Minimize the maximum difference between adjacent elements in an array

Given a non-decreasing array arr[] and an integer K, the task is to remove K elements from the array such that maximum difference between adjacent element is minimum.
Note: K < N – 2

Examples:

Input: arr[] = {3, 7, 8, 10, 14}, K = 2 
Output: 2 
Explanation:
After removing elements A[0] and A[4], 
The maximum difference between adjacent elements is minimum. 
After removing elements, the remaining array is [7, 8, 10] 

Input: arr[] = [12, 16, 22, 31, 31, 38], K = 3 
Output: 6 
Explanation:
After removing elements A[3], A[4] and A[5], 
The maximum difference between adjacent elements is minimum. 
After removing elements, the remaining array is [12, 16, 22] 

Method 1: Brute Force The idea is to generate subsets of the array of size N – K and also compute the maximum difference of the adjacent elements in each subsequence. Finally, find the minimum of such maximum differences.

Below is the implementation of the above approach:

C++

// C++ implementation to find the
// minimum of the maximum difference
// of the adjacent elements after
// removing K elements from the array

#include <bits/stdc++.h>

using namespace std;

// Function to find the minimum
// of the maximum difference of the
// adjacent elements after removing
// K elements from the array
int minimumAdjacentDifference(vector<int> a,
                        int n, int k)
{
    // Initialising the
    // minimum difference
    int minDiff = INT_MAX;

    // Traversing over subsets
    // in iterative manner
    for (int i = 0; i < (1 << n); i++) {

        // Number of elements to
        // be taken in the subset
        // ON bits of i represent
        // elements not to be removed
        int cnt = __builtin_popcount(i);

        // If the removed
        // set is of size k
        if (cnt == n - k) {

            // Creating the new array
            // after removing elements
            vector<int> temp;
            for (int j = 0; j < n; j++) {
                if ((i & (1 << j)) != 0)
                    temp.push_back(a[j]);
            }
            // Maximum difference of adjacent
            // elements of remaining array
            int maxDiff = INT_MIN;
            for (int j = 0; j < temp.size() - 1; j++) {
                maxDiff = max(maxDiff,
                   temp[j + 1] - temp[j]);
            }
            minDiff = min(minDiff, maxDiff);
        }
    }
    return minDiff;
}

// Driver Code
int main()
{
    int n = 5;
    int k = 2;

    vector<int> a= { 3, 7, 8, 10, 14 };

    cout << minimumAdjacentDifference(a, n, k);
    return 0;
}

Java

// Java implementation to find the
// minimum of the maximum difference
// of the adjacent elements after
// removing K elements from the array
import java.util.*;

class GFG{

    // Function to find the minimum
    // of the maximum difference of the
    // adjacent elements after removing
    // K elements from the array
    static int minimumAdjacentDifference(int a[],
                            int n, int k)
    {
        // Initialising the
        // minimum difference
        int minDiff = Integer.MAX_VALUE;

        // Traversing over subsets
        // in iterative manner
        for (int i = 0; i < (1 << n); i++) {

            // Number of elements to
            // be taken in the subset
            // ON bits of i represent
            // elements not to be removed
            int cnt = Integer.bitCount(i);

            // If the removed
            // set is of size k
            if (cnt == n - k) {

                // Creating the new array
                // after removing elements
                 Vector<Integer> temp = new Vector<Integer>();
                for (int j = 0; j < n; j++) {
                    if ((i & (1 << j)) != 0)
                        temp.add(a[j]);
                }

                // Maximum difference of adjacent
                // elements of remaining array
                int maxDiff = Integer.MIN_VALUE;
                for (int j = 0; j < temp.size() - 1; j++) {
                    maxDiff = Math.max(maxDiff,
                    temp.get(j + 1) - temp.get(j));
                }
                minDiff = Math.min(minDiff, maxDiff);
            }
        }
        return minDiff;
    }

    // Driver Code
    public static void main(String args[])
    {
        int n = 5;
        int k = 2;

        int a[] = { 3, 7, 8, 10, 14 };

