Partition array into two subsets with minimum Bitwise XOR between their maximum and minimum

• Last Updated : 30 Apr, 2021

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:
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:

Below is the implementation of above approach:

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 minimumint 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 Codeint main(){    // Given array arr[]    int arr[] = { 3, 1, 2, 6, 4 };     // Size of array    int N = sizeof(arr) / sizeof(arr);     // Function Call    cout << splitArray(arr, N);    return 0;}

Java

 // java program for the above approachimport 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 minimumstatic 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 Codepublic 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 minimumdef 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 Codeif __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 approachusing 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 minimumstatic 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 Codepublic 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


Output:
1

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

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