将数组分成两个子集，它们的最大值和最小值之间具有最小的按位异或

给定大小为N的数组arr[] ，任务是将数组拆分为两个子集，使得第一个子集的最大值和第二个子集的最小值之间的按位异或最小。

例子：

Input: arr[] = {3, 1, 2, 6, 4} 
Output: 1 
Explanation: 
Splitting the given array in two subsets {1, 3}, {2, 4, 6}. 
The maximum of the first subset is 3 and the minimum of the second subset is 2. 
Therefore, their bitwise XOR is equal to 1.

Input: arr[] = {2, 1, 3, 2, 4, 3} 
Output: 0

方法：想法是找到数组中的两个元素，使得两个数组元素之间的按位异或最小。请按照以下步骤解决问题：

• 初始化一个变量，比如minXOR ，以存储一个子集的最大元素和另一个子集的最小元素之间的按位异或的最小可能值。
• 按升序对数组arr[]进行排序。
• 遍历数组并更新 minXOR = min(minXOR, arr[i] ^ arr[i – 1])。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to split the array into two subset
// such that the Bitwise XOR between the maximum
// of one subset and minimum of other is minimum
int splitArray(int arr[], int N)
{
    // Sort the array in
    // increasing order
    sort(arr, arr + N);
 
    int result = INT_MAX;
 
    // Calculating the min Bitwise XOR
    // between consecutive elements
    for (int i = 1; i < N; i++) {
        result = min(result,
                     arr[i] - arr[i - 1]);
    }
 
    // Return the final
    // minimum Bitwise XOR
    return result;
}
 
 
// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 3, 1, 2, 6, 4 };
 
    // Size of array
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    cout << splitArray(arr, N);
    return 0;
}

Java

// java program for the above approach
import java.util.*;
class GFG{
 
// Function to split the array into two subset
// such that the Bitwise XOR between the maximum
// of one subset and minimum of other is minimum
static int splitArray(int arr[], int N)
{
    // Sort the array in
    // increasing order
    Arrays.sort(arr);
 
    int result = Integer.MAX_VALUE;
 
    // Calculating the min Bitwise XOR
    // between consecutive elements
    for (int i = 1; i < N; i++)
    {
        result = Math.min(result,
                          arr[i] - arr[i - 1]);
    }
 
    // Return the final
    // minimum Bitwise XOR
    return result;
}
 
// Driver Code
public static void main(String[] args)
{
    // Given array arr[]
    int arr[] = { 3, 1, 2, 6, 4 };
 
    // Size of array
    int N = arr.length;
 
    // Function Call
    System.out.print(splitArray(arr, N));
}
}

Python3

# Python3 program for the above approach
 
# Function to split the array into two subset
# such that the Bitwise XOR between the maximum
# of one subset and minimum of other is minimum
def splitArray(arr, N):
     
    # Sort the array in increasing
    # order
    arr = sorted(arr)
 
    result = 10 ** 9
 
    # Calculating the min Bitwise XOR
    # between consecutive elements
    for i in range(1, N):
        result = min(result, arr[i] ^ arr[i - 1])
 
    # Return the final
    # minimum Bitwise XOR
    return result
 
# Driver Code
if __name__ == '__main__':
     
    # Given array arr[]
    arr = [ 3, 1, 2, 6, 4 ]
 
    # Size of array
    N = len(arr)
 
    # Function Call
    print(splitArray(arr, N))
 
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
class GFG{
 
// Function to split the array into two subset
// such that the Bitwise XOR between the maximum
// of one subset and minimum of other is minimum
static int splitArray(int []arr, int N)
{
    // Sort the array in increasing order
    Array.Sort(arr);
 
    int result = Int32.MaxValue;
 
    // Calculating the min Bitwise XOR
    // between consecutive elements
    for (int i = 1; i < N; i++)
    {
        result = Math.Min(result,
                          arr[i] ^ arr[i - 1]);
    }
 
    // Return the final
    // minimum Bitwise XOR
    return result;
}
 
// Driver Code
public static void Main()
{
    // Given array arr[]
    int []arr = { 3, 1, 2, 6, 4 };
 
    // Size of array
    int N = arr.Length;
 
    // Function Call
    Console.Write(splitArray(arr, N));
}
}

Javascript

 
// Javascript program for the above approach
 
// Function to split the array into two subset
// such that the Bitwise XOR between the maximum
// of one subset and minimum of other is minimum
function splitArray(arr, N)
{
    // Sort the array in
    // increasing order
    arr.sort();
 
    let result = Number.MAX_VALUE;
 
    // Calculating the min Bitwise XOR
    // between consecutive elements
    for (let i = 1; i < N; i++) {
        result = Math.min(result,
                     arr[i] - arr[i - 1]);
    }
 
    // Return the final
    // minimum Bitwise XOR
    return result;
}
 
 
// Driver Code
    // Given array arr[]
    let arr = [ 3, 1, 2, 6, 4 ];
 
    // Size of array
    let N = arr.length;
 
    // Function Call
    document.write(splitArray(arr, N));
 

输出：

1

时间复杂度： O(N * log N)
辅助空间： O(1)

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