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Magnetic Force Between Two Balls

Difficulty: Medium

Problem Description

In the universe Earth C-137, Rick discovered a special form of magnetic force between two balls if they are put in his newly 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. The 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.

Key Insights

  • The problem can be interpreted as a distribution problem where we want to maximize the minimum distance between the balls.
  • Binary search can be applied to find the maximum possible minimum magnetic force.
  • The key is to check if we can place m balls in such a way that the minimum distance between any two balls is at least a certain value.

Space and Time Complexity

Time Complexity: O(n log(max_distance)), where n is the number of baskets and max_distance is the difference between the maximum and minimum positions in the position array.
Space Complexity: O(1), as we are using a constant amount of extra space.

Solution

The solution involves a binary search for the maximum minimum distance. We first sort the position array. Then, we use binary search to find the largest minimum distance d such that we can place all m balls while maintaining at least d distance between any two balls. The approach involves a helper function that checks if it's feasible to place the balls given a distance d.

Code Solutions

def maxDistance(position, m):
    position.sort()  # Sort the basket positions
    left, right = 1, position[-1] - position[0]  # Possible minimum distance range

    def canPlaceBalls(distance):
        count = 1  # Place the first ball in the first basket
        last_position = position[0]
        
        for i in range(1, len(position)):
            if position[i] - last_position >= distance:  # Check distance condition
                count += 1
                last_position = position[i]  # Update last_position
                if count == m:  # Successfully placed all balls
                    return True
        return False

    # Binary search to find the maximum minimum distance
    while left <= right:
        mid = (left + right) // 2
        if canPlaceBalls(mid):  # If we can place balls with 'mid' distance
            left = mid + 1  # Try for a larger distance
        else:
            right = mid - 1  # Try for a smaller distance

    return right  # The largest minimum distance found
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