        System.out.println(minimumAdjacentDifference(a, n, k));
    }
}


// This code is contributed by AbhiThakur

Python3

# Python3 implementation to find the
# minimum of the maximum difference
# of the adjacent elements after
# removing K elements from the array
import sys

INT_MAX = sys.maxsize;
INT_MIN = -(sys.maxsize - 1)

# Function to find the minimum
# of the maximum difference of the
# adjacent elements after removing
# K elements from the array
def minimumAdjacentDifference(a, n, k) :

    # Initialising the
    # minimum difference
    minDiff = INT_MAX;

    # Traversing over subsets
    # in iterative manner
    for i in range( 1<<n) :

        # Number of elements to
        # be taken in the subset
        # ON bits of i represent
        # elements not to be removed
        cnt = bin(i).count('1');

        # If the removed
        # set is of size k
        if (cnt == n - k) :

            # Creating the new array
            # after removing elements
            temp = [];
            for j in range(n) :
                if ((i & (1 << j)) != 0) :
                    temp.append(a[j]);

            # Maximum difference of adjacent
            # elements of remaining array
            maxDiff = INT_MIN;

            for j in range(len(temp) - 1) :
                maxDiff = max(maxDiff, temp[j + 1] - temp[j]);

            minDiff = min(minDiff, maxDiff);

    return minDiff;

# Driver Code
if __name__ == "__main__" :

    n = 5;
    k = 2;

    a = [ 3, 7, 8, 10, 14 ];

    print(minimumAdjacentDifference(a, n, k));

# This code is contributed by AnkitRai01

C#

// C# implementation to find the
// minimum of the maximum difference
// of the adjacent elements after
// removing K elements from the array
using System;
using System.Collections.Generic;

class GFG{

    // Function to find the minimum
    // of the maximum difference of the
    // adjacent elements after removing
    // K elements from the array
    static int minimumAdjacentDifference(int []a,
                            int n, int k)
    {
        // Initialising the
        // minimum difference
        int minDiff = int.MaxValue;

        // Traversing over subsets
        // in iterative manner
        for (int i = 0; i < (1 << n); i++) {

            // Number of elements to
            // be taken in the subset
            // ON bits of i represent
            // elements not to be removed
            int cnt = countSetBits(i);

            // If the removed
            // set is of size k
            if (cnt == n - k) {

                // Creating the new array
                // after removing elements
                 List<int> temp = new List<int>();
                for (int j = 0; j < n; j++) {
                    if ((i & (1 << j)) != 0)
                        temp.Add(a[j]);
                }

                // Maximum difference of adjacent
                // elements of remaining array
                int maxDiff = int.MinValue;
                for (int j = 0; j < temp.Count - 1; j++) {
                    maxDiff = Math.Max(maxDiff,
                    temp[j + 1] - temp[j]);
                }
                minDiff = Math.Min(minDiff, maxDiff);
            }
        }
        return minDiff;
    }
     static int countSetBits(int x)
     {
         int setBits = 0;
         while (x != 0) {
             x = x & (x - 1);
             setBits++;
         }
         return setBits;
     }
    // Driver Code
    public static void Main(String []args)
    {
        int n = 5;
        int k = 2;

        int []a = { 3, 7, 8, 10, 14 };

        Console.WriteLine(minimumAdjacentDifference(a, n, k));
    }
}

// This code is contributed by sapnasingh4991

Javascript

<script>

// Javascript implementation to find the
// minimum of the maximum difference
// of the adjacent elements after
// removing K elements from the array
function countSetBits(x)
{
    let setBits = 0;
    while (x != 0)
    {
        x = x & (x - 1);
        setBits++;
    }
    return setBits;
}

// Function to find the minimum
// of the maximum difference of the
// adjacent elements after removing
// K elements from the array
function minimumAdjacentDifference(a, n, k)
{

    // Initialising the
    // minimum difference
    let minDiff = Number.MAX_VALUE;

    // Traversing over subsets
    // in iterative manner
    for(let i = 0; i < (1 << n); i++)
    {

        // Number of elements to
        // be taken in the subset
        // ON bits of i represent
        // elements not to be removed
        let cnt = countSetBits(i);

        // If the removed
        // set is of size k
        if (cnt == n - k)
        {

