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;
} |
// 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)