Place k elements such that minimum distance is maximized - GeeksforGeeks

Place k elements such that minimum distance is maximized - GeeksforGeeks
Given an array representing n positions along a straight line. Find k (where k <= n) elements from the array such that the minimum distance between any two (consecutive points among the k points) is maximized.

Input : arr[] = {1, 2, 8, 4, 9}
            k = 3
Output : 3
Largest minimum distance = 3
3 elements arranged at positions 1, 4 and 8, 
Resulting in a minimum distance of 3

A Naive Solution is to consider all subsets of size 3 and find minimum distance for every subset. Finally return the largest of all minimum distances.

An Efficient Solution is based on Binary Search. We first sort the array. Now we know maximum possible value result is arr[n-1] – arr[0] (for k = 2). We do binary search for maximum result for given k. We start with middle of maximum possible result. If middle is a feasible solution, we search on right half of mid. Else we search is left half. To check feasibility, we place k elements under given mid distance.

https://stackoverflow.com/questions/27971223/finding-largest-minimum-distance-among-k-objects-in-n-possible-distinct-position

1. Let's do a binary search over the answer.
2. For a fixed answer x, we can check whether it is feasible or not using a simple linear greedy algorithm(pick the first element and then iterate over the rest of the array adding the current element if the distance between it and the last picked element is greater than or equal to x). In the end, we just need to check that the number of picked elements is at least k.

The time complexity is O(n * log MAX_A), where MAX_A is the maximum element of the array.

Here is a pseudo code for this algorithm:

def isFeasible(positions, dist, k):
    taken = 1
    last = positions[0]
    for i = 1 ... positions.size() - 1:
        if positions[i] - last >= dist:
            taken++
            last = positions[i]
    return taken >= k

def solve(positions, k):
    low = 0 // definitely small enough
    high = maxElement(positions) - minElement(positions) + 1 // definitely too big
    while high - low > 1:
        mid = (low + high) / 2
        if isFeasible(positions, mid, k):
            low = mid
        else:
            high = mid
    return low

// Returns true if it is possible to arrange

// k elements of arr[0..n-1] with minimum distance

// given as mid.

bool isFeasible(int mid, int arr[], int n, int k)

{

    // Place first element at arr[0] position

    int pos = arr[0];

    // Initialize count of elements placed.

    int elements = 1;

    // Try placing k elements with minimum

    // distance mid.

    for (int i=1; i<n; i++)

    {

        if (arr[i] - pos >= mid)

        {

            // Place next element if its

            // distance from the previously

            // placed element is greater

            // than current mid

            pos = arr[i];

            elements++;

            // Return if all elements are placed

            // successfully

            if (elements == k)

              return true;

        }

    }

    return 0;

}

// Returns largest minimum distance for k elements

// in arr[0..n-1]. If elements can't be placed,

// returns -1.

int largestMinDist(int arr[], int n, int k)

{

    // Sort the positions

    sort(arr,arr+n);

    // Initialize result.

    int res = -1;

    // Consider the maximum possible distance

    int left = arr[0], right = arr[n-1];

    // Do binary search for largest minimum distance

    while (left < right)

    {

        int mid = (left + right)/2;

        // If it is possible to place k elements

        // with minimum distance mid, search for

        // higher distance.

        if (isFeasible(mid, arr, n, k))

        {

            // Change value of variable max to mid iff

            // all elements can be successfully placed

            res = max(res, mid);

            left = mid + 1;

        }

        // If not possible to place k elements, search

        // for lower distance

        else

            right = mid;

    }

    return res;

}

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