从阵列中删除一次后最大程度地减小最大最小差异

给定大小为n≥3的数组arr [] ，任务是在删除一个元素后从数组中找到最大和最小元素之间的最小可能差。
例子：

Input: arr[] = {1, 2, 3} 
Output: 1 
Removing 1 will give 3 – 2 = 1 
Removing 2, 3 – 1 = 2 
And removing 3 will result in 2 – 1 = 1
Input: arr[] = {1, 2, 4, 3, 4} 
Output: 2 
 

天真的方法：很明显，要影响差异，只需要删除最小或最大元素即可。

• 对数组进行排序。
• 删除最小值，存储diff1 = arr [n – 1] – arr [1] 。
• 删除最大值，然后diff2 = arr [n – 2] – arr [0] 。
• 最后打印min(diff1，diff2) 。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the minimum required difference
int findMinDifference(int arr[], int n)
{
    // Sort the given array
    sort(arr, arr + n);
 
    // When minimum element is removed
    int diff1 = arr[n - 1] - arr[1];
 
    // When maximum element is removed
    int diff2 = arr[n - 2] - arr[0];
 
    // Return the minimum of diff1 and diff2
    return min(diff1, diff2);
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, 4, 3, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << findMinDifference(arr, n);
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
 
class solution
{
 
// Function to return the minimum required difference
static int findMinDifference(int arr[], int n)
{
    // Sort the given array
    Arrays.sort(arr);
 
    // When minimum element is removed
    int diff1 = arr[n - 1] - arr[1];
 
    // When maximum element is removed
    int diff2 = arr[n - 2] - arr[0];
 
    // Return the minimum of diff1 and diff2
    return Math.min(diff1, diff2);
}
 
// Driver Code
public static void  main(String args[])
{
    int arr[] = { 1, 2, 4, 3, 4 };
    int n = arr.length;
 
    System.out.print(findMinDifference(arr, n));
 
}
}
// This code is contributed by
// Sanjit_Prasad

Python3

# Python3 implementation of the approach
 
# Function to return the minimum
# required difference
def findMinDifference(arr, n) :
 
    # Sort the given array
    arr.sort()
 
    # When minimum element is removed
    diff1 = arr[n - 1] - arr[1]
 
    # When maximum element is removed
    diff2 = arr[n - 2] - arr[0]
 
    # Return the minimum of diff1 and diff2
    return min(diff1, diff2)
 
# Driver Code
if __name__ == "__main__" :
 
    arr = [ 1, 2, 4, 3, 4 ]
    n = len(arr)
 
    print(findMinDifference(arr, n))
 
# This code is contributed by Ryuga

C#

// C# implementation of the approach
using System;
 
public class GFG{
     
// Function to return the minimum required difference
static int findMinDifference(int []arr, int n)
{
    // Sort the given array
    Array.Sort(arr);
 
    // When minimum element is removed
    int diff1 = arr[n - 1] - arr[1];
 
    // When maximum element is removed
    int diff2 = arr[n - 2] - arr[0];
 
    // Return the minimum of diff1 and diff2
    return Math.Min(diff1, diff2);
}
 
// Driver Code
    static public void Main (){
     
    int []arr = { 1, 2, 4, 3, 4 };
    int n = arr.Length;
 
    Console.Write(findMinDifference(arr, n));
 
}
}
// This code is contributed by Sachin..

---

Javascript

    // Javascript implementation of the approach
     
    // Function to return the minimum required difference
    function findMinDifference(arr, n)
    {
        // Sort the given array
        arr.sort();
       
        // When minimum element is removed
        let diff1 = arr[n - 1] - arr[1];
       
        // When maximum element is removed
        let diff2 = arr[n - 2] - arr[0];
       
        // Return the minimum of diff1 and diff2
        return Math.min(diff1, diff2);
    }
     
    let arr = [ 1, 2, 4, 3, 4 ];
    let n = arr.length;
   
    document.write(findMinDifference(arr, n));
     

输出：

2

高效方法：为了从数组中找到min ， secondMin ， max和secondMax元素。我们不需要对数组进行排序，可以在单个数组遍历中完成。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include
using namespace std;
 
