Given an array arr[] of n numbers and a number K, find the number of subsets of arr[] having XOR of elements as K
Examples :
Input: arr[] = {6, 9, 4,2}, k = 6
Output: 2
The subsets are {4, 2} and {6}
Input: arr[] = {1, 2, 3, 4, 5}, k = 4
Output: 4
The subsets are {1, 5}, {4}, {1, 2, 3, 4}
and {2, 3, 5}
We strongly recommend that you click here and practice it, before moving on to the solution.
Brute Force approach O(2n): One naive approach is to generate all the 2n subsets and count all the subsets having XOR value K, but this approach will not be efficient for large values of n.
Meet in the Middle Approach O(2n/2):
An optimization over the naïve approach. We split the arrays and then generate subsets, thereby reducing the time complexity significantly than Brute Force Solution
Let us go over how it works.
arr[] = {6 , 9 , 4 , 2}
Let us split this array in two halves.
arr1[] = {6 , 9}
arr2[] = {4 , 2}
Generating all possible XOR for arr1 Generating all possible XOR for arr2
6 -> 0110 4 -> 0100
9 -> 1001 2 -> 0010
6 ^ 9 -> 1111 4 ^ 2 -> 0110
0 -> 0000 0 -> 0000
K=6 (0110)
arr1[i] ^ unknown = K
unknown = K ^arr1[i]
So, find the count of this unknown from arr2. Use a frequency map.
Follow the steps below to implement the above idea:
1) First partition the array in half and then generate the subsets XOR for each partition and store the frequency.
2) Iterate on the XOR values of partition P1 and search for its corresponding XOR complement value in part2 P2. If XOR value K is to be achieved by the subset, and x is the XOR subset in part1 P1 with y frequency, then K ^ x is the XOR complement value in part2 P2. The total number of subsets will be P1[x]*P2[K^x].
3) Return the answer after the loop terminates.
//Author - RainX (ABHIJIT ROY, NIT AGARTALA) #include <bits/stdc++.h> using namespace std;
// Function to generate all subsets and their XOR values void generate(vector< int >& arr, int curr, int n, unordered_map< int , int >& XOR, int xorSubset) {
if (curr == n) {
XOR[xorSubset]++; // Store XOR values in the map
return ;
}
generate(arr, curr + 1, n, XOR, xorSubset ^ arr[curr]); // Include current element in XOR
generate(arr, curr + 1, n, XOR, xorSubset); // Exclude current element from XOR
} // Function to count subsets with XOR equal to K int subsetXOR(vector< int >& arr, int N, int K) {
unordered_map< int , int > P1, P2; // Maps for first and second half subsets' XOR
generate(arr, 0, N / 2, P1, 0); // Generate XOR for first half
generate(arr, N / 2, N, P2, 0); // Generate XOR for second half
int cnt = 0; // Counter for subsets with XOR equal to K
for ( auto x : P1) {
int find = K ^ x.first; // Find corresponding value in second half
cnt += (P2[find] * x.second); // Increment count with the product of occurrences
}
return cnt; // Return count of subsets with XOR equal to K
} int main() {
vector< int > arr = {1, 2, 3, 4, 5}; // Given array
int k = 4; // XOR value to search for
int N = 5; // Size of the array
cout << "Count of subsets is " << subsetXOR(arr, N, k); // Output the count of subsets
return 0;
} // Code by RainX (Abhijit Roy, NIT AGARTALA) |
//Author: Amar Singh IIT BHU Varanasi import java.util.*;
class Main {
// Function to generate all subsets and their XOR values
static void generate(List<Integer> arr, int curr, int n, Map<Integer, Integer> XOR, int xorSubset) {
if (curr == n) {
XOR.put(xorSubset, XOR.getOrDefault(xorSubset, 0 ) + 1 ); // Store XOR values in the map
return ;
}
generate(arr, curr + 1 , n, XOR, xorSubset ^ arr.get(curr)); // Include current element in XOR
generate(arr, curr + 1 , n, XOR, xorSubset); // Exclude current element from XOR
}
// Function to count subsets with XOR equal to K
static int subsetXOR(List<Integer> arr, int N, int K) {
Map<Integer, Integer> P1 = new HashMap<>();
Map<Integer, Integer> P2 = new HashMap<>();
generate(arr, 0 , N / 2 , P1, 0 ); // Generate XOR for the first half
generate(arr, N / 2 , N, P2, 0 ); // Generate XOR for the second half
int cnt = 0 ; // Counter for subsets with XOR equal to K
for ( int x : P1.keySet()) {
int find = K ^ x; // Find corresponding value in the second half
if (P2.containsKey(find)) {
cnt += P2.get(find) * P1.get(x); // Increment count with the product of occurrences
}
}
return cnt; // Return count of subsets with XOR equal to K
}
public static void main(String[] args) {
