Minimum Limit of Balls in a Bag

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ArraysBinary Search

You are given an integer array nums where the i^th bag contains nums[i] balls. You are also given an integer maxOperations.

You can perform the following operation at most maxOperations times:

Take any bag of balls and divide it into two new bags with a positive number of balls.
- For example, a bag of 5 balls can become two new bags of 1 and 4 balls, or two new bags of 2 and 3 balls.

Your penalty is the maximum number of balls in a bag. You want to minimize your penalty after the operations.

Return the minimum possible penalty after performing the operations.

Example 1:

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.

Example 2:

Input: nums = [2,4,8,2], maxOperations = 4
Output: 2
Explanation:
- Divide the bag with 8 balls into two bags of sizes 4 and 4. [2,4,8,2] -> [2,4,4,4,2].
- Divide the bag with 4 balls into two bags of sizes 2 and 2. [2,4,4,4,2] -> [2,2,2,4,4,2].
- Divide the bag with 4 balls into two bags of sizes 2 and 2. [2,2,2,4,4,2] -> [2,2,2,2,2,4,2].
- Divide the bag with 4 balls into two bags of sizes 2 and 2. [2,2,2,2,2,4,2] -> [2,2,2,2,2,2,2,2].
The bag with the most number of balls has 2 balls, so your penalty is 2, and you should return 2.

Constraints:

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

Try coding it in LeetCode

Minimum Limit of Balls in a Bag

Medium

Asked by:

11 views

Topics:

ArraysBinary Search

You are given an integer array nums where the i^th bag contains nums[i] balls. You are also given an integer maxOperations.

You can perform the following operation at most maxOperations times:

Take any bag of balls and divide it into two new bags with a positive number of balls.
- For example, a bag of 5 balls can become two new bags of 1 and 4 balls, or two new bags of 2 and 3 balls.

Your penalty is the maximum number of balls in a bag. You want to minimize your penalty after the operations.

Return the minimum possible penalty after performing the operations.

Example 1:

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.

Example 2:

Input: nums = [2,4,8,2], maxOperations = 4
Output: 2
Explanation:
- Divide the bag with 8 balls into two bags of sizes 4 and 4. [2,4,8,2] -> [2,4,4,4,2].
- Divide the bag with 4 balls into two bags of sizes 2 and 2. [2,4,4,4,2] -> [2,2,2,4,4,2].
- Divide the bag with 4 balls into two bags of sizes 2 and 2. [2,2,2,4,4,2] -> [2,2,2,2,2,4,2].
- Divide the bag with 4 balls into two bags of sizes 2 and 2. [2,2,2,2,2,4,2] -> [2,2,2,2,2,2,2,2].
The bag with the most number of balls has 2 balls, so your penalty is 2, and you should return 2.

Constraints:

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

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on the size of the `nums` array and the value of `maxOperations`? What is the range of values for the number of balls in each bag?
Can the number of balls in any bag be zero or negative?
If `maxOperations` is zero, should I return the largest number of balls in the initial state of the bags?
Is there always a valid solution, or are there cases where it's impossible to split the bags enough to minimize the maximum number of balls given `maxOperations`?
Could you provide a specific example with a small `nums` array and a small `maxOperations` value to illustrate the splitting process and the expected output?

Brute Force Solution

Approach

The brute force way to find the minimum limit is to try every possible limit value. For each potential limit, we check if we can split all the bags such that no part is larger than the limit.

Here's how the algorithm would work step-by-step:

Start with a very small limit and go up to a very large limit (perhaps the size of the largest bag).
For each possible limit, figure out how many pieces each bag would need to be broken into so that no piece is bigger than the limit.
Add up the total number of pieces we would need across all bags.
If the total number of pieces needed is less than or equal to the allowed number of operations, then this limit works.
Otherwise, this limit is too small and doesn't work.
Keep track of the smallest limit that works.
After checking all possible limits, the smallest limit we found that works is the answer.

