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The problem can be solved efficiently using a binary search on the minimum magnetic force. Start by sorting the basket positions, then perform a binary search to find the maximum valid minimum force. The basic idea is to check if it is possible to place all the balls with at least a certain minimum distance (force) between any two of them. If it is possible, try for a larger minimum distance, otherwise try smaller.
Time Complexity: O(n log(n) + log(maxDistance) * n), where n is the number of baskets.
Space Complexity: O(1)
1#include <stdio.h>
2#include <stdlib.h>
3
4int compare(const void *a, const void *b) {
5 return (*(int*)a - *(int*)b);
6}
7
8int canPlaceBalls(int *position, int n, int m, int minDist) {
9 int count = 1, lastPosition = position[0];
10 for (int i = 1; i < n; i++) {
11 if (position[i] - lastPosition >= minDist) {
12 count++;
13 lastPosition = position[i];
14 if (count >= m)
15 return 1;
16 }
17 }
18 return 0;
19}
20
21int maxDistance(int* position, int n, int m) {
22 qsort(position, n, sizeof(int), compare);
23 int left = 1, right = position[n - 1] - position[0];
24 int answer = 0;
25 while (left <= right) {
26 int mid = (left + right) / 2;
27 if (canPlaceBalls(position, n, m, mid)) {
28 answer = mid;
29 left = mid + 1;
30 } else {
31 right = mid - 1;
32 }
33 }
34 return answer;
35}
36
37int main() {
38 int position[] = {1, 2, 3, 4, 7};
39 int n = 5;
40 int m = 3;
41 printf("%d\n", maxDistance(position, n, m));
42 return 0;
43}This C solution implements a binary search method to find the maximum minimum force. We first sort the basket positions, then binary search for the largest possible minimum distance using a helper function `canPlaceBalls` which checks if we can place all balls with at least `minDist` distance apart.
This approach involves using a binary search for determining the optimal force, aided by a greedy strategy to verify if a particular minimum force is attainable. By iterating over the sorted positions, balls are placed as far apart as possible greedily.
Time Complexity: O(n log(n) + n log(maxDistance)), where n is the number of baskets.
Space Complexity: O(1)
1#include <iostream>
#include <vector>
#include <algorithm>
bool canPlaceBalls(const std::vector<int>& position, int m, int dist) {
int count = 1, last_position = position[0];
for (size_t i = 1; i < position.size(); ++i) {
if (position[i] - last_position >= dist) {
count++;
last_position = position[i];
if (count == m) {
return true;
}
}
}
return false;
}
int maxDistance(std::vector<int>& position, int m) {
std::sort(position.begin(), position.end());
int left = 1, right = position.back() - position.front();
while (left < right) {
int mid = left + (right - left + 1) / 2;
if (canPlaceBalls(position, m, mid)) {
left = mid;
} else {
right = mid - 1;
}
}
return left;
}
int main() {
std::vector<int> position = {5, 4, 3, 2, 1, 1000000000};
int m = 2;
std::cout << maxDistance(position, m) << std::endl;
return 0;
}The C++ solution follows a binary search augmented by a greedy placing method to determine the largest minimum force. The `canPlaceBalls` method applies a linear traversal to determine if the midpoint force is feasible.