List<Integer> arr = Arrays.asList( 1 , 2 , 3 , 4 , 5 ); // Given array
int K = 4 ; // XOR value to search for
int N = arr.size(); // Size of the array
System.out.println( "Count of subsets is " + subsetXOR(arr, N, K)); // Output the count of subsets
}
} //Author: Amar Singh IIT BHU Varanasi |
# Author: Amar Singh IIT BHU Varanasi def generate(arr, curr, n, XOR, xorSubset):
if curr = = n:
XOR[xorSubset] = XOR.get(xorSubset, 0 ) + 1 # Store XOR values in the map
return
generate(arr, curr + 1 , n, XOR, xorSubset ^ arr[curr]) # Include current element in XOR
generate(arr, curr + 1 , n, XOR, xorSubset) # Exclude current element from XOR
def subsetXOR(arr, K):
N = len (arr)
P1, P2 = {}, {}
generate(arr, 0 , N / / 2 , P1, 0 ) # Generate XOR for the first half
generate(arr, N / / 2 , N, P2, 0 ) # Generate XOR for the second half
cnt = 0 # Counter for subsets with XOR equal to K
for x in P1:
find = K ^ x # Find corresponding value in the second half
cnt + = P2.get(find, 0 ) * P1[x] # Increment count with the product of occurrences
return cnt # Return count of subsets with XOR equal to K
arr = [ 1 , 2 , 3 , 4 , 5 ] # Given array
k = 4 # XOR value to search for
print ( "Count of subsets is" , subsetXOR(arr, k)) # Output the count of subsets
# Author: Amar Singh IIT BHU Varanasi |
using System;
using System.Collections.Generic;
class Program
{ // Function to generate all subsets and their XOR values
static void Generate(List< int > arr, int curr, int n, Dictionary< int , int > XOR, int xorSubset)
{
if (curr == n)
{
if (XOR.ContainsKey(xorSubset))
XOR[xorSubset]++;
else
XOR[xorSubset] = 1;
return ;
}
Generate(arr, curr + 1, n, XOR, xorSubset ^ arr[curr]); // Include current element in XOR
Generate(arr, curr + 1, n, XOR, xorSubset); // Exclude current element from XOR
}
// Function to count subsets with XOR equal to K
static int SubsetXOR(List< int > arr, int K)
{
int N = arr.Count;
Dictionary< int , int > P1 = new Dictionary< int , int >();
Dictionary< int , int > P2 = new Dictionary< int , int >();
Generate(arr, 0, N / 2, P1, 0); // Generate XOR for the first half
Generate(arr, N / 2, N, P2, 0); // Generate XOR for the second half
int cnt = 0; // Counter for subsets with XOR equal to K
foreach ( var x in P1)
{
int find = K ^ x.Key; // Find corresponding value in the second half
if (P2.ContainsKey(find))
cnt += P2[find] * x.Value; // Increment count with the product of occurrences
}
return cnt; // Return count of subsets with XOR equal to K
}
static void Main()
{
List< int > arr = new List< int > { 1, 2, 3, 4, 5 }; // Given array
int k = 4; // XOR value to search for
Console.WriteLine( "Count of subsets is " + SubsetXOR(arr, k)); // Output the count of subsets
}
} |
function generate(arr, curr, n, XOR, xorSubset) {
if (curr === n) {
XOR[xorSubset] = (XOR[xorSubset] || 0) + 1; // Store XOR values in the map
return ;
}
generate(arr, curr + 1, n, XOR, xorSubset ^ arr[curr]); // Include current element in XOR
generate(arr, curr + 1, n, XOR, xorSubset); // Exclude current element from XOR
} function subsetXOR(arr, K) {
const N = arr.length;
const P1 = {};
const P2 = {};
generate(arr, 0, Math.floor(N / 2), P1, 0); // Generate XOR for the first half
generate(arr, Math.floor(N / 2), N, P2, 0); // Generate XOR for the second half
let cnt = 0; // Counter for subsets with XOR equal to K
for (let x in P1) {
x = parseInt(x);
const find = K ^ x; // Find corresponding value in the second half
if (P2[find] !== undefined) {
cnt += P2[find] * P1[x]; // Increment count with the product of occurrences
}
}
return cnt; // Return count of subsets with XOR equal to K
} const arr = [1, 2, 3, 4, 5]; // Given array
const k = 4; // XOR value to search for
const N = 5; // Size of the array
console.log( "Count of subsets is " + subsetXOR(arr, k)); // Output the count of subsets
|
Count of subsets is 4
Time Complexity: O(2n/2): where n is the size of the array
Auxiliary Space: O(2n/2): where n is the size of the array, this space is used as auxiliary stack space for recursion
Dynamic Programming Approach O(n*m):
We define a number m such that m = pow(2,(log2(max(arr))+1)) – 1. This number is actually the maximum value any XOR subset will acquire. We get this number by counting bits in largest number. We create a 2D array dp[n+1][m+1], such that dp[i][j] equals to the number of subsets having XOR value j from subsets of arr[0…i-1].
