Category: LeetCode Programming
Difficulty: Medium
Topics: Array Binary Search

#99 1760. Minimum Limit of Balls in a Bag 🚀🧠

Let’s dive into an exciting problem from LeetCode: 1760. Minimum Limit of Balls in a Bag. 🚀 We’ll explore the problem, break it into manageable parts, and tackle it step by step. Get ready for some problem-solving action with detailed explanations, walkthroughs, and Python code snippets! 🐍✨

📖 Problem Statement

You are given an array nums where each element represents the number of balls in a bag. Additionally, you’re provided an integer maxOperations, which is the maximum number of operations you can perform.

Operations allowed:

  1. Choose a bag with x balls.
  2. Divide it into two bags, such that each bag has a positive number of balls.

Your goal is to minimize the maximum number of balls in a bag (penalty) after performing at most maxOperations.

Constraints:

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

🌟 Examples

Example 1:

Input:
nums = [9], maxOperations = 2
Output: 3

Explanation:

  1. Divide the bag [9] into [6, 3] ➡️ Max = 6.
  2. Divide [6] into [3, 3] ➡️ Max = 3.

The penalty is 3, which is the minimum achievable.

Example 2:

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

Explanation:

  1. Divide [8] into [4, 4] ➡️ Max = 4.
  2. Divide [4] into [2, 2] ➡️ Max = 2.

Continue until all operations are used. Penalty = 2.

💡 Key Insights

  • The penalty is directly related to how large the bags are.
  • Reducing a bag’s size involves splitting, which costs operations.
  • We need to determine the minimum penalty such that all splits are within maxOperations.

This problem can be solved using binary search on the penalty size. Let’s build this step by step! 👷‍♀️

✨ Solution Design

The core idea behind solving this problem lies in minimizing the maximum penalty, which translates to controlling the largest number of balls allowed in any bag after performing the given operations. To design the solution effectively, we can break it down into several logical components and decisions:

1️⃣ Key Observations

  • Minimizing the Penalty:
    • The penalty is the largest bag size after dividing the balls.
    • Smaller penalties require more operations to split large bags into smaller ones.
  • Operation Count:
    • For a bag with x balls and a target maximum size mx, the number of operations required to split the bag is: \(\text{operations} = \left\lfloor \frac{x - 1}{mx} \right\rfloor\) This ensures all sub-bags are of size ≤ mx.
  • Monotonicity of Valid Penalty:
    • If a penalty value p is valid (i.e., you can achieve it within maxOperations), then any penalty larger than p will also be valid.
    • Conversely, if p is invalid, any penalty smaller than p will also be invalid.

This monotonic behavior allows us to apply binary search effectively. 🎯

The problem asks for the minimum valid penalty. Binary search is a natural choice for such minimization problems over a continuous or discrete range with monotonic properties.

  • Reduces the search space logarithmically, making it efficient even for large ranges like 1 to 10⁹.
  • Allows systematic testing of potential penalties (midpoints) while narrowing down the valid range.

3️⃣ Helper Function: check(mx)

The helper function determines whether a given penalty mx is achievable within the allowed maxOperations. It calculates the total number of operations needed to ensure all bags have a size ≤ mx.

How It Works:

  1. Iterate through each bag size x in nums.
  2. Calculate the number of operations needed to split x into sub-bags of size ≤ mx: \(\text{operations for bag } x = \left\lfloor \frac{x - 1}{mx} \right\rfloor\) This formula ensures that any remainder balls are included in one of the sub-bags.
  3. Sum up the operations for all bags and compare with maxOperations:
    • If the total operations ≤ maxOperations, then mx is a valid penalty.
    • Otherwise, mx is invalid, and we need to try larger penalties.

Example:

Suppose nums = [10, 15, 20] and maxOperations = 5.
For mx = 7, calculate operations:

  • Bag 10: \(\left\lfloor \frac{10 - 1}{7} \right\rfloor = 1\)
  • Bag 15: \(\left\lfloor \frac{15 - 1}{7} \right\rfloor = 2\)
  • Bag 20: \(\left\lfloor \frac{20 - 1}{7} \right\rfloor = 2\)
    Total operations = 1 + 2 + 2 = 5, which is valid.

