Leetcode 1760: Minimum Limit of Balls in a Bag

Input: The input consists of an integer array `nums`, representing the number of balls in each bag, and an integer `maxOperations`, which is the maximum number of operations you can perform.

Example: Input: nums = [7, 10], maxOperations = 3

Constraints:

• 1 <= nums.length <= 10^5

• 1 <= maxOperations, nums[i] <= 10^9

Output: Return the minimum possible maximum number of balls in any bag after performing the operations, modulo 10^9 + 7.

Example: Output: 4

Constraints:

• The result should be modulo 10^9 + 7.

Goal: The goal is to minimize the maximum number of balls in any bag after performing the allowed number of operations.

Steps:

• 1. Perform a binary search over the possible maximum number of balls in a bag, starting from 1 to the maximum value in the `nums` array.

• 2. For each candidate maximum, count the number of operations needed to achieve this maximum for all bags.

• 3. If the number of operations required is less than or equal to `maxOperations`, it is possible to achieve this maximum, so continue searching for a smaller maximum.

• 4. Return the minimum maximum number of balls that can be achieved.

Goal: Ensure that the solution handles large inputs and works efficiently for arrays of length up to 10^5.

Steps:

• The solution must handle arrays of length up to 10^5 efficiently.

• The solution should perform binary search to minimize the maximum number of balls.

Assumptions:

• All elements in the `nums` array are positive integers.

• The number of operations `maxOperations` is positive and within bounds.

• Input: Input: nums = [7, 10], maxOperations = 3

• Explanation: You can divide the bag with 10 balls into two bags with 5 and 5 balls. The maximum number of balls in any bag is 5. With one more operation, you can divide a bag of 7 balls into two bags with 4 and 3 balls. The final maximum number of balls in a bag is 4.

• Input: Input: nums = [2, 3, 5], maxOperations = 2

• Explanation: You can divide the bag with 5 balls into two bags with 3 and 2 balls, and then divide the bag with 3 balls into two bags with 2 and 1 balls. The maximum number of balls in any bag is 2.

Approach: The solution involves binary search over the possible maximum number of balls, using a helper function to count how many operations are needed to ensure no bag exceeds this maximum.

Observations:

• The problem is about minimizing the maximum value in a list of bags by splitting them into smaller bags.

• Binary search is an ideal approach to efficiently find the minimum possible maximum number of balls.

• We can try various maximum values and check how many operations are required to achieve that value. The smallest possible maximum that requires no more than `maxOperations` is our answer.

Steps:

• 1. Initialize a binary search range from 1 to the largest number in the `nums` array.

• 2. For each middle value in the binary search, count how many operations are needed to ensure no bag exceeds that value.

• 3. If the number of operations is less than or equal to `maxOperations`, search for smaller maximum values.

• 4. If not, search for larger maximum values.

• 5. Return the result modulo 10^9 + 7.

Empty Inputs:

• If `nums` is empty, return 0 since there are no bags.

Large Inputs:

• Ensure that the binary search and the operation counting function work efficiently for large arrays.

Special Values:

• If all bags contain the same number of balls, no operations are needed, and the penalty is the number of balls in any bag.

Constraints:

• The solution should be efficient for arrays of length up to 10^5.

int minimumSize(vector<int>& nums, int mxOps) {
    
    int l = 1, r = *max_element(nums.begin(), nums.end());
    // int total_bags = n + 2 * mxOps - 1;
    
    while(l < r) {
        int mid = l + (r - l) / 2;
        int cnt = 0;
        for(int a: nums)
            cnt += (a - 1) / mid;
        if(cnt <= mxOps)
            r = mid;
        else
            l = mid + 1;
    }
    return l;
}

1 : Function Declaration

int minimumSize(vector<int>& nums, int mxOps) {

Declares a function to minimize the largest bag size given a maximum number of operations.

2 : Variable Initialization

    int l = 1, r = *max_element(nums.begin(), nums.end());

Initializes the binary search range with `l` as the smallest possible size and `r` as the maximum number in the array.

3 : Binary Search Loop

    while(l < r) {

Begins a binary search loop to find the minimum maximum bag size.

4 : Mid Calculation

        int mid = l + (r - l) / 2;

Calculates the middle point to test as the potential maximum bag size.

5 : Variable Initialization

        int cnt = 0;

Initializes the operation counter to track the total number of splits required.

6 : For Loop

        for(int a: nums)

Iterates through each element in the array to calculate the operations required for the current `mid` size.

7 : Operation Calculation

            cnt += (a - 1) / mid;

Calculates the number of operations needed to reduce the current element to the `mid` size or smaller.

8 : Condition Check

        if(cnt <= mxOps)

Checks if the total operations required is within the allowed limit.

9 : Range Update

            r = mid;

Narrows the search range by setting the upper bound to `mid` when the condition is met.

10 : Else Block

        else

Handles the case where the current `mid` size is too small to meet the operation limit.

11 : Range Update

            l = mid + 1;

Narrows the search range by increasing the lower bound when the condition is not met.

12 : Return Statement

    return l;

Returns the smallest possible maximum bag size after applying the allowed operations.

Best Case: O(n log(max(nums)))

Average Case: O(n log(max(nums)))

Worst Case: O(n log(max(nums)))

Description: The time complexity is driven by the binary search and the operation counting function, resulting in O(n log(max(nums))) where `n` is the length of the array and `max(nums)` is the maximum number of balls in any bag.

Best Case: O(1)

Worst Case: O(1)

Description: The space complexity is constant, O(1), since we only need a few variables to track the current state.

Leetcode 1760: Minimum Limit of Balls in a Bag

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