将数组拆分为两个子数组，使它们的最大值之差最小

给定一个由整数组成的数组arr[] ，任务是将给定的数组拆分为两个子数组，使得它们的最大元素之间的差异最小。

例子：

Input: arr[] = {7, 9, 5, 10} 
Output: 1 
Explanation: 
The subarrays are {5, 10} and {7, 9} with the difference between their maximums = 10 – 9 = 1.
Input: arr[] = {6, 6, 6} 
Output: 0  

方法：
我们可以观察到，我们需要将数组拆分为两个子数组，使得：

• 如果最大元素在数组中出现多次，则它需要在两个子数组中至少出现一次。
• 否则，最大的和第二大的元素应该出现在不同的子数组中。

这确保了两个子阵列的最大元素之间的差异最大化。
因此，我们需要对数组进行排序，然后最大的 2 个元素之间的差，即 arr[n – 1] 和 arr[n – 2]，就是所需的答案。
下面是上述方法的实现：

C++

// C++ Program to split a given
// array such that the difference
// between their maximums is minimized.
 
#include 
using namespace std;
 
int findMinDif(int arr[], int N)
{
    // Sort the array
    sort(arr, arr + N);
 
    // Return the difference
    // between two highest
    // elements
    return (arr[N - 1] - arr[N - 2]);
}
 
// Driver Program
int main()
{
 
    int arr[] = { 7, 9, 5, 10 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    cout << findMinDif(arr, N);
    return 0;
}

Java

// Java Program to split a given array
// such that the difference between
// their maximums is minimized.
import java.util.*;
 
class GFG{
 
static int findMinDif(int arr[], int N)
{
     
    // Sort the array
    Arrays.sort(arr);
     
    // Return the difference between
    // two highest elements
    return (arr[N - 1] - arr[N - 2]);
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 7, 9, 5, 10 };
    int N = arr.length;
 
    System.out.println(findMinDif(arr, N));
}
}
 
// This code is contributed by offbeat

Python3

# Python3 Program to split a given
# array such that the difference
# between their maximums is minimized.
def findMinDif(arr, N):
     
    # Sort the array
    arr.sort()
 
    # Return the difference
    # between two highest
    # elements
    return (arr[N - 1] - arr[N - 2])
 
# Driver Program
arr = [ 7, 9, 5, 10 ]
N = len(arr)
print(findMinDif(arr, N))
 
# This code is contributed by yatinagg

C#

// C# Program to split a given array
// such that the difference between
// their maximums is minimized.
using System;
class GFG{
 
static int findMinDif(int []arr, int N)
{
     
    // Sort the array
    Array.Sort(arr);
     
    // Return the difference between
    // two highest elements
    return (arr[N - 1] - arr[N - 2]);
}
 
// Driver code
public static void Main()
{
    int []arr = { 7, 9, 5, 10 };
    int N = arr.Length;
 
    Console.Write(findMinDif(arr, N));
}
}
 
// This code is contributed by Code_Mech

Javascript

// javascript Program to split a given array
// such that the difference between
// their maximums is minimized.
 
    function findMinDif(arr , N) {
 
        // Sort the array
        arr.sort((a,b)=>a-b);
 
        // Return the difference between
        // two highest elements
        return (arr[N - 1] - arr[N - 2]);
    }
 
    // Driver code
     
        var arr = [ 7, 9, 5, 10 ];
        var N = arr.length;
 
        document.write(findMinDif(arr, N));
 
// This code contributed by gauravrajput1

输出：

1

时间复杂度： O(N*log(N))，N是数组的元素个数。

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