Leetcode #1760: Minimum Limit of Balls in a Bag
In this guide, we solve Leetcode #1760 Minimum Limit of Balls in a Bag in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given an integer array nums where the ith bag contains nums[i] balls. You are also given an integer maxOperations.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Array, Binary Search
Intuition
The problem structure suggests a monotonic decision, which makes binary search a natural fit.
By halving the search space each step, we reach the answer efficiently.
Approach
Search either directly on a sorted array or on the answer space using a check function.
Each check is fast, and the logarithmic search keeps the overall runtime low.
Steps:
- Define the search bounds.
- Check the mid point condition.
- Narrow the bounds until convergence.
Example
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.
Python Solution
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
Complexity
The time complexity is , where and are the length and the maximum value of the array , respectively. The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.