            // Creating the new array
            // after removing elements
              let temp = [];
            for(let j = 0; j < n; j++)
            {
                if ((i & (1 << j)) != 0)
                    temp.push(a[j]);
            }

            // Maximum difference of adjacent
            // elements of remaining array
            let maxDiff = Number.MIN_VALUE;
            for(let j = 0; j < temp.length - 1; j++)
            {
                maxDiff = Math.max(
                    maxDiff, temp[j + 1] - temp[j]);
            }
            minDiff = Math.min(minDiff, maxDiff);
        }
    }
    return minDiff;
}

// Driver code
let n = 5;
let k = 2;
let a = [ 3, 7, 8, 10, 14 ];

document.write(minimumAdjacentDifference(a, n, k));

// This code is contributed by divyesh072019

</script>

Output:

2

Time Complexity: O(2^N * N)
Auxiliary Space: O(N) 
Method 2: Optimal approach

• On careful observation, it can be noted that, if removal of element is done from somewhere in between the array (i.e not the end elements), then the maximum difference of remaining elements can only increase or remain the same. 
  For Example:

Let the given array be {1, 5, 6},

If we remove the element 5(not the end element), 
then the maximum difference will always increase.

Therefore, It is always better to remove end elements.

• This means that the resulting array after removing K elements will be a subarray of the original array of size N – K.
• Hence, We can iterate over all the subarrays of size N – K and for each subarray find the maximum difference between adjacent elements. Finally, find the minimum of all the maximum differences of adjacent elements.

Below is the implementation of the above approach:

C++

// C++ implementation to find the
// minimum of the maximum difference
// of the adjacent elements after
// removing K elements from the array

#include <bits/stdc++.h>

using namespace std;

// Function to find the minimum
// of the maximum difference of the
// adjacent elements after removing
// K elements from the array
int minimumAdjacentDifference(vector<int> a,
                        int n, int k)
{
    // Initialising the
    // minimum difference
    int minDiff = INT_MAX;

    // Iterating over all
    // subarrays of size n-k
    for (int i = 0; i <= k; i++) {

        // Maximum difference after
        // removing elements
        int maxDiff = INT_MIN;
        for (int j = 0; j < n - k - 1; j++) {
            for (int p = i; p <= i + j; p++) {
                maxDiff = max(maxDiff,
                     a[p + 1] - a[p]);
            }
        }
        // Minimum Adjacent Difference
        minDiff = min(minDiff, maxDiff);
    }
    return minDiff;
}

// Driver Code
int main()
{
    int n = 5;
    int k = 2;

    vector<int> a = { 3, 7, 8, 10, 14 };

    cout << minimumAdjacentDifference(a, n, k);
    return 0;
}

Java

// Java implementation to find the
// minimum of the maximum difference
// of the adjacent elements after
// removing K elements from the array
class GFG {

    // Function to find the minimum
    // of the maximum difference of the
    // adjacent elements after removing
    // K elements from the array
    static int minimumAdjacentDifference(int a[],
                            int n, int k)
    {
        // Initialising the
        // minimum difference
        int minDiff = Integer.MAX_VALUE;

        // Iterating over all
        // subarrays of size n-k
        for (int i = 0; i <= k; i++) {

            // Maximum difference after
            // removing elements
            int maxDiff = Integer.MIN_VALUE;
            for (int j = 0; j < n - k - 1; j++) {
                for (int p = i; p <= i + j; p++) {
                    maxDiff = Math.max(maxDiff,
                        a[p + 1] - a[p]);
                }
            }

            // Minimum Adjacent Difference
            minDiff = Math.min(minDiff, maxDiff);
        }
        return minDiff;
    }

    // Driver Code
    public static void main (String[] args)
    {
        int n = 5;
        int k = 2;

        int []a = { 3, 7, 8, 10, 14 };

        System.out.println(minimumAdjacentDifference(a, n, k));
    }
}

// This code is contributed by Yash_R

Python3

# Python3 implementation to find the
# minimum of the maximum difference
# of the adjacent elements after
# removing K elements from the array
import sys

INT_MAX = sys.maxsize;
INT_MIN = -(sys.maxsize - 1);

