Minimize the maximum difference between the heights - GeeksforGeeks

Minimize the maximum difference between the heights - GeeksforGeeks
Given heights of n towers and a value k. We need to either increase or decrease height of every tower by k (only once) where k > 0. The task is to minimize the difference between the heights of the longest and the shortest tower after modifications, and output this difference.

1. Sort array in increasing order
2. Initialize maximum and minimum elements.
   maxe = arr[n-1], mine = arr[0]
3. If k is more than difference between maximum and minimum, add/subtract k to all elements as k cannot decrease the difference. Example {6, 4}, k = 10.
4. In sorted array, update first and last elements.
   arr[0] += k; // arr[0] is minimum and k is +ve
   arr[n-1] -= k; // arr[n-1] is maximum and k is -ve
5. Initialize mac and min of modified array (only two elements have been finalized)
   new_max = max(arr[0], arr[n-1]), new_min = min(arr[0], arr[n-1])
6. Finalize middle n-2 elements. Do following for every element arr[j] where j lies from 1 to n-2.
   • If current element is less than min of modified array, add k.
   • Else If current element is more than max of modified array, subtract k.
   • arr[j] is between new_min and new_max.
     1. If arr[j] is closer to new_max, subtract k
     2. Else add k to arr[j].
7. Update new_max and new_min if required
   new_max = max(arr[j], new_max), new_min = min(arr[j], new_min)
8. Returns difference between new_max and new_min
   return (new_max – new_min);

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

{

    // There should be at least two elements

    if (n <= 1)

        return 0;

    // Sort array in increasing order

    sort(arr, arr+n);

    // Initialize maximum and minimum

    int maxe = arr[n-1];

    int mine = arr[0];

    // If k is more than difference between maximum

    // and minimum, add/subtract k to all elements

    // as k cannot decrease the difference

    if (k >= maxe - mine)

    {

        for (int i=0; i<n; i++)

            arr[i] += k; // Subtract would also work

        return (maxe - mine);

    }

    // In sorted array, first element is minimum

    // and last is maximum, we must add k to minium

    // and subtract k from maximum

    arr[0]   += k;

    arr[n-1] -= k;

    // Initialize mac and min of modified array (only

    // two elements have been finalized)

    int new_max = max(arr[0], arr[n-1]);

    int new_min = min(arr[0], arr[n-1]);

    // Finalize middle n-2 elements

    for (int j=1; j<n-1; j++)

    {

        // If current element is less than min of

        // modified array, add k.

        if (arr[j] < new_min)

            arr[j] += k;

        // If current element is more than max of

        // modified array, subtract k.

        else if (arr[j] > new_max)

            arr[j] -= k;

        // arr[j] is between new_min and new_max

        // If arr[j] is closer to new_max, subtract k

        else if ((arr[j] - new_min) > (new_max - arr[j]))

            arr[j] -= k;

        // Else add k

        else

            arr[j] += k;

        // Update new_max and new_min if required

        new_max = max(arr[j], new_max);

        new_min = min(arr[j], new_min);

    }

    // Returns difference between new_max and new_min

    return (new_max - new_min);

}

This approach mentioned above is doing sorting just for getting min and max for that sorting is not required so just modify the above code and it will take O(n) time.

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