1552. Magnetic Force Between Two Balls


Problem Description

In a universe known as Earth C-137, Rick has invented a new type of basket which exerts a magnetic force between balls placed inside it. He has n such baskets, each located at a specific position delineated by the array position[i]. Morty, on the other hand, has m balls that he needs to place into these baskets. The goal is to distribute the balls in such a way that the minimum magnetic force between any two balls is maximized.

Magnetic force here is simply the absolute difference in the positions (|x - y|) of two different balls. Given an integer array position representing the basket positions and an integer m representing the number of balls Morty has, the task is to return the maximum minimum magnetic force that can be achieved between any two balls placed in the baskets.

Intuition

The intuition behind the solution involves utilizing binary search to efficiently find the maximum minimum magnetic force. Since we want to maximize the minimum distance between any two balls, we can search for the best distance by examining the feasible range of distances. By sorting the positions of the baskets initially, we structure our problem space in a way that enables binary search.

The process works as follows:

  • We know that the minimum possible force is 1 (when two balls are next to each other) and the maximum possible force is position[-1] minus position[0] (assuming we place the balls in the first and last basket).
  • We use binary search to probe the middle value (mid) of our current range as a potential candidate for our maximum minimum force.
  • For each mid value tried, we use the check function to see whether we can place all m balls into baskets such that no two balls are less than mid distance apart. Starting from the first basket, we place a ball and then find the next basket that is at least mid distance away to place the next ball. We continue this process until we either place all m balls or run out of baskets.
  • If we can place all m balls with at least mid distance between them, it means that a larger distance could also be feasible, and we shift our search range upwards.
  • If we cannot place all m balls with at least mid distance, it means we need to look for a smaller minimum force, and we shift our search range downwards.
  • This process continues until we pinpoint the largest minimum distance that can accommodate all m balls, which is our desired maximum minimum force.

The idea is akin to finding the "sweet spot" in terms of distance, which is the largest minimum distance that still allows for all m balls to be distributed in the baskets without violating the minimum distance restriction.

Learn more about Binary Search and Sorting patterns.

Solution Approach

The solution employs a binary search algorithm, a commonly used pattern for problems where one must find an element in a sorted array or, like in our case, when trying to decide on the most suitable value within a certain range. Here's how it works step by step:

  • Sorting: First, we sort the position array. This allows us to apply binary search since we can now reason about the baskets' positions relative to each other.

  • Initializing Binary Search: We set variables left to 1 (the minimum possible force if balls are next to each other) and right to position[-1] - position[0] (the maximum possible force with all balls spanning the complete range of baskets). These will be our search boundaries.

  • Binary Search Loop: We enter a loop to perform the binary search. The condition for continuing the loop is left < right, meaning we still have a range to explore.

  • Mid Calculation: Inside the loop, mid is calculated with the expression (left + right + 1) >> 1, which is effectively (left + right + 1) / 2, but using bitwise right shift to do integer division by two. Adding 1 before dividing ensures that mid is rounded up, preventing an infinite loop in certain conditions.

  • Check Function: The check function is the heart of our binary search. It tries to place balls in the baskets starting with the first basket and moving to the right, ensuring that each new ball is at least mid distance apart from the previous. Note that cnt starts at 1 because we place the first ball in the first basket without checking the distance.

    • If it finds a position that is mid or more away from the last filled basket, it places a ball there (cnt is incremented).
    • The loop continues until we either run out of baskets or we have placed all m balls.
    • The function returns True if all m balls were placed (cnt >= m), meaning the mid value is a feasible minimum force.
  • Binary Search Decision: After each call to the check function:

    • If check(mid) returns True, it implies that mid is a valid minimum force and could possibly be increased, so we set left = mid.
    • If check(mid) returns False, it means mid is too large to fit all balls, so we decrease our search range by setting right = mid - 1.
  • Termination:

    • The loop terminates when left equals right, meaning we found the maximum left for which check(left) is True.
    • Finally, we return left, which now represents the maximum minimum force between any two balls according to the constraints.

Using binary search, we've effectively reduced what could be a very time-consuming search across all possible distances to a much faster logarithmic time complexity process. By iteratively halving our search space, we home in on the optimal solution without having to explicitly evaluate every possible distribution of balls among baskets.

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Example Walkthrough

Let's illustrate the solution approach using a small example. Suppose position = [1, 2, 4, 7, 8, 12], representing the positions of baskets, and Morty has m = 3 balls to place.

  1. Sort the position array: The array is already sorted: [1, 2, 4, 7, 8, 12].

  2. Initialize Binary Search: We set left = 1 and right = 12 - 1 = 11 as the furthest possible distance any two balls could have.

  3. Binary Search Loop: We now enter a loop where we will try to find the maximum minimum distance (mid) between balls that still allows us to place all m balls.

