Minimum Limit of Balls in a Bag

Minimum Limit of Balls in a Bag

1. What is this problem about?

The "Minimum Limit of Balls in a Bag" interview question is an optimization problem that deals with distributing resources to meet a specific goal. You are given an array of integers, where each integer represents the number of balls in a bag. You are also given a maximum number of "operations" you can perform.

In one operation, you can take any bag of balls and split it into two new bags. For example, a bag with 10 balls can be split into bags of 3 and 7, or 5 and 5. Your goal is to perform at most maxOperations such that the size of the largest bag (the "penalty") is as small as possible.

2. Why is this asked in interviews?

This is a classic "Optimization over a Range" problem that tests a candidate's ability to use Binary Search on Answer. Companies like Microsoft, Google, and Meta use it to evaluate:

• Optimization Intuition: Recognizing that while finding the exact minimum penalty is hard, checking if a specific penalty is possible is much easier.
• Search Space Management: Defining the range for the binary search (usually from 1 to the maximum bag size).
• Efficiency: Achieving a solution in $O(n \log(\max\_val))$ time.

3. Algorithmic pattern used

The primary pattern is Binary Search on the Answer.

1. The search space for the "minimum possible penalty" is [1, max(nums)].
2. For a middle value mid in this range, we ask: "Can we achieve a maximum bag size of mid using at most maxOperations?"
3. To achieve a max size of mid, a bag with x balls needs to be split ceil(x / mid) - 1 times.
4. We sum these required operations for all bags. If the total is $\le$ maxOperations, then mid is a possible penalty, and we try to find an even smaller one in the left half. Otherwise, we look in the right half.

4. Example explanation

Bags: [9], Max Operations: 2.

• Try mid = 3: To make a bag of 9 into bags $\le 3$:
  • $9 \rightarrow 3, 6$ (1 op)
  • $6 \rightarrow 3, 3$ (1 op) Total ops = 2. $2 \le 2$ (Max Ops). So 3 is possible.
• Try mid = 2: To make a bag of 9 into bags $\le 2$:
  • $9 \rightarrow 2, 7$ (1 op)
  • $7 \rightarrow 2, 5$ (1 op)
  • $5 \rightarrow 2, 3$ (1 op)
  • $3 \rightarrow 2, 1$ (1 op) Total ops = 4. $4 > 2$. Impossible. The minimum penalty is 3.

5. Common mistakes candidates make

• Greedy Approach: Trying to always split the largest bag in half. This doesn't always lead to the global minimum and is harder to manage.
• Off-by-one in Operations: Calculating splits as x / mid instead of (x - 1) / mid. A bag of size 6 needs 1 split to reach size 3, not 2.
• Integer Division: Using standard division instead of ceiling logic or the (x - 1) // mid trick.

6. Interview preparation tip

Practice the "Binary Search on Answer" pattern. It’s applicable whenever you need to find the "minimum maximum" or "maximum minimum" of something and you have a clear monotonic property (if a penalty of 10 is possible, a penalty of 11 is also certainly possible).