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# 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:
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:
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   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 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 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 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 temp = new Vector();                 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<

## 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 temp = new List();                 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



Output:

2

Time Complexity: O(2N * 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   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 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 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



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   using namespace std;   // Function to find the minimum // different in the subarrays // of size K in the array int findKMin(vector 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 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 a,                           int n, int k) {       // Create the difference array     vector 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 a= { 3, 7, 8, 10, 14 };       cout << minimumAdjacentDifference(a, n, k);     return 0; }

## 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

## Javascript



Output:

2

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

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