# Function to find the minimum
# of the maximum difference of the
# adjacent elements after removing
# K elements from the array
def minimumAdjacentDifference(a, n, k) :

    # Initialising the
    # minimum difference
    minDiff = INT_MAX;

    # Iterating over all
    # subarrays of size n-k
    for i in range(k + 1) :

        # Maximum difference after
        # removing elements
        maxDiff = INT_MIN;
        for j in range( n - k - 1) :
            for p in range(i, i + j + 1) :
                maxDiff = max(maxDiff, a[p + 1] - a[p]);

        # Minimum Adjacent Difference
        minDiff = min(minDiff, maxDiff);

    return minDiff;

# Driver Code
if __name__ == "__main__" :

    n = 5;
    k = 2;

    a = [ 3, 7, 8, 10, 14 ];

    print(minimumAdjacentDifference(a, n, k));

# This code is contributed by AnkitRai01

C#

// C# implementation to find the
// minimum of the maximum difference
// of the adjacent elements after
// removing K elements from the array
using System;

class GFG {

    // Function to find the minimum
    // of the maximum difference of the
    // adjacent elements after removing
    // K elements from the array
    static int minimumAdjacentDifference(int []a,
                            int n, int k)
    {
        // Initialising the
        // minimum difference
        int minDiff = int.MaxValue;

        // Iterating over all
        // subarrays of size n-k
        for (int i = 0; i <= k; i++) {

            // Maximum difference after
            // removing elements
            int maxDiff = int.MinValue;
            for (int j = 0; j < n - k - 1; j++) {
                for (int p = i; p <= i + j; p++) {
                    maxDiff = Math.Max(maxDiff,
                        a[p + 1] - a[p]);
                }
            }

            // Minimum Adjacent Difference
            minDiff = Math.Min(minDiff, maxDiff);
        }
        return minDiff;
    }

    // Driver Code
    public static void Main (string[] args)
    {
        int n = 5;
        int k = 2;

        int []a = { 3, 7, 8, 10, 14 };

        Console.WriteLine(minimumAdjacentDifference(a, n, k));
    }
}

// This code is contributed by Yash_R

Javascript

<script>

// JavaScript implementation to find the
// minimum of the maximum difference
// of the adjacent elements after
// removing K elements from the array


    // Function to find the minimum
    // of the maximum difference of the
    // adjacent elements after removing
    // K elements from the array
    function minimumAdjacentDifference(a,n,k)
    {
        // Initialising the
        // minimum difference
        let minDiff = Number.MAX_VALUE;

        // Iterating over all
        // subarrays of size n-k
        for (let i = 0; i <= k; i++) {

            // Maximum difference after
            // removing elements
            let maxDiff = Number.MIN_VALUE;
            for (let j = 0; j < n - k - 1; j++) {
                for (let p = i; p <= i + j; p++) {
                    maxDiff = Math.max(maxDiff,
                        a[p + 1] - a[p]);
                }
            }

            // Minimum Adjacent Difference
            minDiff = Math.min(minDiff, maxDiff);
        }
        return minDiff;
    }

    // Driver Code

        let n = 5;
        let k = 2;

        let a = [ 3, 7, 8, 10, 14 ];

        document.write(minimumAdjacentDifference(a, n, k));



// This code is contributed by sravan

</script>

Output:

2

Time complexity: O(N * K²)
Auxiliary Space: O(1) 

Method 3: Efficient Approach

• Using the idea from Method 2, we need to find the minimum of maximum adjacent element differences of all subarrays of size N – K. If we create a difference array, i.e. an array of differences of adjacent elements of the initial array, then all we need to do is find the minimum element of the maximum of all subarrays of size N – K – 1 of this difference array (as this maximum will represent the maximum adjacent difference of original array’s subarray of size N – K). 
  
• For performing this operation we can use the sliding window method using the double-ended queue. Refer to Sliding Window Maximum (Maximum of all subarrays of size K) for this approach.

