# Minimize the difference between the maximum and minimum values of the modified array

Given an array A of n integers and and integer X. You may choose any integer between , and add k to A[i] for each . The task is to find the smallest possible difference between the maximum value of A and the minimum value of A after updating the array A.

Examples:

Input: arr[] = {1, 3, 6}, x = 3
Output: 0
New array is [3, 3, 3] or [4, 4, 4].

Input: arr[] = {0, 10}, x = 2
Output: 6
New array is [2, 8] i.e add 2 to a and subtract -2 from a.


## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Let A be the original array. Towards trying to minimize max(A) – min(A), let’s try to minimize max(A) and maximize min(A) separately.

The smallest possible value of max(A) is max(A) – K, as the value max(A) cannot go lower. Similarly, the largest possible value of min(A) is min(A) + K. So the quantity max(A) – min(A) is at least ans = (max(A) – K) – (min(A) + K).

We can attain this value, by the following modifications:

• If A[i] <= min(A) + K, then A[i] = min(A) + K
• Else, if A[i] >= max(A) – K, then A[i] = max(A) – K

If ans < 0, the best answer we could have is ans = 0, also using the same modification.

Below is the implementation of above approach.

 // C++ program to find the minimum difference.  #include  using namespace std;     // Function to return required minimum difference  int minDiff(int n, int x, int A[])  {      int mn = A, mx = A;         // finding minimum and maximum values      for (int i = 0; i < n; ++i) {          mn = min(mn, A[i]);          mx = max(mx, A[i]);      }         // returning minimum possible difference      return max(0, mx - mn - 2 * x);  }     // Driver program  int main()  {         int n = 3, x = 3;      int A[] = { 1, 3, 6 };         // function to return the answer      cout << minDiff(n, x, A);         return 0;  }

 // Java program to find the minimum difference.     import java.util.*;  class GFG  {             // Function to return required minimum difference      static int minDiff(int n, int x, int A[])      {          int mn = A, mx = A;                 // finding minimum and maximum values          for (int i = 0; i < n; ++i) {              mn = Math.min(mn, A[i]);              mx = Math.max(mx, A[i]);          }                 // returning minimum possible difference          return Math.max(0, mx - mn - 2 * x);      }             // Driver program      public static void main(String []args)      {                 int n = 3, x = 3;          int A[] = { 1, 3, 6 };                 // function to return the answer          System.out.println(minDiff(n, x, A));                        }     }     // This code is contributed by ihritik

 # Python program to find the minimum difference.            # Function to return required minimum difference  def minDiff( n,  x,  A):          mn =  A       mx =  A          # finding minimum and maximum values      for i in range(0,n):           mn = min( mn,  A[ i])            mx = max( mx,  A[ i])                  # returning minimum possible difference      return max(0,  mx -  mn - 2 *  x)              # Driver program     n = 3  x = 3  A = [1, 3, 6 ]      # function to return the answer  print(minDiff( n,  x,  A))     # This code is contributed by ihritik

 // C# program to find the minimum difference.     using System;  class GFG  {             // Function to return required minimum difference      static int minDiff(int n, int x, int []A)      {          int mn = A, mx = A;                 // finding minimum and maximum values          for (int i = 0; i < n; ++i) {              mn = Math.Min(mn, A[i]);              mx = Math.Max(mx, A[i]);          }                 // returning minimum possible difference          return Math.Max(0, mx - mn - 2 * x);      }             // Driver program      public static void Main()      {                 int n = 3, x = 3;          int []A = { 1, 3, 6 };                 // function to return the answer          Console.WriteLine(minDiff(n, x, A));                    }  }     // This code is contributed by ihritik

 

Output:
0


Time Complexity: O(n)

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Improved By : ihritik

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