Magnetic Force Between Two Balls

Medium

Apple

1 view

Topics:

ArraysBinary SearchGreedy Algorithms

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:

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:

position = [5,4,3,2,1,1000000000], m = 2

Output: 999999999

Explanation: We can use baskets 1 and 1000000000.

Can you provide an efficient algorithm to solve this problem?

Magnetic Force Between Two Balls

Medium

Apple

1 view

Topics:

ArraysBinary SearchGreedy Algorithms

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:

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:

position = [5,4,3,2,1,1000000000], m = 2

Output: 999999999

Explanation: We can use baskets 1 and 1000000000.

Can you provide an efficient algorithm to solve this problem?

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of possible values for the positions in the input array `position`?
Can the input array `position` contain duplicate values?
Is it guaranteed that `m` (the number of balls) will always be less than or equal to the length of the `position` array?
If no possible minimum magnetic force can be achieved (perhaps due to invalid input), what value should I return?
Are the positions in the `position` array guaranteed to be sorted, or do I need to sort them first?

Brute Force Solution

Approach

To find the largest magnetic force, we'll try every possible arrangement of balls within the given positions. We calculate the minimum distance between any two balls for each arrangement and keep track of the largest minimum distance we've found so far.

Here's how the algorithm would work step-by-step:

First, pick any possible number of ball positions from the available positions.
Then, arrange the balls in every possible combination using the positions we picked in the previous step.
For each of these combinations, calculate the smallest distance between any two of the balls.
Compare the smallest distance of the current combination to the largest minimum distance we've seen so far.
If the smallest distance of the current combination is larger, update the largest minimum distance.
Repeat for all possible combinations of ball positions.
The final largest minimum distance is our answer.

Code Implementation

def max_magnetic_force_brute_force(positions, number_of_balls):
    max_min_distance = 0
    number_of_positions = len(positions)

    # Iterate through all possible combinations of ball positions
    for i in range(1 << number_of_positions):
        if bin(i).count('1') != number_of_balls:
            continue

        ball_positions = []
        for j in range(number_of_positions):
            if (i >> j) & 1:
                ball_positions.append(positions[j])

        # Handle the base case where there are less than 2 balls selected.
        if len(ball_positions) < 2:
            continue

        min_distance = float('inf')

        # Calculate min distance for current positions.
        for first_ball_index in range(len(ball_positions)):
            for second_ball_index in range(first_ball_index + 1, len(ball_positions)):
                distance = abs(ball_positions[first_ball_index] - ball_positions[second_ball_index])
                min_distance = min(min_distance, distance)

        # Update the max min distance.
        max_min_distance = max(max_min_distance, min_distance)

    return max_min_distance

Big(O) Analysis

Time Complexity

O(2^n * C(n,m) * m) – The algorithm iterates through all possible subsets of positions to place the balls, which takes O(2^n) time where n is the number of positions. For each subset, if we choose m positions (number of balls), we arrange the balls in C(n,m) ways. Finding the minimum distance between m balls takes O(m) time, where m is less than or equal to n. Therefore, in the worst case, the total time complexity becomes O(2^n * C(n,m) * m).

Space Complexity

O(N) – The described solution explores all possible combinations of ball positions. This implies storing subsets of the input positions. In the worst-case scenario, the algorithm might temporarily store a subset of size N, where N is the number of available positions, to calculate the minimum distance. Therefore, the auxiliary space used for storing these subsets grows linearly with the input size N. Thus, the space complexity is O(N).

Optimal Solution

Approach

The goal is to maximize the smallest distance between any two balls. Instead of checking every possible arrangement, we'll use a method to efficiently search for the largest possible minimum distance that still allows all the balls to fit within the given space.

Here's how the algorithm would work step-by-step:

Think about the smallest possible distance between the balls and the largest possible distance. The answer we're looking for must be somewhere in between.
Pick a distance somewhere in the middle of that range and try to place the balls so that the minimum distance between any two balls is at least that chosen distance.
To see if we can place the balls with that minimum distance, put the first ball at the first position, then place the next ball at the furthest possible spot (greater or equal to our minimum distance) and repeat for all balls.
If we can fit all the balls using that minimum distance, that means the actual answer might be even larger, so try a bigger distance.
If we can't fit all the balls, the actual answer must be smaller, so try a smaller distance.
Keep narrowing down the range until you find the largest minimum distance that still allows you to place all the balls. This is like guessing a number between 1 and 100 and being told 'higher' or 'lower' after each guess.

Code Implementation

def max_magnetic_force(positions, number_of_balls):
    positions.sort()
    left_bound = 1
    right_bound = positions[-1] - positions[0]

    result = -1

    while left_bound <= right_bound:
        mid_value = left_bound + (right_bound - left_bound) // 2

        # Check if we can place balls with min dist of mid_value
        if can_place_balls(positions, number_of_balls, mid_value):

            # If we can, try a larger distance
            result = mid_value
            left_bound = mid_value + 1

        else:
            # If we cannot, try a smaller distance
            right_bound = mid_value - 1

    return result

def can_place_balls(positions, number_of_balls, min_distance):
    balls_placed = 1
    last_position = positions[0]

    for i in range(1, len(positions)):
        # Place ball if the distance from last ball is enough
        if positions[i] - last_position >= min_distance:
            balls_placed += 1
            last_position = positions[i]

            # No need to check further           if balls_placed == number_of_balls:
                return True

    return False

Big(O) Analysis

Time Complexity

O(n log(m)) – The solution employs a binary search algorithm to find the maximum possible minimum distance. The search space is defined by 'm', which represents the difference between the largest and smallest positions in the input array. For each iteration of the binary search, we attempt to place 'n' balls with a given minimum distance. Placing the balls involves iterating through the sorted array once, taking O(n) time. Since the binary search reduces the search space by half at each step, it takes O(log(m)) iterations. Therefore, the overall time complexity is O(n log(m)).

Space Complexity

O(1) – The algorithm's space complexity is determined by the extra variables needed during the binary search and the placement simulation. The plain English explanation describes placing balls iteratively, updating the last placed position, and checking if all balls can be placed. This requires a constant number of variables, like the last placed position and the count of placed balls, irrespective of the input size (positions of the balls and number of balls). Therefore, the space used is constant.

Edge Cases

Case	How to Handle
positions array is null or empty	Return 0 immediately since no balls exist to calculate force.
m is less than or equal to 0	Return 0 immediately since we need at least one group of balls.
m is greater than the number of positions	Return 0 immediately as it's impossible to place more balls than available positions.
positions array contains duplicate values	The binary search and greedy approach will still function correctly, although the minimal magnetic force may involve balls at the same physical location if allowed, handle as a valid input.
positions array contains very large values leading to integer overflow when calculating mid	Use long data type for mid calculation to avoid potential overflow issues.
Positions are not sorted in ascending order	Sort the positions array in ascending order as a preprocessing step.
All positions are clustered very closely together, requiring very high precision to determine if a force is achievable.	The binary search lower and upper bound should be integers, handle by increasing steps until a valid force is found.
The optimal distance is extremely large, close to the maximum integer value, possibly leading to an infinite loop.	Ensure the binary search range is appropriately limited to avoid infinite looping.