1760. Minimum Limit of Balls in a Bag

Problem Description

Imagine you have a collection of bags, and each bag is filled with a certain number of balls. These bags are represented by an array, where each element of the array corresponds to the number of balls in a bag, for example, nums[i] is the number of balls in the i-th bag.

You're given a specific number of operations that you can perform, denoted by maxOperations. In a single operation, you can choose any one of your bags and split its contents into two new bags. The new bags must contain a positive number of balls, meaning each bag must have at least one ball.

The "penalty" is defined as the largest number of balls in any single bag. The goal is to minimize this penalty at the end of all your operations.

For example, if you start with a bag of 5 balls, you could split it into bags containing 3 and 2 balls, respectively. If this is your only bag, your initial penalty is 5 (since it's the only bag), but after the operation, it's reduced to 3.

You need to determine the minimum possible penalty you can achieve after performing at most maxOperations operations.

Intuition

To find the minimum possible penalty, we can utilize a binary search approach. The binary search targets the potential penalties rather than directly searching through the elements of the nums array.

Firstly, we need to establish the search range for the possible penalties. The lower bound is 1, since we cannot have bags with zero balls, and the maximum possible penalty is the largest number of balls in a bag from the input array, max(nums).

During each step of the binary search, we check if a proposed penalty (midpoint of the range) can be achieved with at most maxOperations operations. To check this, we compute the number of operations required to ensure that no bag has more balls than the proposed penalty. For each bag, the number of necessary operations is the number of times we have to split the bag so that each resulting bag has a number of balls less than or equal to the proposed penalty.

If we can achieve the proposed penalty with maxOperations or fewer operations, it means we could possibly do better and thus should search the lower half (reduce the penalty). If not, we need to look for a solution in the upper half (increasing the penalty).

The solution provided uses the bisect_left function to perform the binary search and the check function as a custom condition to decide the direction of the search. The search ends when the bisect_left finds the minimum penalty that satisfies the condition specified by the check function.

Learn more about Binary Search patterns.

Solution Approach

The solution to this problem makes use of a classic algorithmic pattern known as binary search. Binary search is a divide-and-conquer algorithm that quickly locates an item in a sorted list by repeatedly dividing the search interval in half.

Here's the step-by-step solution approach:

1. Define a Check Function: We need a function, check(mx), that returns True if we can make sure that all bags have at most mx balls using no more than maxOperations operations. This function calculates the number of operations needed to reduce the number of balls in each bag to mx or less. It does this by taking each count of balls x in nums and dividing it by mx (after subtracting 1 to avoid an off-by-one error), summing these values up for all bags and comparing the result to maxOperations.

2. Binary Search: We then search for the smallest integer within the range of 1 to max(nums) that can serve as our potential minimum penalty. Within bisect_left, which is the binary search function provided by Python, we use the check function as a key for the binary search, thus guiding the direction of our search:

   • If check returns True for a proposed penalty mx, it means that it is possible to achieve this penalty with maxOperations operations or fewer, and we should continue searching towards a smaller penalty.
   • If check returns False, we have to move towards a larger penalty.

3. Execute the Binary Search with bisect_left: The bisect_left function is called with three arguments:

   • The range range(1, max(nums)), which is our search space for the penalty.
   • The value True which we're trying to find, meaning where our check function results in True.
   • The check function itself, which takes the place of the key argument, allowing bisect_left to use the check function's return value to decide on the search direction.

4. Return the Result: Finally, after the binary search is conducted, we add 1 to the result because bisect_left returns the position where True can be inserted to maintain sorted order, but since our range starts at 1 (not 0), we need to adjust for the zero-based index returned by bisect_left.

The actual code implementation is as follows:

1def minimumSize(self, nums: List[int], maxOperations: int) -> int:
2    def check(mx: int) -> bool:
3        return sum((x - 1) // mx for x in nums) <= maxOperations
4
5    return bisect_left(range(1, max(nums)), True, key=check) + 1

In this code:

• minimumSize is the function that takes nums and maxOperations as input and returns the minimum penalty.
• The check function is nested inside the minimumSize function and is used to determine whether a given maximum number of balls per bag (mx) is achievable under the operation constraints.
• bisect_left performs the binary search and finds the optimal penalty while minimizing the number of operations used.

This code effectively uses binary search to navigate the potential solution space and efficiently arrives at the minimum possible penalty.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach for the problem described. Suppose we have the array of bags nums = [9, 7, 8], and we're allowed maxOperations = 2.

The problem is asking for the minimal penalty after performing at most 2 operations, with the penalty being defined as the largest number of balls in any single bag.

Step-by-Step Solution Approach:

1. Define a Check Function: Our check function receives an integer mx and calculates whether we can ensure all bags have mx or fewer balls using up to maxOperations.

2. Binary Search: We use binary search to find the smallest penalty. We start with a lower bound of 1 (bag can't have less than 1 ball) and upper bound max(nums) which is 9 in this case.

3. Execute the Binary Search with bisect_left: We search for the boundary where our check function starts to return True.

4. Return the Result: Once we find the boundary, it represents the minimum possible penalty we can achieve which is compliant with our operation constraints.

Walkthrough on the Example:

• Lower bound: 1
• Upper bound: 9
• Search space: [1, 2, 3, ... 8, 9]

Now we check with binary search:

• Let's check with mx = 5. This would require dividing the bag with 9 balls into two bags [5, 4] requiring 1 operation, the bag with 7 balls into two bags [5, 2], requiring 1 operation, and no operations for the bag with 8 balls as it already satisfies the condition. In total, we use 2 operations which are equal to maxOperations.

  The result of the check function is True because we can achieve this with 2 operations. But since we might minimize the penalty further, we continue.

• Narrow the search and check mx = 4. With this attempted penalty, we would need to split the bag with 9 balls into [5,4] and [1,4], which uses 2 operations and thus exceeds maxOperations.

  The result of check is False because we've exceeded the allowed number of operations. So, we've gone too far, and the penalty must be higher than 4.

Between 4 and 5, our binary search would choose 5 as the lower True value in the search space, which can now be considered our best possible penalty.

Final Output:

With a maxOperations of 2, the minimum possible penalty we can achieve is 5 as it's the lowest penalty value that returns a True value in our check function without exceeding the allowed number of operations. Thus, the minimumSize function would return 5 + 1 to adjust for the range starting at 1, for a result of 6.

However, in this example, as the penalty of 5 hasn't used all the available operations and is actually possible with the given operations, we do not need to add 1, and the final output is indeed 5.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the code is determined by two factors: the computation within the check function and the binary search using bisect_left. The check function is called for each step in the binary search.

Binary Search: The use of bisect_left implies a binary search over a range determined by the values in nums. Since the range is from 1 to the maximum value in nums, this range can be represented as N, where N = max(nums). The binary search will therefore take O(log N) steps to complete as it narrows down the search range by half with each iteration.

Check Function: The check function is called for each step of the binary search and iterates over all elements in the list nums. If we have M elements in nums, each call to check is O(M) since it potentially goes through the entire list once.

Combining these two factors, the overall time complexity is O(M * log N) where M is the length of the list nums and N is the value of the maximum element in nums.

Space Complexity

The space complexity of the given code is mainly influenced by the space used to store the input nums.

Input Storage: The list nums itself takes up O(M) space, where M is the number of elements in nums. Additional Storage: The space used for the binary search itself is constant, as it operates on a range and does not allocate additional memory proportional to the length of nums or the size of the maximum element in nums.

Thus, the overall space complexity is O(M), as the check function and binary search operations do not use additional space that scales with the size of the input.

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