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Problem 1760 Minimum Limit of Balls in a Bag

Problem Statement

Link - Problem 1760

Question

You are given an integer array nums where the ith 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

Solution

class Solution {
public:
    int minimumSize(vector<int>& nums, int max_operations) {
        int left = 1;
        int right = 0;

        for (auto num : nums) {
            right = max(right, num);
        }

        while (left < right) {
            int middle = (left + right) / 2;

            if (isPossible(middle, nums, max_operations)) {
                right = middle;
            }
            else {
                left = middle + 1;
            }
        }

        return left;
    }

private:
    bool isPossible(int maxBallsInBag, vector<int>& nums, int max_operations) {
        int total_operations = 0;

        for (int num : nums) {
            int operations = ceil(num / (double)maxBallsInBag) - 1;
            total_operations += operations;

            if (total_operations > max_operations)
                return false;
        }

        return true;
    }
};

Complexity Analysis

| Algorithm     | Time Complexity | Space Complexity |
| ------------- | --------------- | ---------------- |
| Binary Search | O(n log m)      | O(1)             |

Explanation

Intial Thoughts

The goal is to minimize the maximum number of balls in a bag. Given an operation to divide a bag into two with a positive number of balls, think about the strategy for dividing the bags. The operation seems to reduce the maximum number of balls by repeatedly dividing the largest bag. Consider using a search strategy, like binary search, to efficiently find the smallest possible maximum. Identify the key factors that influence the minimum penalty, including the range of possible maximum balls and the total number of operations allowed.

Intuitive Analysis

First, determine the feasible range for the minimum penalty, from the smallest possible maximum (1 ball in a bag) to the largest possible maximum (the size of the largest bag without any operations). Next, design a search space to explore this range and identify a suitable criterion to decide on the feasibility of a given maximum. This criterion is whether the total number of operations required to achieve or improve upon the given maximum is within the allowed limit. Perform a binary search to efficiently narrow down the search space and converge on the optimal solution.

1. Intuition

• The problem can be solved using a binary search approach to find the minimum possible penalty.
• The key insight is to find the maximum number of balls in a bag that can be achieved with the given max_operations.
• We can start by initializing the left and right pointers to 1 and the maximum number in the nums array, respectively.
• We then perform a binary search to find the minimum possible penalty.
• The isPossible function checks if it’s possible to achieve a certain maximum number of balls in a bag with the given max_operations.
• The function returns true if the total operations required to achieve the maximum number of balls in a bag is less than or equal to max_operations.
• We use the ceil function to calculate the number of operations required to divide a bag into two bags of sizes maxBallsInBag and num - maxBallsInBag.

2. Implementation

• The minimumSize function initializes the left and right pointers and performs a binary search to find the minimum possible penalty.
• The while loop continues until the left pointer is less than the right pointer.
• Inside the loop, we calculate the middle value and check if it’s possible to achieve the maximum number of balls in a bag with the given max_operations using the isPossible function.
• If it’s possible, we update the right pointer to the middle value; otherwise, we update the left pointer to the middle value plus one.
• The isPossible function calculates the total operations required to achieve the maximum number of balls in a bag and checks if it’s less than or equal to max_operations.
• The ceil function is used to calculate the number of operations required to divide a bag into two bags of sizes maxBallsInBag and num - maxBallsInBag.
• The total_operations variable keeps track of the total operations required to achieve the maximum number of balls in a bag.

Complexity Analysis

Time Complexity:

• The minimumSize function iterates through the nums array once to find the initial right value, resulting in a linear complexity of O(n).
• The while loop then performs binary search, dividing the search space roughly in half at each step. This results in a logarithmic complexity of O(log m), where m represents the range of possible values. Since the loop iterates nums in each iteration, the total time complexity becomes O(n log m).
• The isPossible function iterates through nums once, performing a constant number of operations for each element. Its complexity is O(n) but does not dominate the overall time complexity due to the logarithmic term from the binary search.

Space Complexity:

• The minimumSize function uses a constant amount of space to store variables like left, right, and middle, regardless of the input size.
• The isPossible function also uses a constant amount of space to store variables like total_operations and does not allocate any additional memory.
• The input array nums is not modified, and no additional data structures are used, resulting in a space complexity of O(1).

Footnote

This question is rated as Medium difficulty.

Hints

Let’s change the question if we know the maximum size of a bag what is the minimum number of bags you can make

note that as the maximum size increases the minimum number of bags decreases so we can binary search the maximum size

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