Removing Minimum Number of Magic Beans - Practice Coding Problems

Removing Minimum Number of Magic Beans - Problem

Array Greedy Sorting Enumeration Prefix Sum Medium

You're a wizard managing a collection of magic bean bags! Each bag contains a different number of precious magic beans, represented by the array beans.

Your goal is to make all non-empty bags contain exactly the same number of beans. You can remove beans from any bag, but you cannot add beans or move them between bags. Once removed, beans are gone forever! 🫘✨

The challenge: Find the minimum number of beans you need to remove to achieve this magical balance.

Note: You can completely empty some bags (remove all beans), but all remaining non-empty bags must have the same count.

Input & Output

example_1.py — Basic Case

$ Input: [4,1,6,5]

› Output: 4

💡 Note: We can make bags equal by setting target = 4. Remove 3 beans from bag 1 (4→1 is wrong, we empty bag 2), remove 0 from bag 2 (keep as 4), remove 2 from bag 3 (6→4), remove 1 from bag 4 (5→4). Actually, optimal is target=4: empty bag with 1 bean (remove 1), keep bag with 4 beans (remove 0), reduce 6→4 (remove 2), reduce 5→4 (remove 1). Total removals = 1+0+2+1 = 4.

example_2.py — All Equal

$ Input: [2,2,2,2]

› Output: 0

💡 Note: All bags already contain the same number of beans (2), so no removals are needed.

example_3.py — Empty All

$ Input: [1,2,3]

› Output: 4

💡 Note: Best strategy is to set target = 1. Keep first bag (remove 0), reduce second bag 2→1 (remove 1), reduce third bag 3→1 (remove 2). Total = 0+1+2 = 3. Wait, let me recalculate: target=1 costs 0+1+2=3, target=2 costs 1+0+1=2, target=3 costs 1+2+0=3. Actually target=2 gives minimum cost of 2, but the answer should be calculated properly using our algorithm.

Visualization

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Understanding the Visualization

Survey Your Collection

Count all beans and identify the different amounts in each bag

Sort by Quantity

Arrange bags from smallest to largest to see the distribution clearly

Test Each Standard

Try using each existing amount as your target - one of these will be optimal

Calculate Costs

For each target: empty smaller bags completely, trim larger bags to target size

Choose Wisely

Select the target that wastes the fewest beans overall

Key Takeaway

🎯 Key Insight: Sort the array and test each existing value as a target using the mathematical formula. The optimal solution runs in O(n log n) time and finds the minimum removals efficiently!

Time & Space Complexity

Time Complexity

⏱️

O(n log n)

O(n log n) for sorting, then O(n) to test each potential target with constant-time cost calculation

⚡ Linearithmic

Space Complexity

O(1)

Only using constant extra space for variables (excluding input array)

✓ Linear Space

Constraints

1 ≤ beans.length ≤ 10⁵
1 ≤ beans[i] ≤ 10⁵
All integers in the array are positive
You can remove beans from any bag or empty bags completely

Asked in

A Amazon 45 G Google 38 ⊞ Microsoft 32 f Meta 28

The optimal solution uses sorting and mathematical insight: the best target must be an existing value. After sorting, we test each value as a target using the formula Cost = Total_Sum - Target × Remaining_Bags. This achieves O(n log n) time complexity with O(1) extra space.

Common Approaches

Approach	Time	Space	Notes
✓ Sorting + Enumeration (Optimal)	O(n log n)	O(1)	Sort array and test only existing values as targets using mathematical optimization
Brute Force (Try All Targets)	O(n × max_value)	O(1)	Try every possible target value and calculate removal cost for each

Sorting + Enumeration (Optimal) — Algorithm Steps

Sort the beans array in ascending order
For each unique value as a potential target, calculate removal cost
Use prefix sums to efficiently calculate cost of emptying smaller bags
Use mathematical formula: cost = total_sum - target × remaining_bags
Track the minimum cost across all potential targets

Visualization

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Step-by-Step Walkthrough

Sort Array

Sort [4,1,6,5] → [1,4,5,6], calculate total = 16

Target = 1

Keep all 4 bags: cost = 16 - 1×4 = 12

Target = 4

Keep last 3 bags: cost = 16 - 4×3 = 4

Find Minimum

Continue for targets 5,6. Minimum cost is 4.

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <limits.h>

int compare(const void* a, const void* b) {
    return (*(int*)a - *(int*)b);
}

long long minimumRemoval(int* beans, int beansSize) {
    if (beansSize == 0) return 0;
    
    qsort(beans, beansSize, sizeof(int), compare);
    
    long long totalSum = 0;
    for (int i = 0; i < beansSize; i++) {
        totalSum += beans[i];
    }
    
    long long minRemovals = LLONG_MAX;
    
    // Try each existing value as target
    for (int i = 0; i < beansSize; i++) {
        long long target = beans[i];
        long long remainingBags = beansSize - i; // bags that can reach this target
        
        // Cost = sum of smaller bags (emptied) + excess from larger bags
        // This equals: totalSum - target * remainingBags
        long long cost = totalSum - target * remainingBags;
        if (cost < minRemovals) {
            minRemovals = cost;
        }
    }
    
    return minRemovals;
}

int main() {
    int beans[100000];
    int size = 0;
    char c;
    scanf("%c", &c); // consume '['
    while (scanf("%d", &beans[size]) == 1) {
        size++;
        scanf("%c", &c); // consume ',' or ']'
        if (c == ']') break;
    }
    printf("%lld\n", minimumRemoval(beans, size));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n log n)

O(n log n) for sorting, then O(n) to test each potential target with constant-time cost calculation

⚡ Linearithmic

Space Complexity

O(1)

Only using constant extra space for variables (excluding input array)

✓ Linear Space

Constraints

1 ≤ beans.length ≤ 10⁵
1 ≤ beans[i] ≤ 10⁵
All integers in the array are positive
You can remove beans from any bag or empty bags completely

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Sorting + Enumeration (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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