Removing Minimum Number of Magic Beans
Asked at DE Shaw
Problem
Removing Minimum Number of Magic Beans gives you bags of beans; you may remove beans so that every non-empty bag ends up with the same count. Minimize the number of beans removed. Sorting reveals the structure: pick a target equal to one of the bag sizes, empty the smaller bags, and trim the larger ones.
Asked At
| Company | Difficulty | |
|---|---|---|
| DE Shaw | Medium | View all DE Shaw questions → |
How to Think About It
The final common count should be one of the original bag sizes — any other value removes strictly more beans than rounding it up to the next bag size.
Key insight: sort the bags. If the target is beans[i], every bag before i is emptied and every bag from i onward is trimmed to beans[i]. The beans kept are beans[i] * (n - i).
Removed beans = total - beans[i] * (n - i). Minimize removal by maximizing kept beans.
Walkthrough for [4,1,6,5]: sorted [1,4,5,6], total 16. Kept: 1*4 = 4, 4*3 = 12, 5*2 = 10, 6*1 = 6. Best kept 12 -> removed 4.
Use 64-bit integers: 10^5 bags of 10^5 beans each sum to 10^10.
Optimal Approach
Step 1: Sort beans; total = sum(beans).
Step 2: kept = max(beans[i] * (n - i) for i in range(n)).
Step 3: Return total - kept.
Time: O(n log n). Space: O(1) extra.
What Trips People Up in Real Interviews
Assuming you must keep every bag non-empty. Emptying a bag completely is allowed and often optimal.
Trying every target value up to 10^5 without noticing only the bag sizes matter.
32-bit overflow on the total in C++/Java.
Forgetting to sort before using n - i as the count of bags at least as large.
Solution Code
def minimumRemoval(beans):
beans.sort()
n = len(beans)
total = sum(beans)
kept = max(b * (n - i) for i, b in enumerate(beans))
return total - keptFrequently Asked Questions
What is the Removing Minimum Number of Magic Beans problem?
Removing Minimum Number of Magic Beans gives you bags of beans; you may remove beans so that every non-empty bag ends up with the same count. Minimize the number of beans removed. Sorting reveals the structure: pick a target equal to one of the bag sizes, empty the smaller bags, and trim the larger ones.
How do you solve Removing Minimum Number of Magic Beans?
The optimal approach is described in detail above, including step-by-step walkthroughs, complexity analysis, and solution code in Python. Scroll up to the "Optimal Approach" section.
What companies ask Removing Minimum Number of Magic Beans?
Removing Minimum Number of Magic Beans is asked at DE Shaw. It is a medium difficulty problem.
What are common mistakes on Removing Minimum Number of Magic Beans?
- Assuming you must keep every bag non-empty. Emptying a bag completely is allowed and often optimal.
- Trying every target value up to `10^5` without noticing only the bag sizes matter.
- 32-bit overflow on the total in C++/Java.
- Forgetting to sort before using `n - i` as the count of bags at least as large.