Problem Description

You have an array nums where nums[i] represents the number of balls in the i-th bag. You can perform up to maxOperations operations to split bags.

In each operation, you can:

• Choose any bag with balls and divide it into two new bags
• Each new bag must contain a positive number of balls
• For example, a bag with 5 balls can be split into bags of 1 and 4 balls, or 2 and 3 balls

Your penalty is defined as the maximum number of balls in any single bag after all operations. The goal is to minimize this penalty.

The problem asks you to find the minimum possible penalty (the smallest possible maximum bag size) that can be achieved using at most maxOperations splits.

For instance, if you have bags [9] and maxOperations = 2, you could:

• Split the bag of 9 into 6 and 3
• Split the bag of 6 into 3 and 3
• Result: bags of [3, 3, 3], giving a penalty of 3

The solution uses binary search to find the minimum penalty. It searches for the smallest maximum bag size that can be achieved within the operation limit. For each candidate maximum size mx, it calculates how many splits are needed: for a bag with x balls to have at most mx balls per bag, you need (x - 1) // mx splits. If the total splits needed is within maxOperations, that maximum size is achievable.

Intuition

The key insight is recognizing that this is an optimization problem where we want to find the minimum possible maximum value. When we think about the relationship between the penalty (maximum bag size) and the number of operations needed, we notice an important pattern: as we allow larger maximum bag sizes, fewer operations are required.

Consider a bag with 9 balls. If we want the maximum to be 3, we need to split it into pieces of at most 3 balls each, requiring 2 operations. But if we allow the maximum to be 5, we only need 1 operation (splitting into 5 and 4). This inverse relationship suggests binary search could work.

The challenge becomes: given a target maximum bag size, how many operations do we need? For a bag with x balls to be split into bags of at most mx balls each, we need ceil(x/mx) - 1 operations. This can be written as (x - 1) // mx operations. We sum this across all bags to get the total operations needed.

Since there's a monotonic relationship (smaller maximum requires more operations, larger maximum requires fewer operations), we can binary search on the answer. We search for the smallest maximum bag size where the required operations don't exceed maxOperations.

The search space is from 1 (minimum possible bag size) to max(nums) (no operations needed). For each candidate maximum in our binary search, we check if it's achievable within the operation limit. The smallest achievable maximum is our answer.

Learn more about Binary Search patterns.

Solution Approach

The solution implements a binary search to find the minimum possible maximum bag size. Here's how the implementation works:

Binary Search Setup:

• Left boundary l = 1: The minimum possible bag size is 1 ball
• Right boundary r = max(nums): If we don't perform any operations, the maximum is the largest original bag

Helper Function check(mx): This function determines if we can achieve a maximum bag size of mx within maxOperations:

• For each bag with x balls, we calculate how many splits are needed: (x - 1) // mx
• This formula works because to split x balls into bags of at most mx balls, we need ceil(x/mx) - 1 operations
• Sum all required operations and check if the total is ≤ maxOperations

Binary Search Process: The code uses Python's bisect_left with a custom key function:

bisect_left(range(1, max(nums) + 1), True, key=check) + 1

This searches for the leftmost position where check(mx) returns True. The search works as follows:

1. For each middle value mid in the range [1, max(nums)]
2. If check(mid) returns True, the value mid is achievable, so we search for smaller values
3. If check(mid) returns False, we need a larger maximum, so we search higher values
4. The process continues until we find the smallest value where check returns True

Time Complexity: O(n * log(max(nums))) where n is the length of the array. We perform binary search over the range [1, max(nums)], and for each iteration, we check all n bags.

Space Complexity: O(1) as we only use a constant amount of extra space.

The + 1 at the end adjusts for the fact that bisect_left returns an index starting from 0, while our range starts from 1.

Example Walkthrough

Let's walk through the solution with nums = [7, 17] and maxOperations = 2.

Step 1: Initialize Binary Search Range

• Left boundary: l = 1 (minimum possible bag size)
• Right boundary: r = max([7, 17]) = 17 (maximum without any operations)
• We're searching for the smallest maximum bag size in range [1, 17]

Step 2: Binary Search Process

First Iteration - Try mid = 9:

• Check if we can achieve maximum bag size of 9 with ≤ 2 operations
• Bag with 7 balls: needs (7-1)//9 = 0 operations (already ≤ 9)
• Bag with 17 balls: needs (17-1)//9 = 1 operation (split into 9 and 8)
• Total operations needed: 0 + 1 = 1 ≤ 2 ✓
• Since this works, try smaller maximum (search left half): range becomes [1, 8]

