Magnetic Force Between Two Balls

class Solution {
public:
    bool isPossibleToPlace(int minGap, vector<int> &positions, int count) {
        int lastPlaced = positions[0];
        int placed = 1;
    
        for (int index = 1; index < positions.size() && placed < count; ++index) {
            if (positions[index] - lastPlaced >= minGap) {
                placed++;
                lastPlaced = positions[index];
            }
        }
        return placed == count;
    }
    
    int maximalDistance(vector<int> &positions, int count) {
        int result = 0;
        int size = positions.size();
        sort(positions.begin(), positions.end());
    
        int minDist = 1;
        int maxDist = ceil(positions[size - 1] / (count - 1.0));
        while (minDist <= maxDist) {
            int mid = minDist + (maxDist - minDist) / 2;
            if (isPossibleToPlace(mid, positions, count)) {
                result = mid;
                minDist = mid + 1;
            } else {
                maxDist = mid - 1;
            }
        }
        return result;
    }
};

This solution in C++ aims to determine the maximum minimal distance ('maximal distance') between any two of the count balls positioned on a linear track, represented by the vector positions. The function maximalDistance computes this distance by employing a binary search strategy combined with greedy placement.

• The code contains two primary functions:

  • isPossibleToPlace(minGap, positions, count): Checks if it's possible to place the specified number of balls such that the minimum gap between any two consecutive balls is at least minGap.

  • maximalDistance(positions, count): Determines the maximal distance possible under the placement constraint.

• In maximalDistance, iteratively adjust your search range for the minimal distance:

  1. Sort the positions vector for sequential consideration.
  2. Initialize the searching boundaries minDist and maxDist. The maximum bound maxDist is initially set as the potential maximal gap considering equal distribution along the distances, determined by floor(positions.back() / (count - 1.0)).
  3. Use a binary search mechanism:
     • Compute the midpoint mid of current minDist and maxDist.
     • Use isPossibleToPlace to verify if it’s feasible to position the balls with at least mid distance apart.
     • If yes, it means we can potentially increase the minimum gap, so update result to mid and increase minDist to search for a possibly larger minimum gap.
     • If not, decrease maxDist to search with tighter constraints.

This approach leverages the efficiency of binary search to find the optimal placement solution quickly in conjunction with a greedy checking mechanism, ensuring that the balls are spaced as far apart as possible according to the provided conditions. The sorted nature of positions and iterative adjustment of distance boundaries help efficiently pinpoint the maximal minimal distance satisfying the placement conditions.

class Solution {
    
    private boolean isPossible(int dist, int[] positions, int count) {
        int lastPos = positions[0];
        int placedBalls = 1;
    
        for (int i = 1; i < positions.length && placedBalls < count; ++i) {
            int current = positions[i];
            if (current - lastPos >= dist) {
                placedBalls += 1;
                lastPos = current;
            }
        }
        return placedBalls == count;
    }
    
    public int findMaxDistance(int[] positions, int count) {
        int result = 0;
        int len = positions.length;
        Arrays.sort(positions);
        int low = 1;
        int high = (int) Math.ceil(positions[len - 1] / (count - 1.0));
    
        while (low <= high) {
            int mid = low + (high - low) / 2;
            if (isPossible(mid, positions, count)) {
                result = mid;
                low = mid + 1;
            } else {
                high = mid - 1;
            }
        }
        return result;
    }
}

The given Java solution solves for the maximum distance possible between any two balls placed in certain positions on a linear scale. This is achieved under the constraint that the balls are placed in such a way that the minimum distance between any two adjacent balls is maximized. The implementation is a classic application of binary search to optimize placement.

• Starts by sorting the array of positions.
• Uses binary search to determine the maximum minimum distance that can be maintained between the balls:
  • Initializes variables low to 1 (since distance cannot be zero) and high such that it assumes that balls are placed at equal maximum intervals.
  • Continually adjusts the range based on the binary search middle value (mid) by checking if it's possible to place all count balls with at least mid distance between each.
  • The method isPossible assists in checking placement feasibility by iterating through positions and ensuring that each subsequent ball is placed at least dist away from the last placed ball.
• Returns the maximum feasible dist as result when the binary search concludes.

Here, findMaxDistance is the main function that employs isPossible to successively guess and check distances using binary search logic, refining the balls' placement until the optimal solution is found. This approach efficiently narrows down the search space, making the solution both effective and optimal for the outlined problem.

class Solution:
    def canDistribute(self, minimum_gap, positions, count):
        last_position = positions[0]
        placed_count = 1
    
        for index in range(1, len(positions)):
            current_position = positions[index]
            if current_position - last_position >= minimum_gap:
                placed_count += 1
                last_position = current_position
            if placed_count == count:
                return True
        return False
    
    def findMaxDistance(self, positions: List[int], count: int) -> int:
        max_distance = 0
        positions.sort()
        start = 1
        end = positions[-1] // (count - 1) + 1
    
        while start <= end:
            mid = start + (end - start) // 2
            if self.canDistribute(mid, positions, count):
                max_distance = mid
                start = mid + 1
            else:
                end = mid - 1
        return max_distance

The Python solution solves the problem of finding the maximum minimum distance between any two of the placed balls on a line with certain positions. The approach uses binary search and a helper function to determine feasibility.

Solution Steps:

1. Define the helper function canDistribute to check if it's possible to place the specified number of balls with at least the given minimum distance apart. Implement this function to iterate over sorted positions of balls, maintaining a tally of successfully placed balls.

2. Sort the list of ball positions initially to streamline the distance comparisons.

3. Set the initial search range for binary search: start as 1 (minimum possible distance) and end calculated by dividing the distance spanned by the positions by (count - 1) and adding 1 for safety.

4. Conduct the binary search:

   • Calculate the midpoint mid which represents the current guess for the minimum distance.
   • Use canDistribute to check if it's possible to place all balls such that they are at least mid units apart.
   • If canDistribute returns True, update max_distance to mid and explore larger distances by setting start to mid + 1.
   • Otherwise, reduce the search range to smaller distances by setting end to mid - 1.

5. The result, max_distance, will contain the largest value of the minimum gap where all balls can still be placed properly.

The algorithm employs sorting followed by binary search, making it efficient for this kind of optimization problem where direct computation would be less feasible. This method efficiently narrows down the range of possible distances, quickly converging on the optimal solution.