Magnetic Force Between Two Balls
Binary Search
Asked at 9 companies
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Community-reported · source · updated 2026-10-08. ● asked in the last 6 months.
Go step by step: try it yourself before the video and the visual. Progress saves on this device.
0 / 130 pointsSolved (Medium) 0/40Reviews 0/50Mastered 0/25Practice steps 0/15
Step 1
Know these before you start
Practise each one here, or tick what you already know
Builds on: Two pointers from opposite ends · When sorting first unlocks the pattern
Step 2
Try it on paper
5–10 minutes, no code yet
In the wild · git bisect: To find the commit that broke a build, git halves the range of commits on every test, so even 1,000 commits take about 10 checks.
About this problem
Cut a sorted search space in half each step. Also used on answer ranges, not only arrays.
Read the full statement and examples on LeetCode, try it for 20–30 minutes, then come back for the walkthrough.
Examples
Full statement on LeetCode ↗Example 1
- Input
- position = [1, 2, 3, 4, 7]m = 3
- Output
- 3
Example 2
- Input
- position = [1, 2]m = 2
- Output
- 1
Example 3
- Input
- position = [1, 5, 9, 13]m = 4
- Output
- 4
Code it here
Tab indents · ⌘/Ctrl+Enter runs · Esc leaves the editor
Step 6
Get it accepted
All test cases pass
Step 7
Review it so it sticks
Re-solve from memory after
The step that matters most: without reviews most of a solved problem fades within weeks. Three short re-solves, each just as you start to forget, are what make it yours.
Your notes
Three lines in your own words. They come back with this problem's reviews, so you can check yourself after recalling the solution.
Solve it yourself before you look
Reading the solution first feels like progress, but it makes the next similar problem — and the interview version — much harder, because you skipped the part where you figure it out. Give it an honest 20–30 minutes. Stuck? Re-read the pattern, watch the concept video, or try the brute force first.
Hidden: optimal solution
Solution code
Optimal solutionSort the positions and on the answer, the smallest gap. For a gap g, place balls greedily from the left whenever the next position is at least g away; if all m fit, try a bigger gap.
class Solution:
def maxDistance(self, position: List[int], m: int) -> int:
position.sort()
def fits(gap: int) -> bool:
placed, last = 1, position[0]
for p in position[1:]:
if p - last >= gap:
placed += 1
last = p
return placed >= m
lo, hi = 1, position[-1] - position[0]
while lo < hi:
mid = (lo + hi + 1) // 2 # round up: we keep mid when it fits
if fits(mid):
lo = mid
else:
hi = mid - 1
return loReference solution, tested in Python, Java and C++.
Other ways to solve it
- Thought of another approach (slower, or cleverer)? Code it here: the same tests check any approach.
- Community solutions on LeetCode ↗ for many more versions, in many languages. Read them after you've solved it.