Second Iteration - Try mid = 4:

• Check if we can achieve maximum bag size of 4 with ≤ 2 operations
• Bag with 7 balls: needs (7-1)//4 = 1 operation (split into 4 and 3)
• Bag with 17 balls: needs (17-1)//4 = 4 operations (need to split into 5 pieces)
• Total operations needed: 1 + 4 = 5 > 2 ✗
• This doesn't work, need larger maximum (search right half): range becomes [5, 8]

Third Iteration - Try mid = 6:

• Check if we can achieve maximum bag size of 6 with ≤ 2 operations
• Bag with 7 balls: needs (7-1)//6 = 1 operation (split into 6 and 1)
• Bag with 17 balls: needs (17-1)//6 = 2 operations (split into 6, 6, and 5)
• Total operations needed: 1 + 2 = 3 > 2 ✗
• Still doesn't work, search right half: range becomes [7, 8]

Fourth Iteration - Try mid = 7:

• Check if we can achieve maximum bag size of 7 with ≤ 2 operations
• Bag with 7 balls: needs (7-1)//7 = 0 operations (already = 7)
• Bag with 17 balls: needs (17-1)//7 = 2 operations (split into 7, 7, and 3)
• Total operations needed: 0 + 2 = 2 ≤ 2 ✓
• This works, but check if smaller is possible: range becomes [7, 7]

Result: The minimum possible penalty is 7.

We can verify: With 2 operations, we split the 17-ball bag twice to get [7, 7, 3], combined with the original 7-ball bag gives us [7, 7, 7, 3]. The maximum is 7.

Solution Implementation

Time and Space Complexity

The time complexity is O(n × log M), where n is the length of the array nums and M is the maximum value in the array nums.

This complexity arises from:

• Binary search over the range [1, max(nums)], which has M elements, taking O(log M) iterations
• For each binary search iteration, the check function is called, which iterates through all n elements in nums to compute sum((x - 1) // mx for x in nums), taking O(n) time
• Total: O(n) × O(log M) = O(n × log M)

The space complexity is O(1).

The algorithm only uses a constant amount of extra space for variables in the check function and the binary search operation. The range object in Python is a lazy iterator that doesn't create the entire list in memory, so it doesn't contribute to space complexity.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Split Calculation Formula

Pitfall: A common mistake is using the wrong formula to calculate the number of operations needed to split a bag. Developers might incorrectly use:

• x // mx (forgetting to subtract 1)
• (x + mx - 1) // mx (this gives the number of resulting bags, not operations)
• x // mx - 1 (off by one when x is divisible by mx)

Why the formula (x - 1) // mx works:

• To split x balls into bags of at most mx balls each, you need ceil(x/mx) total bags
• Since you start with 1 bag, you need ceil(x/mx) - 1 operations
• The formula (x - 1) // mx is equivalent to ceil(x/mx) - 1

Example:

• With 9 balls and max size 3: (9-1)//3 = 8//3 = 2 operations needed
• This creates 3 bags of 3 balls each (9 → 6+3 → 3+3+3)

2. Binary Search Boundary Issues

Pitfall: Setting incorrect search boundaries or updating them improperly:

• Starting with left = 0 instead of left = 1 (bag size must be positive)
• Using left = mid instead of left = mid + 1 when the current max_size is not achievable
• Using right = mid - 1 instead of right = mid when the current max_size is achievable

Correct approach:

while left < right:
    mid = (left + right) // 2
    if can_achieve_max_size(mid):
        right = mid  # Keep mid as a candidate
    else:
        left = mid + 1  # Mid is too small, exclude it

3. Integer Overflow in Other Languages

Pitfall: When translating to languages like C++ or Java, the sum of operations might overflow for large inputs.

Solution: Use appropriate data types:

// C++ example
long long total_operations = 0;  // Use long long instead of int
for (int bag_size : nums) {
    total_operations += (bag_size - 1) / max_size;
}

4. Misunderstanding the Problem Goal

Pitfall: Confusing the objective - trying to minimize the total number of balls or the number of bags, rather than minimizing the maximum bag size.

Clarification: The goal is to minimize the penalty, which is the size of the largest bag after all operations. We're looking for the smallest possible value of the maximum bag size.

5. Edge Case Handling

Pitfall: Not considering edge cases properly:

• When maxOperations = 0 (no splits allowed, answer is max(nums))
• When all bags have size 1 (no splits needed, answer is 1)
• When maxOperations is very large (answer approaches 1)

Robust solution includes these checks implicitly through the binary search range and operation counting.