LeetCode #1760 — MEDIUM

Minimum Limit of Balls in a Bag

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given an integer array nums where the i^th bag contains nums[i] balls. You are also given an integer maxOperations.

You can perform the following operation at most maxOperations times:

Take any bag of balls and divide it into two new bags with a positive number of balls.
- For example, a bag of 5 balls can become two new bags of 1 and 4 balls, or two new bags of 2 and 3 balls.

Your penalty is the maximum number of balls in a bag. You want to minimize your penalty after the operations.

Return the minimum possible penalty after performing the operations.

Example 1:

Input: nums = [9], maxOperations = 2
Output: 3
Explanation: 
- Divide the bag with 9 balls into two bags of sizes 6 and 3. [9] -> [6,3].
- Divide the bag with 6 balls into two bags of sizes 3 and 3. [6,3] -> [3,3,3].
The bag with the most number of balls has 3 balls, so your penalty is 3 and you should return 3.

Example 2:

Input: nums = [2,4,8,2], maxOperations = 4
Output: 2
Explanation:
- Divide the bag with 8 balls into two bags of sizes 4 and 4. [2,4,8,2] -> [2,4,4,4,2].
- Divide the bag with 4 balls into two bags of sizes 2 and 2. [2,4,4,4,2] -> [2,2,2,4,4,2].
- Divide the bag with 4 balls into two bags of sizes 2 and 2. [2,2,2,4,4,2] -> [2,2,2,2,2,4,2].
- Divide the bag with 4 balls into two bags of sizes 2 and 2. [2,2,2,2,2,4,2] -> [2,2,2,2,2,2,2,2].
The bag with the most number of balls has 2 balls, so your penalty is 2, and you should return 2.

Constraints:

1 <= nums.length <= 10⁵
1 <= maxOperations, nums[i] <= 10⁹

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums where the ith bag contains nums[i] balls. You are also given an integer maxOperations. You can perform the following operation at most maxOperations times: Take any bag of balls and divide it into two new bags with a positive number of balls. For example, a bag of 5 balls can become two new bags of 1 and 4 balls, or two new bags of 2 and 3 balls. Your penalty is the maximum number of balls in a bag. You want to minimize your penalty after the operations. Return the minimum possible penalty after performing the operations.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Binary Search

Example 1

[9]
2

Example 2

[2,4,8,2]
4

Core Insight

What unlocks the optimal approach

Let's change the question if we know the maximum size of a bag what is the minimum number of bags you can make
note that as the maximum size increases the minimum number of bags decreases so we can binary search the maximum size

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1760: Minimum Limit of Balls in a Bag
class Solution {
    public int minimumSize(int[] nums, int maxOperations) {
        int l = 1, r = Arrays.stream(nums).max().getAsInt();
        while (l < r) {
            int mid = (l + r) >> 1;
            long s = 0;
            for (int x : nums) {
                s += (x - 1) / mid;
            }
            if (s <= maxOperations) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    }
}

// Accepted solution for LeetCode #1760: Minimum Limit of Balls in a Bag
func minimumSize(nums []int, maxOperations int) int {
	r := slices.Max(nums)
	return 1 + sort.Search(r, func(mx int) bool {
		mx++
		s := 0
		for _, x := range nums {
			s += (x - 1) / mx
		}
		return s <= maxOperations
	})
}

# Accepted solution for LeetCode #1760: Minimum Limit of Balls in a Bag
class Solution:
    def minimumSize(self, nums: List[int], maxOperations: int) -> int:
        def check(mx: int) -> bool:
            return sum((x - 1) // mx for x in nums) <= maxOperations

        return bisect_left(range(1, max(nums) + 1), True, key=check) + 1

// Accepted solution for LeetCode #1760: Minimum Limit of Balls in a Bag
impl Solution {
    pub fn minimum_size(nums: Vec<i32>, max_operations: i32) -> i32 {
        let mut l = 1;
        let mut r = *nums.iter().max().unwrap();

        while l < r {
            let mid = (l + r) / 2;
            let mut s: i64 = 0;

            for &x in &nums {
                s += ((x - 1) / mid) as i64;
            }

            if s <= max_operations as i64 {
                r = mid;
            } else {
                l = mid + 1;
            }
        }

        l
    }
}

// Accepted solution for LeetCode #1760: Minimum Limit of Balls in a Bag
function minimumSize(nums: number[], maxOperations: number): number {
    let [l, r] = [1, Math.max(...nums)];
    while (l < r) {
        const mid = (l + r) >> 1;
        const s = nums.map(x => ((x - 1) / mid) | 0).reduce((a, b) => a + b);
        if (s <= maxOperations) {
            r = mid;
        } else {
            l = mid + 1;
        }
    }
    return l;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.