Find maximum xor of k elements in an array
Given an array arr[] of N integers and an integer K. The task is to find the maximum xor subset of size K of the given array.
Examples:
Input: arr[] = {2, 5, 4, 1, 3, 7, 6, 8}, K = 3
Output: 15
We obtain 15 by selecting 2, 5, 8
Input: arr[] = {3, 4, 7, 7, 9}, K = 3
Output: 14
Naive approach: Iterate over all subsets of size K of the array and find maximum xor among them.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int Max_Xor( int arr[], int n, int k)
{
int maxXor = INT_MIN;
for ( int i = 0; i < (1 << n); i++) {
if (__builtin_popcount(i) == k) {
int cur_xor = 0;
for ( int j = 0; j < n; j++) {
if (i & (1 << j))
cur_xor = cur_xor ^ arr[j];
}
maxXor = max(maxXor, cur_xor);
}
}
return maxXor;
}
int main()
{
int arr[] = { 2, 5, 4, 1, 3, 7, 6, 8 };
int n = sizeof (arr) / sizeof ( int );
int k = 3;
cout << Max_Xor(arr, n, k);
return 0;
}
|
Java
import java.util.*;
class GFG
{
static int Max_Xor( int arr[], int n, int k)
{
int maxXor = Integer.MIN_VALUE;
for ( int i = 0 ; i < ( 1 << n); i++)
{
if (Integer.bitCount(i) == k)
{
int cur_xor = 0 ;
for ( int j = 0 ; j < n; j++)
{
if ((i & ( 1 << j)) == 0 )
cur_xor = cur_xor ^ arr[j];
}
maxXor = Math.max(maxXor, cur_xor);
}
}
return maxXor;
}
public static void main(String[] args)
{
int arr[] = { 2 , 5 , 4 , 1 , 3 , 7 , 6 , 8 };
int n = arr.length;
int k = 3 ;
System.out.println(Max_Xor(arr, n, k));
}
}
|
Python3
MAX = 10000
MAX_ELEMENT = 50
dp = [[[ - 1 for i in range ( MAX )]
for j in range (MAX_ELEMENT)]
for k in range (MAX_ELEMENT)]
def Max_Xor(arr, i, j, mask, n):
if (i > = n):
if (j = = 0 ):
return mask
else :
return 0
if (dp[i][j][mask] ! = - 1 ):
return dp[i][j][mask]
ans = 0
if (j > 0 ):
ans = Max_Xor(arr, i + 1 , j - 1 , mask ^ arr[i], n)
ans = max (ans, Max_Xor(arr, i + 1 , j, mask, n))
dp[i][j][mask] = ans
return ans
arr = [ 2 , 5 , 4 , 1 , 3 , 7 , 6 , 8 ]
n = len (arr)
k = 3
print (Max_Xor(arr, 0 , k, 0 , n))
|
C#
using System;
class GFG
{
static int Max_Xor( int []arr, int n, int k)
{
int maxXor = int .MinValue;
for ( int i = 0; i < (1 << n); i++)
{
if (bitCount(i) == k)
{
int cur_xor = 0;
for ( int j = 0; j < n; j++)
{
if ((i & (1 << j)) == 0)
cur_xor = cur_xor ^ arr[j];
}
maxXor = Math.Max(maxXor, cur_xor);
}
}
return maxXor;
}
static int bitCount( long x)
{
int setBits = 0;
while (x != 0)
{
x = x & (x - 1);
setBits++;
}
return setBits;
}
public static void Main(String[] args)
{
int []arr = { 2, 5, 4, 1, 3, 7, 6, 8 };
int n = arr.Length;
int k = 3;
Console.WriteLine(Max_Xor(arr, n, k));
}
}
|
Javascript
<script>
function Max_Xor(arr, n, k)
{
let maxXor = Number.MIN_VALUE;
for (let i = 0; i < (1 << n); i++) {
if (bitCount(i) == k) {
let cur_xor = 0;
for (let j = 0; j < n; j++) {
if (i & (1 << j))
cur_xor = cur_xor ^ arr[j];
}
maxXor = Math.max(maxXor, cur_xor);
}
}
return maxXor;
}
function bitCount(x)
{
let setBits = 0;
while (x != 0)
{
x = x & (x - 1);
setBits++;
}
return setBits;
}
let arr = [ 2, 5, 4, 1, 3, 7, 6, 8 ];
let n = arr.length;
let k = 3;
document.write(Max_Xor(arr, n, k));
</script>
|
Time Complexity: O(n*2n)
Auxiliary Space: O(1)
Efficient approach: The problem can be solved using dynamic programming. Create a dp table dp[i][j][mask] which stores the maximum xor possible at the ith index (with or without including it) and j denotes the number of remaining elements we can include in our subset of K elements. Mask is the xor of all the elements selected till the ith index.