We fill the dp array as following:
- We initialize all values of dp[i][j] as 0.
- Set value of dp[0][0] = 1 since XOR of an empty set is 0.
-
Iterate over all the values of arr[i] from left to right and for each arr[i], iterate over all the possible values of XOR i.e from 0 to m (both inclusive) and fill the dp array asfollowing:
for i = 1 to n:
for j = 0 to m:
dp[i][j] = dp[i-1][j] + dp[i-1][j^arr[i-1]]
This can be explained as, if there is a subset arr[0…i-2] with XOR value j, then there also exists a subset arr[0…i-1] with XOR value j. also if there exists a subset arr[0….i-2] with XOR value j^arr[i] then clearly there exist a subset arr[0…i-1] with XOR value j, as j ^ arr[i-1] ^ arr[i-1] = j. - Counting the number of subsets with XOR value k: Since dp[i][j] is the number of subsets having j as XOR value from the subsets of arr[0..i-1], then the number of subsets from set arr[0..n] having XOR value as K will be dp[n][K]
// arr dynamic programming solution to finding the number // of subsets having xor of their elements as k #include <bits/stdc++.h> using namespace std;
// Returns count of subsets of arr[] with XOR value equals // to k. int subsetXOR( int arr[], int n, int k)
{ // Find maximum element in arr[]
int max_ele = arr[0];
for ( int i = 1; i < n; i++)
if (arr[i] > max_ele)
max_ele = arr[i];
// Maximum possible XOR value
int m = (1 << ( int )(log2(max_ele) + 1)) - 1;
if (k > m)
return 0;
// The value of dp[i][j] is the number of subsets having
// XOR of their elements as j from the set arr[0...i-1]
int dp[n + 1][m + 1];
// Initializing all the values of dp[i][j] as 0
for ( int i = 0; i <= n; i++)
for ( int j = 0; j <= m; j++)
dp[i][j] = 0;
// The xor of empty subset is 0
dp[0][0] = 1;
// Fill the dp table
for ( int i = 1; i <= n; i++)
for ( int j = 0; j <= m; j++)
dp[i][j]
= dp[i - 1][j] + dp[i - 1][j ^ arr[i - 1]];
// The answer is the number of subset from set
// arr[0..n-1] having XOR of elements as k
return dp[n][k];
} // Driver program to test above function int main()
{ int arr[] = { 1, 2, 3, 4, 5 };
int k = 4;
int n = sizeof (arr) / sizeof (arr[0]);
cout << "Count of subsets is " << subsetXOR(arr, n, k);
return 0;
} // This code is contributed by Aditya Kumar (adityakumar129) |
// arr dynamic programming solution to finding the number // of subsets having xor of their elements as k #include <math.h> #include <stdio.h> // Returns count of subsets of arr[] with XOR value // equals to k. int subsetXOR( int arr[], int n, int k)
{ // Find maximum element in arr[]
int max_ele = arr[0];
for ( int i = 1; i < n; i++)
if (arr[i] > max_ele)
max_ele = arr[i];
// Maximum possible XOR value
int m = (1 << ( int )(log2(max_ele) + 1)) - 1;
if (k > m)
return 0;
// The value of dp[i][j] is the number of subsets having
// XOR of their elements as j from the set arr[0...i-1]
int dp[n + 1][m + 1];
// Initializing all the values of dp[i][j] as 0
for ( int i = 0; i <= n; i++)
for ( int j = 0; j <= m; j++)
dp[i][j] = 0;
// The xor of empty subset is 0
dp[0][0] = 1;
// Fill the dp table
for ( int i = 1; i <= n; i++)
for ( int j = 0; j <= m; j++)
dp[i][j]