// Function to return the minimum required difference
int findMinDifference(int arr[], int n)
{
    int min__, secondMin, max__, secondMax;
 
    min__ = secondMax = (arr[0] < arr[1]) ? arr[0] : arr[1];
    max__ = secondMin = (arr[0] < arr[1]) ? arr[1] : arr[0];
 
    for (int i = 2; i < n; i++)
    {
        // If current element is greater than max
        if (arr[i] > max__)
        {
            // max will become secondMax
            secondMax = max__;
 
            // Update the max
            max__ = arr[i];
        }
 
        // If current element is greater than secondMax
        // but smaller than max
        else if (arr[i] > secondMax)
        {
 
            // Update the secondMax
            secondMax = arr[i];
        }
 
        // If current element is smaller than min
        else if (arr[i] < min__)
        {
 
            // min will become secondMin
            secondMin = min__;
 
            // Update the min
            min__ = arr[i];
        }
 
        // If current element is smaller than secondMin
        // but greater than min
        else if (arr[i] < secondMin) {
                 
            // Update the secondMin
            secondMin = arr[i];
        }
    }
 
    // Minimum of the two possible differences
    int diff = min(max__ - secondMin, secondMax - min__);
    return diff;
}
 
// Driver code
int main()
{
    int arr[] = { 1, 2, 4, 3, 4 };
    int n = sizeof(arr)/sizeof(arr[0]);
    cout << (findMinDifference(arr, n));
}
     
// This code is contributed by
// Shashank_Sharma

Java

// Java implementation of the approach
public class GFG {
 
    // Function to return the minimum required difference
    static int findMinDifference(int arr[], int n)
    {
        int min, secondMin, max, secondMax;
 
        min = secondMax = (arr[0] < arr[1]) ? arr[0] : arr[1];
        max = secondMin = (arr[0] < arr[1]) ? arr[1] : arr[0];
 
        for (int i = 2; i < n; i++) {
 
            // If current element is greater than max
            if (arr[i] > max) {
 
                // max will become secondMax
                secondMax = max;
 
                // Update the max
                max = arr[i];
            }
 
            // If current element is greater than secondMax
            // but smaller than max
            else if (arr[i] > secondMax) {
 
                // Update the secondMax
                secondMax = arr[i];
            }
 
            // If current element is smaller than min
            else if (arr[i] < min) {
 
                // min will become secondMin
                secondMin = min;
 
                // Update the min
                min = arr[i];
            }
 
            // If current element is smaller than secondMin
            // but greater than min
            else if (arr[i] < secondMin) {
 
                // Update the secondMin
                secondMin = arr[i];
            }
        }
 
        // Minimum of the two possible differences
        int diff = Math.min(max - secondMin, secondMax - min);
 
        return diff;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 1, 2, 4, 3, 4 };
        int n = arr.length;
 
        System.out.println(findMinDifference(arr, n));
    }
}

Python3

# Python 3 implementation of the approach
 
# Function to return the minimum
# required difference
def findMinDifference(arr, n):
     
    if(arr[0] < arr[1]):
        min__ = secondMax = arr[0]
    else:
        min__ = secondMax = arr[1]
         
    if(arr[0] < arr[1]):
        max__ = secondMin = arr[1]
    else:
        max__ = secondMin = arr[0]
 
    for i in range(2, n):
         
        # If current element is greater
        # than max
        if (arr[i] > max__):
             
            # max will become secondMax
            secondMax = max__
 
            # Update the max
            max__ = arr[i]
 
        # If current element is greater than
        # secondMax but smaller than max
        elif (arr[i] > secondMax):
             
            # Update the secondMax
            secondMax = arr[i]
 
        # If current element is smaller than min
        elif(arr[i] < min__):
             
            # min will become secondMin
            secondMin = min__
 
            # Update the min
            min__ = arr[i]
 
        # If current element is smaller than
        # secondMin but greater than min
        elif(arr[i] < secondMin):
             
            # Update the secondMin
            secondMin = arr[i]
     
    # Minimum of the two possible
    # differences
    diff = min(max__ - secondMin,
                       secondMax - min__)
    return diff
 