Below is the implementation of the above approach:

C++

// C++ implementation to find the
// minimum of the maximum difference
// of the adjacent elements after
// removing K elements from the array

#include <bits/stdc++.h>

using namespace std;

// Function to find the minimum
// different in the subarrays
// of size K in the array
int findKMin(vector<int> arr, int n, int k)
{
    // Create a Double Ended Queue, Qi
    // that will store indexes
    // of array elements, queue will
    // store indexes of useful elements
    // in every window
    deque<int> Qi(k);

    // Process first k (or first window)
    // elements of array
    int i;
    for (i = 0; i < k; ++i) {
        // For every element,
        // the previous smaller elements
        // are useless so remove them from Qi
        while ((!Qi.empty()) &&
               arr[i] >= arr[Qi.back()])
            Qi.pop_back(); // Remove from rear

        // Add new element at rear of queue
        Qi.push_back(i);
    }

    int minDiff = INT_MAX;

    // Process rest of the elements,
    // i.e., from arr[k] to arr[n-1]
    for (; i < n; ++i) {
        // The element at the front
        // of the queue is the largest
        //  element of previous window
        minDiff = min(minDiff, arr[Qi.front()]);

        // Remove the elements
        // which are out of this window
        while ((!Qi.empty()) && Qi.front() <= i - k)
            Qi.pop_front();

        // Remove all elements smaller
        // than the currently being
        // added element (remove useless elements)
        while ((!Qi.empty()) &&
                arr[i] >= arr[Qi.back()])
            Qi.pop_back();

        // Add current element
        // at the rear of Qi
        Qi.push_back(i);
    }

    // compare the maximum
    // element of last window
    minDiff = min(minDiff, arr[Qi.front()]);
    return minDiff;
}

// Function to find the minimum
// of the maximum difference of the
// adjacent elements after removing
// K elements from the array
int minimumAdjacentDifference(vector<int> a,
                          int n, int k)
{

    // Create the difference array
    vector<int> diff(n-1);
    for (int i = 0; i < n - 1; i++) {
        diff[i] = a[i + 1] - a[i];
    }

    // find minimum of all maximum
    // of subarray sizes n - k - 1
    int answer = findKMin(diff,
                  n - 1, n - k - 1);
    return answer;
}

// Driver Code
int main()
{
    int n = 5;
    int k = 2;

    vector<int> a= { 3, 7, 8, 10, 14 };

    cout << minimumAdjacentDifference(a, n, k);
    return 0;
}

Java

// Java implementation to find the
// minimum of the maximum difference
// of the adjacent elements after
// removing K elements from the array
import java.util.*;
import java.lang.*;

class GFG{

// Function to find the minimum
// different in the subarrays
// of size K in the array
static int findKMin(int arr[], int n, int k)
{

    // Create a Double Ended Queue, Qi
    // that will store indexes
    // of array elements, queue will
    // store indexes of useful elements
    // in every window
    Deque<Integer> Qi = new LinkedList<>();

    // Process first k (or first window)
    // elements of array
    int i;
    for(i = 0; i < k; ++i)
    {

        // For every element,
        // the previous smaller elements
        // are useless so remove them from Qi
        while ((!Qi.isEmpty()) &&
               arr[i] >= arr[Qi.peekLast()])

            // Remove from rear
            Qi.pollLast();

        // Add new element at rear of queue
        Qi.addLast(i);
    }

    int minDiff = Integer.MAX_VALUE;

    // Process rest of the elements,
    // i.e., from arr[k] to arr[n-1]
    for(; i < n; ++i)
    {

        // The element at the front
        // of the queue is the largest
        //  element of previous window
        minDiff = Math.min(minDiff,
                           arr[Qi.peekFirst()]);

        // Remove the elements
        // which are out of this window
        while ((!Qi.isEmpty()) &&
                 Qi.peekFirst() <= i - k)
            Qi.pollFirst();

        // Remove all elements smaller
        // than the currently being
        // added element (remove useless elements)
        while ((!Qi.isEmpty()) &&
                arr[i] >= arr[Qi.peekLast()])
            Qi.pollLast();

        // Add current element
        // at the rear of Qi
        Qi.addLast(i);
    }

    // Compare the maximum
    // element of last window
    minDiff = Math.min(minDiff,
                       arr[Qi.peekFirst()]);
    return minDiff;
}