Note: This approach will only work for smaller arrays due to space requirements for the dp array.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
#define MAX 10000
#define MAX_ELEMENT 50
int dp[MAX_ELEMENT][MAX_ELEMENT][MAX];
int Max_Xor( int arr[], int i, int j, int mask, int n)
{
if (i >= n) {
if (j == 0)
return mask;
else
return 0;
}
if (dp[i][j][mask] != -1)
return dp[i][j][mask];
int ans = 0;
if (j > 0)
ans = Max_Xor(arr, i + 1, j - 1, mask ^ arr[i], n);
ans = max(ans, Max_Xor(arr, i + 1, j, mask, n));
return dp[i][j][mask] = ans;
}
int main()
{
int arr[] = { 2, 5, 4, 1, 3, 7, 6, 8 };
int n = sizeof (arr) / sizeof ( int );
int k = 3;
memset (dp, -1, sizeof (dp));
cout << Max_Xor(arr, 0, k, 0, n);
return 0;
}
|
Java
import java.util.*;
class GFG
{
static int MAX = 10000 ;
static int MAX_ELEMENT = 50 ;
static int [][][] dp = new int [MAX_ELEMENT][MAX_ELEMENT][MAX];
static int Max_Xor( int arr[], int i,
int j, int mask, int n)
{
if (i >= n)
{
if (j == 0 )
return mask;
else
return 0 ;
}
if (dp[i][j][mask] != - 1 )
return dp[i][j][mask];
int ans = 0 ;
if (j > 0 )
ans = Max_Xor(arr, i + 1 , j - 1 ,
mask ^ arr[i], n);
ans = Math.max(ans, Max_Xor(arr, i + 1 , j,
mask, n));
return dp[i][j][mask] = ans;
}
public static void main(String[] args)
{
int arr[] = { 2 , 5 , 4 , 1 , 3 , 7 , 6 , 8 };
int n = arr.length;
int k = 3 ;
for ( int i = 0 ; i < MAX_ELEMENT; i++)
{
for ( int j = 0 ; j < MAX_ELEMENT; j++)
{
for ( int l = 0 ; l < MAX; l++)
dp[i][j][l] = - 1 ;
}
}
System.out.println(Max_Xor(arr, 0 , k, 0 , n));
}
}
|
Python3
MAX = 10000
MAX_ELEMENT = 50
dp = [[[ - 1 for i in range ( MAX )] for j in range (MAX_ELEMENT)] for k in range (MAX_ELEMENT)]
def Max_Xor(arr, i, j, mask, n):
if (i > = n):
if (j = = 0 ):
return mask
else :
return 0
if (dp[i][j][mask] ! = - 1 ):
return dp[i][j][mask]
ans = 0
if (j > 0 ):
ans = Max_Xor(arr, i + 1 , j - 1 , mask ^ arr[i], n)
ans = max (ans, Max_Xor(arr, i + 1 , j, mask, n))
dp[i][j][mask] = ans
return ans
arr = [ 2 , 5 , 4 , 1 , 3 , 7 , 6 , 8 ]
n = len (arr)
k = 3
print (Max_Xor(arr, 0 , k, 0 , n))
|
C#
using System;
class GFG
{
static int MAX = 10000;
static int MAX_ELEMENT = 50;
static int [,,] dp = new int [MAX_ELEMENT, MAX_ELEMENT, MAX];
static int Max_Xor( int [] arr, int i,
int j, int mask, int n)
{
if (i >= n)
{
if (j == 0)
return mask;
else
return 0;
}
if (dp[i,j,mask] != -1)
return dp[i,j,mask];
int ans = 0;
if (j > 0)
ans = Max_Xor(arr, i + 1, j - 1,
mask ^ arr[i], n);
ans = Math.Max(ans, Max_Xor(arr, i + 1, j,
mask, n));
return dp[i,j,mask] = ans;
}
public static void Main ()
{
int [] arr = { 2, 5, 4, 1, 3, 7, 6, 8 };
int n = arr.Length;
int k = 3;
for ( int i = 0; i < MAX_ELEMENT; i++)
{
for ( int j = 0; j < MAX_ELEMENT; j++)
{
for ( int l = 0; l < MAX; l++)
dp[i,j,l] = -1;
}
}
Console.WriteLine(Max_Xor(arr, 0, k, 0, n));
}
}
|
Javascript
const MAX = 10000;
const MAX_ELEMENT = 50;
var dp = [];
for ( var i = 0; i < MAX_ELEMENT; i++)
{
dp[i] = [];
for ( var j = 0; j < MAX_ELEMENT; j++)
{
dp[i][j] = [];
for ( var k = 0; k < MAX_ELEMENT; k++)
{
dp[i][j][k] = -1;
}
}
}
function Max_Xor(arr, i, j, mask, n)
{
if (i >= n)
{
if (j == 0)
return mask;
else
return 0;
}
if (dp[i][j][mask] != -1)
return dp[i][j][mask];
var ans = 0;
if (j > 0)
ans = Max_Xor(arr, i + 1, j - 1, mask ^ arr[i], n);
ans = Math.max(ans, Max_Xor(arr, i + 1, j, mask, n));
dp[i][j][mask] = ans;
return ans;
}
var arr = [2, 5, 4, 1, 3, 7, 6, 8];
var n = arr.length;
var k = 3;
console.log(Max_Xor(arr, 0, k, 0, n));
|
Time Complexity: O(n*n)
Auxiliary Space: O(MAX*MAX_ELEMENT2) where MAX and MAX_ELEMENT are defined constants.
Last Updated :
09 Oct, 2023
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