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The problem is to determine the maximum possible minimum distance between any two balls when they are placed in different positions from the given array.

This problem can be efficiently solved using a binary search on the possible distances. The idea is to find the maximum minimum distance such that we can place all ( m ) balls with at least this distance between any two of them.

1. Binary Search Initialization:

   • Sort the positions array to facilitate the placement of balls in order.
   • Set low to 1, the smallest possible distance.
   • Set high to the difference between the maximum and minimum positions.

2. Binary Search Loop:

   • Calculate mid as the average of low and high.
   • Use a helper function fix to count the number of balls that can be placed with at least mid distance between them.
   • If the number of balls placed is greater than or equal to ( m ), adjust the search to find a potentially larger distance by setting low to mid + 1.
   • If the number of balls placed is less than ( m ), adjust the search to find a smaller distance by setting high to mid - 1.

3. Helper Function fix:

   • This function counts the number of balls that can be placed with at least mid distance between them.
   • It iterates through the sorted positions and places balls only if the current position is at least mid distance from the last placed ball.

class Solution {
    int fix(vector<int> &pos, int m, int mid) {
        int cnt = 1, n = pos.size();
        int curr = pos[0];

        for (int i = 1; i < n; i++) {
            if (abs(curr - pos[i]) >= mid) {
                cnt++;
                curr = pos[i];
            }
        }

        return cnt;
    }

public:
    int maxDistance(vector<int>& pos, int m) {
        int n = pos.size();
        sort(pos.begin(), pos.end());

        int low = 1, high = pos[n - 1] - pos[0];

        while (low <= high) {
            int mid = (low + high) / 2;
            if (fix(pos, m, mid) >= m) {
                low = mid + 1;
            } else {
                high = mid - 1;
            }
        }

        return high;
    }
};

Helper Function fix:

• Initializes cnt to 1 (since we place the first ball at the first position) and curr to the first position.
• Iterates through the pos array, placing a ball only if the current position is at least mid distance away from the last placed ball (curr).
• Returns the total number of balls placed.

Main Function maxDistance:

• Sorts the pos array to facilitate the placement of balls in order.
• Initializes low to 1 and high to the difference between the maximum and minimum positions in pos.
• Uses binary search to find the maximum possible minimum distance where ( m ) balls can be placed.
• Adjusts the search range based on whether placing balls with the current mid distance is feasible.

Efficiency:

• The binary search approach ensures a time complexity of O(N log D), where N is the number of positions and D is the range of distances between the minimum and maximum positions, making it efficient for large inputs.

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