// Function to find the minimum
// of the maximum difference of the
// adjacent elements after removing
// K elements from the array
static int minimumAdjacentDifference(int a[],
                                     int n, int k)
{

    // Create the difference array
    int[] diff = new int[n - 1];
    for(int i = 0; i < n - 1; i++)
    {
        diff[i] = a[i + 1] - a[i];
    }

    // find minimum of all maximum
    // of subarray sizes n - k - 1
    int answer = findKMin(diff,
                          n - 1,
                          n - k - 1);
    return answer;
}

// Driver code
public static void main(String[] args)
{
    int n = 5;
    int k = 2;

    int a[] = { 3, 7, 8, 10, 14 };

    System.out.println(minimumAdjacentDifference(a, n, k));
}
}

// This code is contributed by offbeat

Python3

# Python3 implementation to find the
# minimum of the maximum difference
# of the adjacent elements after
# removing K elements from the array
import sys

# Function to find the minimum
# different in the subarrays
# of size K in the array
def findKMin(arr, n, k):

    # Create a Double Ended Queue, Qi
    # that will store indexes
    # of array elements, queue will
    # store indexes of useful elements
    # in every window
    Qi = []

    # Process first k (or first window)
    # elements of array
    i = 0

    for j in range(k):

        # For every element,
        # the previous smaller elements
        # are useless so remove them from Qi
        while ((len(Qi) != 0) and
                 arr[i] >= arr[Qi[-1]]):
            Qi.pop() # Remove from rear

        # Add new element at rear of queue
        Qi.append(i)
        i += 1

    minDiff = sys.maxsize;

    # Process rest of the elements,
    # i.e., from arr[k] to arr[n-1]
    for j in range(i, n):

        # The element at the front
        # of the queue is the largest
        #  element of previous window
        minDiff = min(minDiff, arr[Qi[0]])

        # Remove the elements
        # which are out of this window
        while ((len(Qi) != 0) and
                  Qi[0] <= i - k):
            Qi.pop(0)

        # Remove all elements smaller
        # than the currently being
        # added element (remove
        # useless elements)
        while ((len(Qi) != 0) and
                 arr[i] >= arr[Qi[-1]]):
            Qi.pop()

        # Add current element
        # at the rear of Qi
        Qi.append(i)
        i += 1

    # Compare the maximum
    # element of last window
    minDiff = min(minDiff, arr[Qi[0]])

    return minDiff

# Function to find the minimum
# of the maximum difference of the
# adjacent elements after removing
# K elements from the array
def minimumAdjacentDifference(a, n, k):

    # Create the difference array
    diff = [0 for i in range(n - 1)]

    for i in range(n - 1):
        diff[i] = a[i + 1] - a[i]

    # Find minimum of all maximum
    # of subarray sizes n - k - 1
    answer = findKMin(diff, n - 1,
                        n - k - 1)
    return answer

# Driver code
if __name__=="__main__":

    n = 5
    k = 2

    a = [ 3, 7, 8, 10, 14 ]

    print(minimumAdjacentDifference(a, n, k))

# This code is contributed by rutvik_56

C#

// C# implementation to find the
// minimum of the maximum difference
// of the adjacent elements after
// removing K elements from the array
using System;
using System.Collections.Generic;

class GFG
{

  // Function to find the minimum
  // different in the subarrays
  // of size K in the array
  static int findKMin(List<int> arr, int n, int k)
  {

    // Create a Double Ended Queue, Qi
    // that will store indexes
    // of array elements, queue will
    // store indexes of useful elements
    // in every window
    List<int> Qi = new List<int>();

    // Process first k (or first window)
    // elements of array
    int i = 0;

    for (int j = 0; j < k; j++)
    {

      // For every element,
      // the previous smaller elements
      // are useless so remove them from Qi
      while ((Qi.Count != 0)
             && (arr[i] >= arr[Qi[Qi.Count - 1]]))
        Qi.RemoveAt(Qi.Count
                    - 1); // Remove from rear

      // Add new element at rear of queue
      Qi.Add(i);
      i += 1;
    }
    int minDiff = Int32.MaxValue;

    // Process rest of the elements,
    // i.e., from arr[k] to arr[n-1]
    for (int j = i; j < n; j++) {