= dp[i - 1][j] + dp[i - 1][j ^ arr[i - 1]];
// The answer is the number of subset from set
// arr[0..n-1] having XOR of elements as k
return dp[n][k];
} // Driver program to test above function int main()
{ int arr[] = { 1, 2, 3, 4, 5 };
int k = 4;
int n = sizeof (arr) / sizeof (arr[0]);
printf ( "Count of subsets is %d" , subsetXOR(arr, n, k));
return 0;
} // This code is contributed by Aditya Kumar (adityakumar129) |
// Java dynamic programming solution to finding the number // of subsets having xor of their elements as k import java.util.*;
import java.io.*;
class GFG {
// Returns count of subsets of arr[] with XOR value
// equals to k.
static int subsetXOR( int [] arr, int n, int k)
{
// Find maximum element in arr[]
int max_ele = arr[ 0 ];
for ( int i = 1 ; i < n; i++)
if (arr[i] > max_ele)
max_ele = arr[i];
// Maximum possible XOR value
int m = ( 1 << ( int )(Math.log(max_ele) / Math.log( 2 ) + 1 )) - 1 ;
if (k > m)
return 0 ;
// The value of dp[i][j] is the number of subsets
// having XOR of their elements as j from the set
// arr[0...i-1]
int [][] dp = new int [n + 1 ][m + 1 ];
// Initializing all the values of dp[i][j] as 0
for ( int i = 0 ; i <= n; i++)
for ( int j = 0 ; j <= m; j++)
dp[i][j] = 0 ;
// The xor of empty subset is 0
dp[ 0 ][ 0 ] = 1 ;
// Fill the dp table
for ( int i = 1 ; i <= n; i++)
for ( int j = 0 ; j <= m; j++)
dp[i][j] = dp[i - 1 ][j] + dp[i - 1 ][j ^ arr[i - 1 ]];
// The answer is the number of subset from set
// arr[0..n-1] having XOR of elements as k
return dp[n][k];
}
// Driver code
public static void main(String arg[])
{
int [] arr = { 1 , 2 , 3 , 4 , 5 };
int k = 4 ;
int n = arr.length;
System.out.println( "Count of subsets is " + subsetXOR(arr, n, k));
}
} // This code is contributed by Aditya Kumar (adityakumar129) |
# Python 3 arr dynamic programming solution # to finding the number of subsets having # xor of their elements as k import math
# Returns count of subsets of arr[] with # XOR value equals to k. def subsetXOR(arr, n, k):
# Find maximum element in arr[]
max_ele = arr[ 0 ]
for i in range ( 1 , n):
if arr[i] > max_ele :
max_ele = arr[i]
# Maximum possible XOR value
m = ( 1 << ( int )(math.log2(max_ele) + 1 )) - 1
if ( k > m ):
return 0
# The value of dp[i][j] is the number
# of subsets having XOR of their elements
# as j from the set arr[0...i-1]
# Initializing all the values
# of dp[i][j] as 0
dp = [[ 0 for i in range (m + 1 )]
for i in range (n + 1 )]
# The xor of empty subset is 0
dp[ 0 ][ 0 ] = 1
# Fill the dp table
for i in range ( 1 , n + 1 ):
for j in range (m + 1 ):
dp[i][j] = (dp[i - 1 ][j] +
dp[i - 1 ][j ^ arr[i - 1 ]])
# The answer is the number of subset
# from set arr[0..n-1] having XOR of
# elements as k
return dp[n][k]
# Driver Code arr = [ 1 , 2 , 3 , 4 , 5 ]
k = 4
n = len (arr)
print ( "Count of subsets is" ,
subsetXOR(arr, n, k))
# This code is contributed # by sahishelangia |
// C# dynamic programming solution to finding the number // of subsets having xor of their elements as k using System;
class GFG
{ // Returns count of subsets of arr[] with // XOR value equals to k. static int subsetXOR( int []arr, int n, int k)
