How do you solve Magnetic Force Between Two Balls on LeetCode?

Magnetic Force Between Two Balls (LeetCode #1552) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log(max_distance)) time complexity and O(1) space complexity. Key insight: Sorting the positions helps in easily calculating distances.

What is the brute force solution for Magnetic Force Between Two Balls?

The brute force approach for Magnetic Force Between Two Balls is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach tries every possible combination of placing the balls in the baskets and calculates the minimum magnetic force for each configuration. This is simple but inefficient as it checks all possible placements. The optimal Optimal Solution approach improves this to O(n log(max_distance)).

What algorithm or data structure is used to solve Magnetic Force Between Two Balls?

Magnetic Force Between Two Balls uses the following concepts: Array, Binary Search, Sorting. The recommended patterns to study are: Binary Search, Greedy Algorithm.

What are the interview tips for Magnetic Force Between Two Balls?

Always consider edge cases, such as the minimum and maximum positions. Explain your thought process clearly while coding. Practice writing both brute-force and optimal solutions to understand the problem deeply.

What are common mistakes when solving Magnetic Force Between Two Balls?

Not sorting the positions before applying binary search. Confusing the conditions for placing balls with minimum force.

#1552

Magnetic Force Between Two Balls

Medium

ArrayBinary SearchSortingBinary SearchGreedy Algorithm

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log(max_distance))
Space	O(1)	O(1)

💡

Intuition

Time O(n log(max_distance))Space O(1)

The optimal solution uses binary search to find the maximum minimum magnetic force. By checking if a certain force can be achieved, we can efficiently narrow down the possible values.

⚙️

Algorithm

3 steps

1Step 1: Sort the positions array.
2Step 2: Use binary search on the possible minimum forces (from 1 to max(position) - min(position)).
3Step 3: For each mid value in binary search, check if it's possible to place m balls with at least mid distance apart using a greedy approach.

solution.py24 lines

1# Full working Python code
2def canPlaceBalls(position, m, min_force):
3    count = 1
4    last_position = position[0]
5    for i in range(1, len(position)):
6        if position[i] - last_position >= min_force:
7            count += 1
8            last_position = position[i]
9            if count == m:
10                return True
11    return False
12
13def maxMinForce(position, m):
14    position.sort()
15    left, right = 1, position[-1] - position[0]
16    result = 0
17    while left <= right:
18        mid = (left + right) // 2
19        if canPlaceBalls(position, m, mid):
20            result = mid
21            left = mid + 1
22        else:
23            right = mid - 1
24    return result

ℹ

Complexity note: The time complexity is O(n log(max_distance)) because we perform a binary search on the distance (which can be at most the difference between the max and min positions) and for each mid value, we check if we can place the balls in O(n) time.

1Sorting the positions helps in easily calculating distances.
2Binary search allows us to efficiently find the maximum minimum force.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.