# 1552. Magnetic Force Between Two Balls

## Description

In the universe Earth C-137, Rick discovered a special form of magnetic force between two balls if they are put in his new invented basket. Rick has n empty baskets, the ith basket is at position[i], Morty has m balls and needs to distribute the balls into the baskets such that the minimum magnetic force between any two balls is maximum.

Rick stated that magnetic force between two different balls at positions x and y is |x - y|.

Given the integer array position and the integer m. Return the required force.

Example 1:

Input: position = [1,2,3,4,7], m = 3
Output: 3
Explanation: Distributing the 3 balls into baskets 1, 4 and 7 will make the magnetic force between ball pairs [3, 3, 6]. The minimum magnetic force is 3. We cannot achieve a larger minimum magnetic force than 3.


Example 2:

Input: position = [5,4,3,2,1,1000000000], m = 2
Output: 999999999
Explanation: We can use baskets 1 and 1000000000.


Constraints:

• n == position.length
• 2 <= n <= 105
• 1 <= position[i] <= 109
• All integers in position are distinct.
• 2 <= m <= position.length

## Solutions

• class Solution {
public int maxDistance(int[] position, int m) {
Arrays.sort(position);
int left = 1, right = position[position.length - 1];
while (left < right) {
int mid = (left + right + 1) >>> 1;
if (check(position, mid, m)) {
left = mid;
} else {
right = mid - 1;
}
}
return left;
}

private boolean check(int[] position, int f, int m) {
int prev = position[0];
int cnt = 1;
for (int i = 1; i < position.length; ++i) {
int curr = position[i];
if (curr - prev >= f) {
prev = curr;
++cnt;
}
}
return cnt >= m;
}
}

• class Solution {
public:
int maxDistance(vector<int>& position, int m) {
sort(position.begin(), position.end());
int left = 1, right = position[position.size() - 1];
while (left < right) {
int mid = (left + right + 1) >> 1;
if (check(position, mid, m))
left = mid;
else
right = mid - 1;
}
return left;
}

bool check(vector<int>& position, int f, int m) {
int prev = position[0];
int cnt = 1;
for (int i = 1; i < position.size(); ++i) {
int curr = position[i];
if (curr - prev >= f) {
prev = curr;
++cnt;
}
}
return cnt >= m;
}
};

• class Solution:
def maxDistance(self, position: List[int], m: int) -> int:
def check(f):
prev = position[0]
cnt = 1
for curr in position[1:]:
if curr - prev >= f:
prev = curr
cnt += 1
return cnt >= m

position.sort()
left, right = 1, position[-1]
while left < right:
mid = (left + right + 1) >> 1

if check(mid):
left = mid
else:
right = mid - 1
return left


• func maxDistance(position []int, m int) int {
sort.Ints(position)
left, right := 1, position[len(position)-1]
check := func(f int) bool {
prev, cnt := position[0], 1
for _, curr := range position[1:] {
if curr-prev >= f {
prev = curr
cnt++
}
}
return cnt >= m
}
for left < right {
mid := (left + right + 1) >> 1
if check(mid) {
left = mid
} else {
right = mid - 1
}
}
return left
}

• /**
* @param {number[]} position
* @param {number} m
* @return {number}
*/
var maxDistance = function (position, m) {
position.sort((a, b) => {
return a - b;
});
let left = 1,
right = position[position.length - 1];
const check = function (f) {
let prev = position[0];
let cnt = 1;
for (let i = 1; i < position.length; ++i) {
const curr = position[i];
if (curr - prev >= f) {
prev = curr;
++cnt;
}
}
return cnt >= m;
};
while (left < right) {
const mid = (left + right + 1) >> 1;
if (check(mid)) {
left = mid;
} else {
right = mid - 1;
}
}
return left;
};


• function maxDistance(position: number[], m: number): number {
position.sort((a, b) => a - b);
let [l, r] = [1, position.at(-1)!];
const count = (f: number): number => {
let cnt = 1;
let prev = position[0];
for (const curr of position) {
if (curr - prev >= f) {
cnt++;
prev = curr;
}
}
return cnt;
};
while (l < r) {
const mid = (l + r + 1) >> 1;
if (count(mid) >= m) {
l = mid;
} else {
r = mid - 1;
}
}
return l;
}