Given an array arr[] of size N, the task is to split the array into two subsets such that the Bitwise XOR between the maximum of the first subset and minimum of the second subset is minimum.

Examples:

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

Approach: The idea is to find the two elements in the array such that the Bitwise XOR between the two array elements is minimum. Follow the steps below to solve the problem:

• Initialize a variable, say minXOR, to store the minimum possible value of Bitwise XOR between the maximum element of one subset and the minimum element of the other subset.
• Sort the array arr[] in ascending order.
• Traverse the array and update minXOR = min(minXOR, arr[i] ^ arr[i - 1]).

Below is the implementation of above approach:
 

// C++ program for the above approach

#include <bits/stdc++.h>
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;
}

// C++ program for the above approach

​

#include <bits/stdc++.h>

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 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 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# 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));
}
}

<script>

// 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));

</script>

1

Time Complexity: O(N * log N) 
Auxiliary Space: O(1)