最小化修改后的数组的最大值和最小值之间的差异

给定一个由n个整数和一个整数X组成的数组A。您可以选择以下任意整数$-X\leq k\leq X$ ，然后将k分别添加到A [i] $0\leq i \leq n-1$ 。的任务是找到A的最大值和更新所述阵列A之后A的最小值之间的最小可能差。

例子：

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[0] and subtract -2 from a[1].

方法：设A为原始数组。为了最小化max(A)– min(A) ，让我们分别尝试最小化max(A)和最大化min(A) 。

max(A)的最小可能值为max(A)– K ，因为max(A)的值不能降低。类似地， min(A)的最大可能值是min(A)+ K。因此，数量max(A)– min(A)至少是ans =(max(A)– K)–(min(A)+ K) 。

我们可以通过以下修改来获得此值：

• 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.

下面是上述方法的实现。

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[0], mx = A[0];
  
    // 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[0], mx = A[0];
      
        // 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[0] 
    mx =  A[0] 
  
    # 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[0], mx = A[0];
      
        // 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

输出：

0

时间复杂度： O(n)