Solution Explanation: Magnetic Force Between Two Balls

This problem asks to find the maximum minimum magnetic force between any two balls when distributing m balls into n baskets located at positions given in the position array. The magnetic force between two balls is the absolute difference of their positions.

The core idea is to use binary search. The problem exhibits monotonicity: a larger minimum force implies fewer balls can be placed. Therefore, we can efficiently search for the maximum minimum force using binary search.

Algorithm:

1. Sort: Sort the position array in ascending order. This is crucial for efficiently checking if a given minimum force is achievable.

2. Binary Search: Perform a binary search on the range of possible minimum forces.

   • The lower bound is 1 (minimum possible force).
   • The upper bound is the maximum difference between consecutive positions in the sorted array (or the difference between the last and first positions for simplicity).

3. Check Feasibility (count function): In each iteration of the binary search, we need to check if a given minimum force (f) is achievable. This is done using a helper function count(f).

   • count(f) iterates through the sorted position array.
   • It keeps track of the last placed ball's position (prev).
   • For each position, it checks if the distance to the last placed ball (curr - prev) is greater than or equal to f. If it is, a ball can be placed at this position, incrementing the cnt (number of balls placed).
   • count(f) returns the total number of balls that can be placed with a minimum force of f.

4. Binary Search Logic:

   • If count(mid) (where mid is the midpoint in the binary search) is greater than or equal to m, it means we can place m balls with a minimum force of at least mid. So, we update the lower bound (l) to mid.
   • Otherwise, we update the upper bound (r) to mid - 1.

5. Result: The final value of l after the binary search represents the maximum minimum magnetic force achievable.

Time Complexity: O(n log n + n log M), where n is the number of baskets and M is the range of positions (approximately the maximum position value). The O(n log n) comes from sorting, and O(n log M) comes from the binary search and the count function called in each iteration.

Space Complexity: O(log n) due to the recursive calls in the binary search (in some implementations). If the sorting algorithm uses in-place sorting, space complexity could be considered O(1)

Code Examples (Python):

import bisect
 
class Solution:
    def maxDistance(self, position: List[int], m: int) -> int:
        def check(f: int) -> bool:
            prev = -float('inf')
            cnt = 0
            for curr in position:
                if curr - prev >= f:
                    prev = curr
                    cnt += 1
            return cnt >= m  #changed to >= for correctness
 
        position.sort()
        l, r = 1, position[-1] - position[0]  # improved upper bound
        ans = 0
        while l <= r:
            mid = (l + r) // 2
            if check(mid):
                ans = mid
                l = mid + 1
            else:
                r = mid - 1
        return ans
 

This Python code directly implements the binary search. The check function efficiently determines feasibility. The upper bound of the binary search is improved to position[-1] - position[0]. This prevents unnecessary iterations.

Other languages (Java, C++, Go, TypeScript, JavaScript) follow a similar structure, adapting the syntax and data structures accordingly. The core logic of binary search and the feasibility check remains consistent.