  4. Mid Calculation: We calculate mid as (left + right + 1) >> 1. Initially, mid = (1 + 11 + 1) >> 1 = 6.

  5. Check Function: We try to place 3 balls in the baskets ensuring that each new ball is at least mid (6) distance apart.

    • Place the first ball at position 1 (since cnt starts at 1).
    • The next ball can be placed no closer than position 7, which is 6 units away from position 1.
    • The third ball can be placed no closer than position 12, which is 5 units away from position 7. However, this distance is less than mid (6).

Since we cannot place the third ball at the required distance of mid, check(mid) returns False.

  1. Binary Search Decision:
    • Because check(mid) was False, we set right = mid - 1 = 6 - 1 = 5.

We repeat steps 4 to 6 with the new value of right. The new mid would then be (1 + 5 + 1) >> 1 = 3.

  1. New Check Function: With mid = 3, we try to place the balls again:
    • Place the first ball at position 1.
    • The next ball can be placed at position 4, which is 3 units away from position 1.
    • The third ball can be placed at position 7, which is 3 units away from position 4.

All m balls are placed with at least mid distance between them, so check(mid) returns True.

  1. Binary Search Decision:
    • As check(mid) was True, we update left = mid = 3.

Next, we search between our updated left and right. The new mid would be (3 + 5 + 1) >> 1 = 4.

Repeating the check with mid = 4, we would find that we can place balls at positions 1, 4, 8, which satisfy the condition, so we update left = mid = 4.

If we continue this process, we'd eventually narrow down the value of left and right until left equals right, at which point we find the optimal solution.

  1. Termination:
    • Eventually left equals right, meaning we found the maximum left for which check(left) is True.
    • For this example, we would find that the maximum minimum distance that can be maintained while placing all m balls is 4.
    • Therefore, we'd return 4 as our answer. This is the maximum minimum magnetic force between any two balls in the given positions.

