How do you solve Minimum Limit of Balls in a Bag on LeetCode?

Minimum Limit of Balls in a Bag (LeetCode #1760) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log(max(nums))) time complexity and O(1) space complexity. Key insight: Binary search helps efficiently narrow down the possible maximum size of balls in a bag.

What is the brute force solution for Minimum Limit of Balls in a Bag?

The brute force approach for Minimum Limit of Balls in a Bag is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves trying every possible way to divide the bags and checking the maximum number of balls left in any bag after each operation. This is simple but inefficient as it explores all combinations. The optimal Optimal Solution approach improves this to O(n log(max(nums))).

What algorithm or data structure is used to solve Minimum Limit of Balls in a Bag?

Minimum Limit of Balls in a Bag uses the following concepts: Array, Binary Search. The recommended patterns to study are: Binary Search, Greedy Algorithms.

What are the interview tips for Minimum Limit of Balls in a Bag?

Always clarify the problem and constraints before diving into coding. Think aloud during the interview to show your thought process. Practice explaining your code and logic clearly, as communication is key.

What are common mistakes when solving Minimum Limit of Balls in a Bag?

Failing to account for edge cases where all bags are already at or below the desired size. Not properly implementing the binary search logic, leading to incorrect bounds.

#1760

Minimum Limit of Balls in a Bag

Medium

ArrayBinary SearchBinary SearchGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log(max(nums)))
Space	O(1)	O(1)

💡

Intuition

Time O(n log(max(nums)))Space O(1)

The optimal solution uses binary search to efficiently find the minimum possible penalty. By guessing a maximum size and checking if we can achieve that with the allowed operations, we can narrow down our search quickly.

⚙️

Algorithm

4 steps

1Step 1: Set low to 1 and high to the maximum number of balls in any bag.
2Step 2: Perform binary search while low is less than or equal to high.
3Step 3: For each mid value, calculate the number of operations needed to ensure no bag exceeds mid.
4Step 4: If operations needed are less than or equal to maxOperations, update result and search lower; otherwise, search higher.

solution.py20 lines

1# Full working Python code
2
3def minPenalty(nums, maxOperations):
4    def canAchieve(maxSize):
5        operations = 0
6        for balls in nums:
7            operations += (balls - 1) // maxSize
8        return operations <= maxOperations
9
10    low, high = 1, max(nums)
11    result = high
12    while low <= high:
13        mid = (low + high) // 2
14        if canAchieve(mid):
15            result = mid
16            high = mid - 1
17        else:
18            low = mid + 1
19    return result
20

ℹ

Complexity note: The time complexity is driven by the binary search over the range of possible maximum sizes (1 to max(nums)), and for each size, we check the number of operations needed, which is O(n).

1Binary search helps efficiently narrow down the possible maximum size of balls in a bag.
2Understanding how to calculate the number of operations needed for a given maximum size is crucial.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.