4️⃣ Binary Search Implementation

The binary search algorithm is used to find the minimum penalty:

  1. Define Search Space:
    • Start with low = 1 (minimum possible penalty).
    • Set high = max(nums) (maximum possible penalty if no operations are performed).
  2. Binary Search Loop:
    • Compute the midpoint: \(mid = \left\lfloor \frac{low + high}{2} \right\rfloor\).
    • Use the helper function check(mid) to determine if mid is a valid penalty:
      • If valid, update high = mid (try smaller penalties).
      • If invalid, update low = mid + 1 (try larger penalties).
  3. Termination:
    • When low == high, the minimum valid penalty is found.

5️⃣ Efficiency of the Design

  • Binary Search:
    • The number of iterations depends on the range of penalties (1 to max(nums)): \(O(\log(\text{max(nums)}))\)
  • Helper Function:
    • For each midpoint, iterate through nums to compute the total operations: \(O(n)\)
  • Overall Complexity:
    • Combining binary search and helper function: \(O(n \cdot \log(\text{max(nums)}))\)

This design ensures the solution is efficient, even for large input sizes (n ≈ 10⁵ and nums[i] ≈ 10⁹).

6️⃣ Edge Cases

While the design is robust, let’s highlight potential edge cases to test its correctness:

  1. Single Bag:
    • nums = [1], maxOperations = 0:
      No operations needed, penalty = 1.
  2. Maximum Ball Count:
    • nums = [10⁹], maxOperations = 1:
      Must split the bag into two. Penalty = 5 \times 10⁸.
  3. All Bags Are Small:
    • nums = [1, 2, 3], maxOperations = 10:
      No need for operations; penalty = 3.
  4. Insufficient Operations:
    • nums = [100, 200], maxOperations = 1:
      Impossible to achieve a penalty ≤ 50. The algorithm should return the best achievable penalty.
  5. Uniform Bag Sizes:
    • nums = [5, 5, 5], maxOperations = 2:
      Optimal splitting distributes operations evenly.

By incorporating these edge cases, the solution ensures robustness across diverse scenarios. ✅

With this comprehensive design in mind, let’s tackle the example walkthrough next! 🛠️

🧑‍💻 Python Implementation

Here’s the Python implementation of the optimized solution:

from typing import List
from bisect import bisect_left

class Solution:
    def minimumSize(self, nums: List[int], maxOperations: int) -> int:
        def check(mx: int) -> bool:
            # Calculate required operations for a given maximum bag size
            return sum((x - 1) // mx for x in nums) <= maxOperations

        # Binary search to find the minimum penalty
        return bisect_left(range(1, max(nums) + 1), True, key=check) + 1

📊 Time Complexity Analysis

  • Binary Search:
    • Search range: 1 to max(nums) ➡️ O(log(max(nums))).
  • Checking Function:
    • Iterates through nums ➡️ O(n).

Overall Time Complexity:
\(O(n \cdot \log(\text{max(nums)}))\)

Space Complexity:
\(O(1)\) (no extra space used).

🔍 Example Walkthrough

Let’s apply the solution to nums = [27, 15, 10], maxOperations = 5:

Step 1: Define the search range

  • Minimum penalty = 1.
  • Maximum penalty = 27 (largest bag size).

Step 2: Binary Search on Penalty

  1. Initial Midpoint:
    Penalty = (1 + 27) // 2 = 14.

  2. Check if Penalty = 14 is valid:

    • Bag [27]: Requires 1 operation ➡️ [14, 13].
    • Bag [15]: Requires 1 operation ➡️ [14, 1].
    • Bag [10]: No operation needed.
      Total = 2 operations (valid).

    Update range: [1, 14].

  1. New Midpoint:
    Penalty = (1 + 14) // 2 = 7.

  2. Check if Penalty = 7 is valid:

    • Bag [27]: Requires 3 operations ➡️ [7, 7, 7, 6].
    • Bag [15]: Requires 2 operations ➡️ [7, 7, 1].
      Total = 5 operations (valid).

    Update range: [1, 7].

  1. New Midpoint:
    Penalty = (1 + 7) // 2 = 4.

  2. Check if Penalty = 4 is valid:

    • Bag [27]: Requires 6 operations (exceeds limit).

    Update range: [5, 7].

Final Result

The minimum penalty is 7. 🎉

🚀 Conclusion

This problem elegantly combines binary search and greedy logic to minimize the penalty for splitting bags of balls. The optimized solution handles large inputs efficiently, ensuring a fast and reliable approach.

Keep tackling challenges like these to sharpen your problem-solving skills! 💪 If you enjoyed this, stay tuned for more insights and tutorials. ✨

Written on December 7, 2024