LeetCode #1552 — MEDIUM

Magnetic Force Between Two Balls

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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 i^th 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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Binary Search

Example 1

[1,2,3,4,7]
3

Example 2

[5,4,3,2,1,1000000000]
2

Core Insight

What unlocks the optimal approach

If you can place balls such that the answer is x then you can do it for y where y < x.
Similarly if you cannot place balls such that the answer is x then you can do it for y where y > x.
Binary search on the answer and greedily see if it is possible.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1552: Magnetic Force Between Two Balls
class Solution {
    private int[] position;

    public int maxDistance(int[] position, int m) {
        Arrays.sort(position);
        this.position = position;
        int l = 1, r = position[position.length - 1];
        while (l < r) {
            int mid = (l + r + 1) >> 1;
            if (count(mid) >= m) {
                l = mid;
            } else {
                r = mid - 1;
            }
        }
        return l;
    }

    private int count(int f) {
        int prev = position[0];
        int cnt = 1;
        for (int curr : position) {
            if (curr - prev >= f) {
                ++cnt;
                prev = curr;
            }
        }
        return cnt;
    }
}

// Accepted solution for LeetCode #1552: Magnetic Force Between Two Balls
func maxDistance(position []int, m int) int {
	sort.Ints(position)
	return sort.Search(position[len(position)-1], func(f int) bool {
		prev := position[0]
		cnt := 1
		for _, curr := range position {
			if curr-prev >= f {
				cnt++
				prev = curr
			}
		}
		return cnt < m
	}) - 1
}

# Accepted solution for LeetCode #1552: Magnetic Force Between Two Balls
class Solution:
    def maxDistance(self, position: List[int], m: int) -> int:
        def check(f: int) -> bool:
            prev = -inf
            cnt = 0
            for curr in position:
                if curr - prev >= f:
                    prev = curr
                    cnt += 1
            return cnt < m

        position.sort()
        l, r = 1, position[-1]
        return bisect_left(range(l, r + 1), True, key=check)

// Accepted solution for LeetCode #1552: Magnetic Force Between Two Balls
struct Solution;

impl Solution {
    fn max_distance(mut position: Vec<i32>, m: i32) -> i32 {
        position.sort_unstable();
        let n = position.len();
        let mut low = std::i32::MAX;
        for w in position.windows(2) {
            low = low.min(w[1] - w[0]);
        }
        let mut high = (position[n - 1] - position[0]) / (m - 1) as i32;
        while low < high {
            let mid = (low + high + 1) / 2;
            let mut prev = position[0];
            let mut count = 1;
            for i in 1..n {
                if position[i] - prev >= mid {
                    count += 1;
                    prev = position[i];
                }
            }
            if count < m {
                high = mid - 1;
            } else {
                low = mid;
            }
        }
        low
    }
}

#[test]
fn test() {
    let position = vec![1, 2, 3, 4, 7];
    let m = 3;
    let res = 3;
    assert_eq!(Solution::max_distance(position, m), res);
    let position = vec![1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
    let m = 4;
    let res = 3;
    assert_eq!(Solution::max_distance(position, m), res);
}

// Accepted solution for LeetCode #1552: Magnetic Force Between Two Balls
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;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n + n × log M)

Space

O(log n)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.