从数组中精确删除一个元素，以使max – min为最小值

给定一个数组$a$由N个正整数组成。该任务是从该数组中精确删除一个元素，以最小化max(a)-min(a)并打印出可能的最小值(max(a)-min(a)) 。

注意： max(a)表示数组中的最大数字$a$ min(a)表示数组中的最小数字$a$ 。
数组中至少有2个元素。

例子：

Input: arr[] = {1, 3, 3, 7}
Output: 2
Remove 7, then max(a) will be 3 and min(a) will be 1.
So our answer will be 3-1 = 2.

Input: arr[] = {1, 1000}
Output: 0
Remove either 1 or 1000, then our answer will 1-1 =0 or
1000-1000=0

简单方法：在这里可以看出，我们总是必须删除数组的最小值或最大值。我们首先对数组进行排序。排序后，如果我们删除最小元素，则差将为a [n-1] – a [1]。如果我们删除最大元素，则差将为a [n-2] – a [0]。我们返回这两个差异中的最小值。

下面是上述方法的实现：

C++

// C++ implementation of the above approach
#include 
using namespace std;
  
// function to calculate max-min
int max_min(int a[], int n)
{
    sort(a, a + n);
  
    return min(a[n - 2] - a[0], a[n - 1] - a[1]);
}
  
// Driver code
int main()
{
    int a[] = { 1, 3, 3, 7 };
    int n = sizeof(a) / sizeof(a[0]);
  
    cout << max_min(a, n);
    return 0;
}

Java

// Java implementation of the above approach
  
import java.util.*;
class GFG
{
    // function to calculate max-min
    static int max_min(int a[], int n)
    {
        Arrays.sort(a);
      
        return Math.min(a[n - 2] - a[0], a[n - 1] - a[1]);
    }
      
    // Driver code
    public static void main(String []args)
    {
        int a[] = { 1, 3, 3, 7 };
        int n = a.length;
      
        System.out.println(max_min(a, n));
      
    }
}
  
// This code is contributed 
// by ihritik

Python3

# Python3 implementation of the
# above approach 
  
# function to calculate max-min 
def max_min(a, n):
    a.sort()
    return min(a[n - 2] - a[0], 
               a[n - 1] - a[1])
  
# Driver code 
a = [1, 3, 3, 7] 
n = len(a)
print(max_min(a, n))
  
# This code is contributed 
# by sahishelangia

C#

// C# implementation of the above approach
  
using System;
class GFG
{
    // function to calculate max-min
    static int max_min(int []a, int n)
    {
        Array.Sort(a);
      
        return Math.Min(a[n - 2] - a[0], a[n - 1] - a[1]);
    }
      
    // Driver code
    public static void Main()
    {
        int []a = { 1, 3, 3, 7 };
        int n = a.Length;
      
        Console.WriteLine(max_min(a, n));
      
    }
}
  
// This code is contributed 
// by ihritik

输出：

2

时间复杂度： O(n log n)

高效方法：
一种有效的方法是执行以下操作。
1)求第一最小值和第二最小值
2)找出第一个最大值和第二个最大值
3)返回以下两个差的最小值。
…..a)第一最大值和第二最小值
…..b)第二个最大值和第一个最小值

C++

// C++ implementation of the above approach
#include 
using namespace std;
  
// function to calculate max-min
int max_min(int a[], int n)
{
    // There should be at-least two elements
    if (n <= 1)
      return INT_MAX;
  
    // To store first and second minimums
    int f_min = a[0], s_min = INT_MAX;
  
    // To store first and second maximums
    int f_max = a[0], s_max = INT_MIN;
  
    for (int i = 1; i= f_max)
        {
           s_max = f_max;
           f_max = a[i];
        }
        else if (a[i] > s_max)
        {
           s_max = a[i];
        }
    }
  
    return min((f_max - s_min), (s_max - f_min));
}
  
// Driver code
int main()
{
    int a[] = { 1, 3, 3, 7 };
    int n = sizeof(a) / sizeof(a[0]);
  
    cout << max_min(a, n);
    return 0;
}

Java

// Java implementation of the above approach
  
class GFG
{
    // function to calculate max-min
    static int max_min(int a[], int n)
    {
        // There should be at-least two elements
        if (n <= 1)
        return Integer.MAX_VALUE;
      
        // To store first and second minimums
        int f_min = a[0], s_min = Integer.MAX_VALUE;
      
        // To store first and second maximums
        int f_max = a[0], s_max = Integer.MIN_VALUE;
      
        for (int i = 1; i= f_max)
            {
            s_max = f_max;
            f_max = a[i];
            }
            else if (a[i] > s_max)
            {
            s_max = a[i];
            }
        }
      
        return Math.min((f_max - s_min), (s_max - f_min));
    }
      
    // Driver code
    public static void main(String []args)
    {
        int a[] = { 1, 3, 3, 7 };
        int n = a.length;
      
        System.out.println(max_min(a, n));
      
    }
  
}
  
// This code is contributed 
// by ihritik

Python3

# Python3 implementation of the 
# above approach 
import sys
  
# function to calculate max-min 
def max_min(a, n) : 
      
    # There should be at-least two elements 
    if (n <= 1) :
        return sys.maxsize
  
    # To store first and second minimums 
    f_min = a[0]
    s_min = sys.maxsize 
  
    # To store first and second maximums 
    f_max = a[0]
    s_max = -(sys.maxsize - 1)
  
    for i in range(n) : 
          
        if (a[i] <= f_min) :
            s_min = f_min
            f_min = a[i]
          
        elif (a[i] < s_min) :
            s_min = a[i]
  
        if (a[i] >= f_max) :
            s_max = f_max
            f_max = a[i]
      
        elif (a[i] > s_max) :
            s_max = a[i]
  
    return min((f_max - s_min), (s_max - f_min))
  
# Driver code 
if __name__ == "__main__" : 
    a = [ 1, 3, 3, 7 ]
    n = len(a)
  
    print(max_min(a, n)) 
  
# This code is contributed by Ryuga

C＃

// C# implementation of the above approach
using System;
class GFG
{
    // function to calculate max-min
    static int max_min(int []a, int n)
    {
        // There should be at-least two elements
        if (n <= 1)
        return Int32.MaxValue;
      
        // To store first and second minimums
        int f_min = a[0], s_min = Int32.MaxValue;
      
        // To store first and second maximums
        int f_max = a[0], s_max = Int32.MinValue;
      
        for (int i = 1; i= f_max)
            {
            s_max = f_max;
            f_max = a[i];
            }
            else if (a[i] > s_max)
            {
            s_max = a[i];
            }
        }
      
        return Math.Min((f_max - s_min), (s_max - f_min));
    }
      
    // Driver code
    public static void Main()
    {
        int []a = { 1, 3, 3, 7 };
        int n = a.Length;
      
        Console.WriteLine(max_min(a, n));
      
    }
  
}
  
// This code is contributed 
// by ihritik

的PHP

= $f_max)
        {
            $s_max = $f_max;
            $f_max = $a[$i];
        }
        else if ($a[$i] > $s_max)
        {
            $s_max = $a[$i];
        }
    }
  
    return min(($f_max - $s_min), 
               ($s_max - $f_min));
}
  
// Driver code
$a = array ( 1, 3, 3, 7 );
$n = sizeof($a);
  
echo(max_min($a, $n));
  
// This code is contributed 
// by Mukul Singh
?>

输出：

2

时间复杂度： O(n)