Solution Implementation

1from typing import List
2
3class Solution:
4    def maxDistance(self, positions: List[int], m: int) -> int:
5        # Helper function to check if we can place m balls with at least 'min_distance' distance apart.
6        def can_place_balls(min_distance: int) -> bool:
7            previous_position = positions[0]
8            count = 1
9            # Iterate through the sorted positions to check
10            for current_position in positions[1:]:
11                if current_position - previous_position >= min_distance:
12                    previous_position = current_position
13                    count += 1
14            # Return True if we can place at least m balls
15            return count >= m
16
17        # Sort the positions to simplify the distance checks
18        positions.sort()
19        # Initialize the binary search bounds
20        left, right = 1, positions[-1] - positions[0]
21      
22        # Binary search to find the largest minimum distance
23        while left < right:
24            # Calculate middle value to test the placement
25            mid = (left + right + 1) // 2  # (left + right + 1) to handle the middle for even length
26          
27            # Adjust bounds based on the ability to place m balls
28            if can_place_balls(mid):
29                left = mid
30            else:
31                right = mid - 1
32      
33        # 'left' now holds the largest minimum distance we can place 'm' balls
34        return left
35
1class Solution {
2    public int maxDistance(int[] positions, int m) {
3        // Sort the positions array to establish ordered distances
4        Arrays.sort(positions);
5      
6        // Set initial search boundary for the maximum distance
7        int left = 1;                 // Minimum possible distance
8        int right = positions[positions.length - 1]; // Maximum possible distance
9      
10        // Use binary search to find the largest minimum distance between m balls
11        while (left < right) {
12            // Calculate the middle value of the current search boundary
13            int mid = (left + right + 1) >>> 1;
14          
15            // Check if it's possible to place m balls with at least 'mid' distance apart
16            if (isFeasible(positions, mid, m)) {
17                left = mid; // If possible, continue the search on the right half
18            } else {
19                right = mid - 1; // If not possible, continue the search on the left half
20            }
21        }
22      
23        // Return the maximum minimum distance found
24        return left;
25    }
26
27    // Helper method to check if m balls can be placed with at least 'distance' units apart
28    private boolean isFeasible(int[] positions, int distance, int m) {
29        // Start from the first position
30        int prevPosition = positions[0];
31        // One ball is already placed at the first position
32        int count = 1;
33      
34        // Iterate through the positions to place the rest of the balls
35        for (int i = 1; i < positions.length; ++i) {
36            int currentPosition = positions[i];
37          
38            // If the current position is at least 'distance' away from the previously placed ball
39            if (currentPosition - prevPosition >= distance) {
40                // Place the ball and move to the next
41                prevPosition = currentPosition;
42                ++count; // Increment the count of placed balls
43            }
44        }
45      
46        // If the count of placed balls is at least m, it's feasible
47        return count >= m;
48    }
49}
50
1class Solution {
2public:
3    int maxDistance(vector<int>& positions, int m) {
4        // Sort the positions vector to ensure that the elements are in increasing order.
5        sort(positions.begin(), positions.end());
6
7        // Initialize the left and right pointers for binary search.
8        // left is the smallest possible distance, right is the largest possible distance.
9        int left = 1;
10        int right = positions.back();
11
12        // Use binary search to find the maximum minimum distance.
13        while (left < right) {
14            // Calculate the middle value of the current search range.
15            int mid = (left + right + 1) / 2;
16
17            // Check if it's possible to place m balls with at least 'mid' distance apart.
18            if (canPlaceBalls(positions, mid, m))
19                left = mid; // If true, we know that we can try a larger minimum distance.
20            else
21                right = mid - 1; // If false, decrease the search range.
22        }
23
24        // The final value of left is the maximum minimum distance we can achieve.
25        return left;
26    }
27
28    bool canPlaceBalls(vector<int>& positions, int minDistance, int m) {
29        // Place the first ball at the first position.
30        int prevPosition = positions[0];
31        // Initialized at one since we already placed one ball.
32        int count = 1;
33
34        // Iterate through the positions starting from the second ball.
35        for (int i = 1; i < positions.size(); ++i) {
36            int currentPosition = positions[i];
37            // Check if placing a ball here would maintain the minimum distance 'minDistance'.
38            if (currentPosition - prevPosition >= minDistance) {
39                prevPosition = currentPosition; // Place the ball.
40                ++count; // Increment the ball count.
41            }
42        }
43
44        // If we can place at least 'm' balls, return true.
45        return count >= m;
46    }
47};
48
1/**
2 * Calculate the maximum distance between m balls placed in sorted positions.
3 * @param {number[]} positions - Array of initial positions of balls (not necessarily sorted).
4 * @param {number} m - Number of balls to be placed.
5 * @returns {number} - The largest minimum distance between any two balls.
6 */
7const maxDistance = (positions: number[], m: number): number => {
8    // First, sort the positions array in ascending order.
9    positions.sort((a, b) => a - b);
10
11    // Define the binary search boundaries.
12    let left: number = 1; // The minimum possible distance.
13    let right: number = positions[positions.length - 1]; // The maximum possible distance, which is between the first and the last position.
14
15    /**
16     * Helper function to check if we can place m balls with at least 'distance' between them.
17     * @param {number} distance - The minimum distance to maintain between any two balls.
18     * @returns {boolean} - True if it's possible to place all m balls with the minimum distance, false otherwise.
19     */
20    const canPlaceBalls = (distance: number): boolean => {
21        let count = 1; // Start by placing the first ball.
22        let prevPosition = positions[0]; // The position of the last placed ball.
23
24        // Loop through each position starting from the second one.
25        for (let i = 1; i < positions.length; ++i) {
26            const currentPosition = positions[i];
27            if (currentPosition - prevPosition >= distance) {
28                // If the current position is at least 'distance' away from the last placed ball, place a new ball here.
29                prevPosition = currentPosition;
30                ++count;
31            }
32        }
33        // If we have placed at least m balls, return true.
34        return count >= m;
35    };
36
37    // Perform a binary search to find the largest minimum distance that we can achieve.
38    while (left < right) {
39        const mid = Math.floor((left + right + 1) / 2); // Calculate the mid-point distance.
40        if (canPlaceBalls(mid)) {
41            // If we can place all balls with at least 'mid' distance between them, try a larger distance.
42            left = mid;
43        } else {
44            // If we cannot place all balls, reduce the distance.
45            right = mid - 1;
46        }
47    }
48
49    // After binary search, 'left' will represent the largest minimum distance we can achieve.
50    return left;
51};
52

Time and Space Complexity

Time Complexity

The time complexity of the maxDistance function is determined by two main operations: sorting the position list and the binary search that is performed to find the maximum distance.

  1. Sorting the position array has a time complexity of O(n log n), where n is the number of elements in the position list.
  2. The binary search runs in O(log(max(position) - min(position))) time since it operates on the range of possible answers, which is determined by the difference between the maximum and minimum elements in the sorted position array. Within each iteration of the binary search, there is a loop that operates in O(n) time to check if a given force f allows placing m magnets. This check is performed every time the binary search narrows the search space.

Thus, the total time complexity of the binary search is O(n log(max(position) - min(position))).

Combining these two operations, the overall time complexity of the maxDistance function is O(n log n) + O(n log(max(position) - min(position))). Since the binary search is dependent on the range of position values which might be larger than n itself, in practice, the dominating term depends on the specific values of position. If max(position) - min(position) is significantly larger than n, the binary search could be the more significant term.

Hence, the final time complexity is O(n log n + n log(max(position) - min(position))).

Space Complexity

The space complexity of the maxDistance function arises from the sorting operation and the additional space used by the check function.

  • Sorting the array in-place has a space complexity of O(1).
  • The check function uses constant space, only storing a few variables like prev, cnt, and curr.
  • No additional data structures that grow with the size of the input are used.

Therefore, the overall space complexity of the function is O(1), which denotes constant space complexity.

Learn more about how to find time and space complexity quickly using problem constraints.


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