      // The element at the front
      // of the queue is the largest
      //  element of previous window
      minDiff = Math.Min(minDiff, arr[Qi[0]]);

      // Remove the elements
      // which are out of this window
      while ((Qi.Count != 0) && (Qi[0] <= i - k))
        Qi.RemoveAt(0);

      // Remove all elements smaller
      // than the currently being
      // added element (remove
      // useless elements)
      while ((Qi.Count != 0)
             && (arr[i] >= arr[Qi[Qi.Count - 1]]))
        Qi.RemoveAt(Qi.Count - 1);

      // Add current element
      // at the rear of Qi
      Qi.Add(i);
      i += 1;
    }

    // Compare the maximum
    // element of last window
    minDiff = Math.Min(minDiff, arr[Qi[0]]);

    return minDiff;
  }

  // Function to find the minimum
  // of the maximum difference of the
  // adjacent elements after removing
  // K elements from the array
  static int minimumAdjacentDifference(int[] a, int n,
                                       int k)
  {

    // Create the difference array
    List<int> diff = new List<int>();

    for (var i = 0; i < n - 1; i++)
      diff.Add(a[i + 1] - a[i]);

    // Find minimum of all maximum
    // of subarray sizes n - k - 1
    var answer = findKMin(diff, n - 1, n - k - 1);
    return answer;
  }


  // Driver code
  public static void Main(string[] args)
  {
    int n = 5;
    int k = 2;

    int[] a = { 3, 7, 8, 10, 14 };

    // Function call
    Console.Write(minimumAdjacentDifference(a, n, k));
  }
}

// This code is contributed by phasing17

Javascript

<script>
// JavaScript implementation to find the
// minimum of the maximum difference
// of the adjacent elements after
// removing K elements from the array

// Function to find the minimum
// different in the subarrays
// of size K in the array
function findKMin(arr, n, k)
{

    // Create a Double Ended Queue, Qi
    // that will store indexes
    // of array elements, queue will
    // store indexes of useful elements
    // in every window
    var Qi = [];

    // Process first k (or first window)
    // elements of array
    var i = 0;

    for (var j = 0; j < k; j++)
    {
        // For every element,
        // the previous smaller elements
        // are useless so remove them from Qi
        while ((Qi.length != 0) && (arr[i] >= arr[Qi[-1]]))
            Qi.pop(); // Remove from rear

        // Add new element at rear of queue
        Qi.push(i);
        i += 1;
    }
    var minDiff = Number.MAX_SAFE_INTEGER;

    // Process rest of the elements,
    // i.e., from arr[k] to arr[n-1]
    for (var j = i; j < n; j++)
    {

        // The element at the front
        // of the queue is the largest
        //  element of previous window
        minDiff = Math.min(minDiff, arr[Qi[0]]);

        // Remove the elements
        // which are out of this window
        while ((Qi.length != 0)  && (Qi[0] <= i - k))
            Qi.shift();

        // Remove all elements smaller
        // than the currently being
        // added element (remove
        // useless elements)
        while ((Qi.length != 0) && (arr[i] >= arr[Qi[Qi.length -1]]))
            Qi.pop();

        // Add current element
        // at the rear of Qi
        Qi.push(i);
        i += 1;
    }
    // Compare the maximum
    // element of last window
    minDiff = Math.min(minDiff, arr[Qi[0]]);

    return minDiff;
}

// Function to find the minimum
// of the maximum difference of the
// adjacent elements after removing
// K elements from the array
function minimumAdjacentDifference(a, n, k)
{

    // Create the difference array
    var diff = [];

    for (var i = 0; i < n - 1; i++)
        diff.push(a[i + 1] - a[i]);

    // Find minimum of all maximum
    // of subarray sizes n - k - 1
    var answer = findKMin(diff, n - 1, n - k - 1);
    return answer;
}

// Driver code
let n = 5;
let k = 2;

let a = [ 3, 7, 8, 10, 14 ];

// function call
document.write(minimumAdjacentDifference(a, n, k));

// This code is contributed by phasing17
</script>

Output:

2

Time complexity: O(N)
Auxiliary Space: O(N)