# Driver code
if __name__ == '__main__':
    arr = [1, 2, 4, 3, 4]
    n = len(arr)
    print(findMinDifference(arr, n))
 
# This code is contributed by
# Surendra_Gangwar

C＃

using System;
                     
// C# implementation of the approach
public class GFG {
 
    // Function to return the minimum required difference
    static int findMinDifference(int []arr, int n)
    {
        int min, secondMin, max, secondMax;
 
        min = secondMax = (arr[0] < arr[1]) ? arr[0] : arr[1];
        max = secondMin = (arr[0] < arr[1]) ? arr[1] : arr[0];
 
        for (int i = 2; i < n; i++) {
 
            // If current element is greater than max
            if (arr[i] > max) {
 
                // max will become secondMax
                secondMax = max;
 
                // Update the max
                max = arr[i];
            }
 
            // If current element is greater than secondMax
            // but smaller than max
            else if (arr[i] > secondMax) {
 
                // Update the secondMax
                secondMax = arr[i];
            }
 
            // If current element is smaller than min
            else if (arr[i] < min) {
 
                // min will become secondMin
                secondMin = min;
 
                // Update the min
                min = arr[i];
            }
 
            // If current element is smaller than secondMin
            // but greater than min
            else if (arr[i] < secondMin) {
 
                // Update the secondMin
                secondMin = arr[i];
            }
        }
 
        // Minimum of the two possible differences
        int diff = Math.Min(max - secondMin, secondMax - min);
 
        return diff;
    }
 
    // Driver code
    public static void Main()
    {
        int []arr = { 1, 2, 4, 3, 4 };
        int n = arr.Length;
 
        Console.WriteLine(findMinDifference(arr, n));
    }
}
// This code is contributed by 29AjayKumar

的PHP

 $max__)
        {
            // max will become secondMax
            $secondMax = $max__;
 
            // Update the max
            $max__ = $arr[$i];
        }
 
        // If current element is greater than secondMax
        // but smaller than max
        else if ($arr[$i] > $secondMax)
        {
 
            // Update the secondMax
            $secondMax = $arr[$i];
        }
 
        // If current element is smaller than min
        else if ($arr[$i] < $min__)
        {
 
            // min will become secondMin
            $secondMin = $min__;
 
            // Update the min
            $min__ = $arr[$i];
        }
 
        // If current element is smaller than secondMin
        // but greater than min
        else if ($arr[$i] < $secondMin)
        {
                 
            // Update the secondMin
            $secondMin = $arr[$i];
        }
    }
 
    // Minimum of the two possible differences
    $diff = min($max__ - $secondMin,
                         $secondMax - $min__);
    return $diff;
}
 
// Driver code
$arr = array( 1, 2, 4, 3, 4 );
$n = count($arr);
print(findMinDifference($arr, $n));
 
// This code is contributed by mits
?>

Java脚本

 
    // Javascript implementation of the approach
     
    // Function to return the minimum required difference
    function findMinDifference(arr, n)
    {
        let min__, secondMin, max__, secondMax;
      
        min__ = secondMax = (arr[0] < arr[1]) ? arr[0] : arr[1];
        max__ = secondMin = (arr[0] < arr[1]) ? arr[1] : arr[0];
      
        for (let i = 2; i < n; i++)
        {
            // If current element is greater than max
            if (arr[i] > max__)
            {
                // max will become secondMax
                secondMax = max__;
      
                // Update the max
                max__ = arr[i];
            }
      
            // If current element is greater than secondMax
            // but smaller than max
            else if (arr[i] > secondMax)
            {
      
                // Update the secondMax
                secondMax = arr[i];
            }
      
            // If current element is smaller than min
            else if (arr[i] < min__)
            {
      
                // min will become secondMin
                secondMin = min__;
      
                // Update the min
                min__ = arr[i];
            }
      
            // If current element is smaller than secondMin
            // but greater than min
            else if (arr[i] < secondMin) {
                      
                // Update the secondMin
                secondMin = arr[i];
            }
        }
      
        // Minimum of the two possible differences
        let diff = Math.min(max__ - secondMin, secondMax - min__);
        return diff;
    }  
     
    let arr = [ 1, 2, 4, 3, 4 ];
    let n = arr.length;
    document.write(findMinDifference(arr, n));
     

输出：

2

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