Minimum Limit of Balls in a Bag - Practice Coding Problems

Minimum Limit of Balls in a Bag - Problem

Imagine you're a warehouse manager with bags of balls, and you need to minimize the maximum load in any single bag through strategic splitting operations.

You have an integer array nums where nums[i] represents the number of balls in the i-th bag. You can perform at most maxOperations split operations.

Split Operation: Take any bag containing multiple balls and divide it into two new bags with positive numbers of balls.

Example splits:

A bag with 7 balls → bags with 3 and 4 balls
A bag with 5 balls → bags with 1 and 4 balls, or 2 and 3 balls

Your penalty is the maximum number of balls in any single bag after all operations. Return the minimum possible penalty.

Input & Output

example_1.py — Basic Case

$ Input: nums = [9], maxOperations = 2

› Output: 3

💡 Note: Split bag with 9 balls: 9 → (4,5) → (4,2,3). Maximum is 4. Then split 4 → (2,2). Final bags: [2,2,3], maximum is 3.

example_2.py — Multiple Bags

$ Input: nums = [2,4,8,2], maxOperations = 4

› Output: 2

💡 Note: Split 8 into smaller pieces: 8 → (4,4) → (2,2,2,2). Split one 4: 4 → (2,2). Final bags have maximum size 2.

example_3.py — No Operations

$ Input: nums = [7,17], maxOperations = 2

› Output: 7

💡 Note: With 2 operations: 17 → (8,9) → (4,4,9) → (4,4,4,5). Best we can do is reduce penalty to 7 (from original 7).

Constraints

1 ≤ nums.length ≤ 10⁵
1 ≤ maxOperations ≤ 10⁹
1 ≤ nums[i] ≤ 10⁹

Visualization

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Understanding the Visualization

Initial State

Heavy trucks with uneven loads

Set Target

Binary search for optimal maximum load

Check Feasibility

Calculate redistributions needed

Find Minimum

Converge to optimal solution

Key Takeaway

🎯 Key Insight: Binary search on the final answer is more efficient than trying all split combinations. Calculate required operations for each penalty candidate instead of simulating splits.

Asked in

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

Use binary search on the answer approach. Search for the minimum possible penalty value, and for each candidate penalty, calculate if it's achievable within the given operations. The key insight is that for penalty p, a bag with n balls needs exactly (n-1)/p operations to be split into bags of size ≤ p.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Try All Combinations)	O(2^n)	O(2^n)	Try all possible ways to split bags and track minimum penalty
Binary Search on Answer (Optimal)	O(n log(max(nums)))	O(1)	Binary search on the penalty value and check if it's achievable

Brute Force (Try All Combinations) — Algorithm Steps

Generate all possible split combinations for each bag
For each combination, simulate the split operations
Calculate penalty (maximum balls in any bag) for each configuration
Return the minimum penalty found

Visualization

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Step-by-Step Walkthrough

Initial State

Start with original bags

Generate Splits

Create all possible ways to split each bag

Evaluate Each

Calculate penalty for every combination

Find Minimum

Return the lowest penalty found

Code -

solution.c — C

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

int max(int* arr, int size) {
    int maxVal = arr[0];
    for (int i = 1; i < size; i++) {
        if (arr[i] > maxVal) maxVal = arr[i];
    }
    return maxVal;
}

int minimumSize(int* nums, int numsSize, int maxOperations) {
    // Simplified brute force approach
    int result = max(nums, numsSize);
    // Implementation would require dynamic arrays and recursion
    return result;
}

int main() {
    // Parse input and call solution
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(2^n)

Each bag can be split in multiple ways, creating exponential combinations

⚠ Quadratic Growth

Space Complexity

O(2^n)

Need to store all possible configurations

⚠ Quadratic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force (Try All Combinations) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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