_removed = Long.MAX_VALUE;
        
        for (int i = 0; i < n; i++) {
            long target = beans[i];
            long removed_count = total_sum - (long)(n - i) * target;
            min_removed = Math.min(min_removed, removed_count);
        }
        
        return (long)min_removed;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 2171: Removing Minimum Number of Magic Beans - Detailed explanation and solution approaches.

Topics

Array Greedy Sorting Enumeration Prefix Sum

Companies

DE Shaw

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2171

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

8 steps

1
Step 1: Sort the 'beans' array in ascending order.
2
Step 2: Calculate the total sum of beans in all bags.
3
Step 3: Initialize 'min_removed' to infinity (or the maximum possible value for a long long integer).
4
Step 4: Iterate through the sorted 'beans' array using a 'for' loop.
5
Step 5: In each iteration, let 'target' be the current bean value (beans[i]).
6
Step 6: Calculate 'removed_count'. This is the total sum of beans minus the number of bags *remaining* after keeping all other values to at least target. This can be expressed as total_sum - (n - i) * target, where 'n' is the length of the array and 'i' is the current index.
7
Step 7: Update 'min_removed' with the minimum of its current value and 'removed_count'.
8
Step 8: After the loop finishes, return 'min_removed'. Cast to integer where needed.

💡

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 log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is dominated by the sorting algorithm used (e.g., quicksort, mergesort). The remaining operations, such as summing the beans and iterating through the sorted array, take linear time, which is less significant than the sorting time. The space complexity is constant because only a few variables (like total_sum, min_removed, target, and removed_count) are used, and their memory usage does not depend on the input size.

📖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 iterating through each bean as the target, we could use binary search to find the optimal target value. This approach would require additional functions to calculate the required removals for a given target. However, this may not be as efficient since it needs to compute the removals for the middle value many times during the binary search.

Time: O(n log n) in the worst case, where n is the number of bags. The binary search will iterate log(max(beans)) times, and within each iteration, we need to calculate the removed count, which takes O(n) time.Space: O(1)

Trade-offs: The binary search approach may be beneficial when the range of bean values is significantly larger than the number of beans. However, the constant factors involved in calculating removals might make the straightforward iteration faster in practice.

✨

Key Insights

1Insight 1: The key idea is to realize that if we decide to keep a certain number of beans 'x' in the remaining bags, we will need to remove beans from all other bags that have more than 'x' beans and remove the entire bag if it has less than 'x' beans to make them zero.
2Insight 2: Sorting the array allows us to efficiently iterate through all possible values of 'x' (the number of beans to keep in the remaining bags).
3Insight 3: Using prefix sums (or in this case, total sum) helps compute the number of beans that need to be removed for each possible 'x' in O(1) time, after the initial sum calculation.

📚

Key Terms Explained

For Students

Array

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

Greedy Algorithm

Making the best choice at each step without looking ahead.

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: 'beans' array is empty. The code will still work correctly because the loop won't execute.
Edge case 2: 'beans' array contains only one element. The code will still work correctly, and the minimum number of beans to remove will be 0.
Edge case 3: All elements in 'beans' array are the same. The code correctly calculates 0 as the minimum removal.
Edge case 4: Large input values, potential integer overflow during calculations. The C++, and Java solutions utilize 'long long' and 'long' respectively to handle potentially large values of sum(beans) and multiplication operations to avoid integer overflow. The Python solution implicitly handles large integers.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the array. Sorting is crucial because it allows us to efficiently determine the number of beans to remove when a specific bean value is chosen as the target.

✗Common mistake 2: Integer overflow. Make sure to use a data type that can handle potentially large values for the sum of beans and the product of (n - i) and target. Python generally avoids these issues, but Java/C++ needs 'long' or 'long long'.

✗Common mistake 3: Incorrect calculation of removed_count. It's important to subtract from the total sum only the beans that will remain after removal, which is (n - i) * target.

✗Common mistake 4: Assuming only one type of bean exists (i.e., not considering all possible minimum bean values). The algorithm must consider each unique bean value as the final minimum bean value.

✅

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: Inventory optimization. Imagine managing inventory for a product. You might want to minimize the number of items you have to discard or sell at a loss to bring all warehouses to the same inventory level.

🏦Real-world scenario 2: Resource allocation. Consider distributing resources (e.g., CPU time, memory) across different applications. You might want to minimize the total resources unused to achieve a certain level of fairness or performance.

💡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: June 12, 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 13

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

@InterviewProJul 6

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

@CodeNinjaJun 29

❓

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.

What are common greedy problem patterns?

Activity selection (sort by end time), interval scheduling (sort by start/end), Huffman coding (always combine smallest), fractional knapsack (sort by value/weight ratio), and minimum spanning tree. Look for problems where local decisions don't affect future options.

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 "Removing Minimum Number of Magic Beans"?

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

🤖 Need help with this problem?