Code Implementation

def minimum_limit_of_balls_in_bag_brute_force(bags, max_operations):
    maximum_bag_size = max(bags)
    minimum_possible_limit = 1

    best_possible_limit = maximum_bag_size

    # Iterate through all possible limits
    for current_limit in range(minimum_possible_limit, maximum_bag_size + 1):

        total_operations_needed = 0

        # Calculate required operations for current limit
        for bag_size in bags:
            total_operations_needed += (bag_size - 1) // current_limit

        # Check if current limit is feasible
        if total_operations_needed <= max_operations:
            # If it is feasible, update best limit
            best_possible_limit = current_limit

            # We can stop here and return
            # because we are going from min to max
            return best_possible_limit

    return best_possible_limit

Big(O) Analysis

Time Complexity

O(m * n) – The outer loop iterates through a range of possible limit values. In the worst case, this range extends from 1 to the size of the largest bag, let's call that m. Inside this outer loop, we iterate through each of the n bags. For each bag, we perform a division operation to determine how many pieces are needed. Thus, the inner loop takes O(n) time. Since the outer loop iterates m times, and the inner loop iterates n times, the total time complexity is O(m * n).

Space Complexity

O(1) – The brute force solution iterates through possible limit values and calculates the number of pieces needed for each bag. It only uses a few integer variables to store the current limit, the number of pieces for each bag, the total number of pieces, and the best limit found so far. The space required for these variables does not depend on the size of the input bags, represented as N, so the space complexity is constant. No additional data structures like lists or hash maps are created that depend on the size of the input.

Optimal Solution

Approach

The fastest way to find the smallest possible maximum bag size is to use a process of intelligent guessing and checking. We repeatedly guess a potential maximum size, then quickly check if it's possible to split all the bags so none are bigger than that guess. We use the results of our checks to refine our guesses to converge on the smallest possible maximum.

Here's how the algorithm would work step-by-step:

First, find the smallest and largest number of balls in any bag.
Imagine the answer we're looking for must be somewhere between the smallest and largest bag sizes.
Make a guess for a potential maximum bag size somewhere in the middle of that range.
Check if it's possible to split all the bags so that none of the resulting bags have more balls than our guess.
If it IS possible to split all bags based on this maximum limit, then we know the actual smallest maximum size must be equal or smaller than our current guess. Try a smaller guess.
If it IS NOT possible to split all bags based on this maximum limit, then we know the actual smallest maximum size must be larger than our current guess. Try a larger guess.
Keep making guesses and adjusting them based on whether they work or not, narrowing the range. Eventually, the low and high guesses will meet, indicating the smallest maximum bag size.

Code Implementation

def minimum_size(nums, max_operations):
    left = 1
    right = max(nums)

    while left < right:
        mid = (left + right) // 2
        operations_needed = 0

        for num in nums:
            operations_needed += (num - 1) // mid

        # Check if the current mid is feasible.
        if operations_needed <= max_operations:
            right = mid
        else:
            left = mid + 1

    # Left and right converge to the minimum size.
    return left

Big(O) Analysis

Time Complexity

O(n log m) – The algorithm performs a binary search to find the minimum possible maximum bag size. The search space is between the smallest and largest bag sizes, where 'm' represents the largest value in the input array bags. For each guess (potential maximum size) within the binary search, the algorithm iterates through the input array 'bags' of size 'n' to check if it's possible to split all bags according to the guess. Therefore, the time complexity of the check is O(n), and since we perform a binary search within the range of 1 to m, we perform log m checks. The total time complexity is thus O(n log m).

Space Complexity

O(1) – The algorithm primarily uses variables to store the minimum and maximum bag sizes, and intermediate values during the binary search process. These variables consume a constant amount of space regardless of the input array's size, denoted as N (the number of bags). No auxiliary data structures like lists, hash maps, or significant recursion are employed; hence the space complexity remains constant. Therefore, the auxiliary space complexity is O(1).

Edge Cases

Case	How to Handle
Empty input array (nums is null or empty)	Return 0 immediately as there are no bags to operate on, indicating an edge case where no balls exist.
maxOperations is 0	Return the maximum element in nums as no operations can be performed, so the largest bag remains unchanged.
nums contains only one element	Binary search needs a lower bound of 1; if maxOperations is greater than 0, return 1; otherwise, return the single element.
All elements in nums are the same.	Binary search will still converge to a result, but the number of operations will either be 0 (if a small enough penalty is chosen) or a larger number if it needs to be reduced.
Large values in nums potentially causing integer overflow during calculations.	Use long data type or appropriate bounds checking within the cost function to prevent overflow and ensure accuracy.
maxOperations is a very large number, allowing for aggressive splitting.	The binary search should converge to a minimum limit even with very large maxOperations efficiently.
A bag already has a small number of balls close to the potentially minimum maximum	The binary search should correctly find the lower bound that balances operation usage and maximum ball count.
Input array contains large number of bags with widely varying counts	The binary search approach will still efficiently narrow down the optimal minimum limit, regardless of the distribution.