1760. Minimum Limit Of Balls In A Bag

Intuition

• The idea is to minimize the maximum size of a bag of balls (x) by repeatedly dividing larger bags into smaller ones, constrained by the number of allowed operations. It can be solved using binary search to determine the minimum possible penalty.

 

Approach

1. Binary Search

• The penalty (x) can range between 1 (minimum) and the maximum value in nums (maximum size of a bag).
• Use binary search to find the smallest x such that the total number of operations required to ensure no bag has more than x balls is less than or equal to maxOperations.

2. Helper Function

• Write a function to calculate the number of operations required to ensure no bag exceeds a given size x.
• If a bag has k balls, and k > x, the number of splits needed is: $operations = \lceil\frac{k}{x}\rceil - 1$

3. Optimal x:

• Perform binary search over the range $\lceil 1, \max(\text{nums}) \rceil$.
• For each x, calculate the total operations needed using the helper function.
• If the total operations are within maxOperations, update the result and try smaller x.

Explanation:

1. Binary Search

• Start with the range $[1, \max(\text{nums})]$.
• For each midpoint (x), check if it is possible to split the bags such that no bag exceeds size x within maxOperations.

2. Helper function

• For each bag, calculate the required operations:
• If k > x, the number of splits needed is: $\lceil \frac{k}{x} \rceil - 1 = (k-1) / x$
• If the total operations exceed maxOperations, x is invalid.

3. Result update

• If x is valid, update the result and try smaller penalties.
• Otherwise, increase x.

Complexity

• Time complexity: $O(n \cdot \log(\max(\text{nums})))$, where n is the length of the array nums.
• Space complexity: $O(1)$,

Code

```go [] func possible(nums []int, x int, maxOperations int) bool { for _, num := range nums { count := (num - 1) / x if maxOperations < count { return false } maxOperations -= count } return true }

func minimumSize(nums []int, maxOperations int) int { left, right := 1, 0 for _, num := range nums { right = max(right, num) }

for left <= right {
    mid := left + (right - left) / 2
    if possible(nums, mid, maxOperations) {
        right = mid - 1
    } else {
        left = mid + 1
    }
}
return left

} rust [] impl Solution { fn possible(nums: &Vec, x: i32, max_operations: i32) -> bool { let mut max_operations = max_operations; for &num in nums { let count = (num - 1) / x; if count > max_operations { return false; } max_operations -= count; } true } pub fn minimum_size(nums: Vec, max_operations: i32) -> i32 { let (mut left, mut right) = (1, 0); for &num in &nums { right = right.max(num); } while left <= right { let mid = left + (right - left) / 2; if Self::possible(&nums, mid, max_operations) { right = mid - 1; } else { left = mid + 1; } } left } } ```