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.

CPP

 // 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

 // 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

Python3

 # 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#

 // 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

PHP



Output:

0

Time Complexity: O(n)

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Improved By : ihritik