#1760

Medium

auto_awesomeBinary search over the valid answer space

LeetCode Problem Workspace

Minimum Limit of Balls in a Bag

Find the minimum penalty (maximum balls in a bag) after performing up to maxOperations to split bags of balls.

Problem Statement

You are given an integer array nums, where the ith element represents the number of balls in the ith bag. You are also given an integer maxOperations that dictates how many times you can split a bag into two smaller bags. The goal is to minimize the maximum number of balls in any bag after performing at most maxOperations splits.

Each time you perform a split on a bag, it is divided into two smaller bags. You can perform this operation on any bag multiple times but are limited by maxOperations. After performing the splits, the penalty is defined as the maximum number of balls in any of the bags. You need to find the minimum penalty possible.

Examples

Example 1

Input: nums = [9], maxOperations = 2

Output: 3

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

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

Solution Approach

Binary Search on Maximum Bag Size

Perform binary search over the possible maximum number of balls that could exist in a bag after splitting. The range of possible values is from 1 to the maximum number of balls in the largest bag in the array. For each candidate size, calculate the number of splits needed to ensure that no bag has more than this number of balls.

Check Feasibility of a Maximum Bag Size

For each candidate size found by the binary search, check whether it's feasible to split all bags such that no bag has more than the candidate size. This is done by iterating through the bags and calculating how many splits each bag requires to reduce its size to the maximum size.

Optimizing the Result with Binary Search

Using binary search, optimize the result by adjusting the possible maximum bag size and narrowing down to the minimum feasible value. This ensures that the solution is efficient and can handle the large input size.

Complexity Analysis

Metric	Value
Time	O(n \log k)
Space	O(1)

The time complexity is O(n log k), where n is the number of bags and k is the maximum number of balls in a bag. The binary search over the range of possible maximum bag sizes takes log k iterations, and for each iteration, we check the feasibility by iterating through the bags, resulting in O(n log k) time complexity. The space complexity is O(1) since only a few extra variables are needed for the binary search and calculations.

What Interviewers Usually Probe

The candidate understands how to apply binary search to a problem involving finding an optimal solution in a range.
The candidate demonstrates understanding of problem constraints and how to balance performance and correctness in large inputs.
The candidate effectively uses the binary search technique to narrow down the feasible solution space and optimize the answer.

Common Pitfalls or Variants

Common pitfalls

Forgetting to check if the candidate maximum bag size can be achieved with the given maxOperations.
Misinterpreting the problem as needing to minimize the number of splits, rather than the maximum bag size after splitting.
Incorrectly handling edge cases such as very small or very large values in the nums array or maxOperations.

Follow-up variants

What if there were additional constraints such as only being able to split bags in a certain ratio?
How would the solution change if the number of splits were unlimited, and you needed to minimize the largest bag size?
What if there were a constraint that you could only split bags evenly or into specific sizes?

FAQ

What is the primary pattern in the 'Minimum Limit of Balls in a Bag' problem?

The primary pattern is binary search over the valid answer space to minimize the maximum number of balls in any bag after performing splits.

How does binary search help in solving the problem?

Binary search helps by narrowing down the possible values of the maximum bag size, allowing you to efficiently find the minimal penalty after the splits.

What is the time complexity of the 'Minimum Limit of Balls in a Bag' problem?

The time complexity is O(n log k), where n is the number of bags and k is the maximum number of balls in a bag.

What happens if maxOperations is larger than the number of bags?

If maxOperations exceeds the number of bags, it may still not be enough to reduce the largest bag size to a smaller value, and you will need to use binary search to determine the smallest feasible maximum bag size.

What are common mistakes when solving this problem?

Common mistakes include misunderstanding the goal of minimizing the maximum bag size instead of the number of splits, and failing to properly calculate the number of splits required for each bag.

terminal

Solution

Solution 1: Binary Search

This problem requires us to minimize the cost, which is the maximum number of balls in a single bag. As the maximum value increases, the number of operations decreases, making it easier to meet the condition.

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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

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