{ // Find maximum element in arr[]
int max_ele = arr[0];
for ( int i = 1; i < n; i++)
if (arr[i] > max_ele)
max_ele = arr[i];
// Maximum possible XOR value
int m = (1 << ( int )(Math.Log(max_ele,2) + 1) ) - 1;
if ( k > m ){
return 0;
}
// The value of dp[i][j] is the number of subsets having
// XOR of their elements as j from the set arr[0...i-1]
int [,]dp= new int [n+1,m+1];
// Initializing all the values of dp[i][j] as 0
for ( int i = 0; i <= n; i++)
for ( int j = 0; j <= m; j++)
dp[i, j] = 0;
// The xor of empty subset is 0
dp[0, 0] = 1;
// Fill the dp table
for ( int i = 1; i <= n; i++)
for ( int j = 0; j <= m; j++)
dp[i, j] = dp[i-1, j] + dp[i-1, j^arr[i-1]];
// The answer is the number of subset from set
// arr[0..n-1] having XOR of elements as k
return dp[n, k];
} // Driver code
static public void Main ()
{
int []arr = {1, 2, 3, 4, 5};
int k = 4;
int n = arr.Length;
Console.WriteLine ( "Count of subsets is " + subsetXOR(arr, n, k));
}
} // This code is contributed by jit_t. |
<script> // Javascript dynamic programming solution
// to finding the number of subsets
// having xor of their elements as k
// Returns count of subsets of arr[] with
// XOR value equals to k.
function subsetXOR(arr, n, k)
{
// Find maximum element in arr[]
let max_ele = arr[0];
for (let i = 1; i < n; i++)
if (arr[i] > max_ele)
max_ele = arr[i];
// Maximum possible XOR value
let m = (1 << parseInt(Math.log(max_ele) /
Math.log(2) + 1, 10) ) - 1;
if (k > m)
{
return 0;
}
// The value of dp[i][j] is the number
// of subsets having XOR of their
// elements as j from the set arr[0...i-1]
let dp = new Array(n + 1);
// Initializing all the values of dp[i][j] as 0
for (let i = 0; i <= n; i++)
{
dp[i] = new Array(m + 1);
for (let j = 0; j <= m; j++)
{
dp[i][j] = 0;
}
}
// The xor of empty subset is 0
dp[0][0] = 1;
// Fill the dp table
for (let i = 1; i <= n; i++)
for (let j = 0; j <= m; j++)
dp[i][j] = dp[i - 1][j] +
dp[i - 1][j ^ arr[i - 1]];
// The answer is the number of
// subset from set arr[0..n-1]
// having XOR of elements as k
return dp[n][k];
}
let arr = [ 1, 2, 3, 4, 5 ];
let k = 4;
let n = arr.length;
document.write( "Count of subsets is " + subsetXOR(arr, n, k));
</script> |
<?php // PHP arr dynamic programming // solution to finding the number // of subsets having xor of their // elements as k // Returns count of subsets of // arr[] with XOR value equals to k. function subsetXOR( $arr , $n , $k )
{ // Find maximum element in arr[]
$max_ele = $arr [0];
for ( $i = 1; $i < $n ; $i ++)
if ( $arr [ $i ] > $max_ele )
$max_ele = $arr [ $i ];
// Maximum possible XOR value
$m = (1 << (int)(log( $max_ele ,
2) + 1) ) - 1;
if ( $k > $m ){
return 0;
}
// The value of dp[i][j] is the
// number of subsets having
// XOR of their elements as j
// from the set arr[0...i-1]
// Initializing all the
// values of dp[i][j] as 0
for ( $i = 0; $i <= $n ; $i ++)
for ( $j = 0; $j <= $m ; $j ++)
$dp [ $i ][ $j ] = 0;
// The xor of empty subset is 0
$dp [0][0] = 1;
// Fill the dp table
for ( $i = 1; $i <= $n ; $i ++)
for ( $j = 0; $j <= $m ; $j ++)
$dp [ $i ][ $j ] = $dp [ $i - 1][ $j ] +
$dp [ $i - 1][ $j ^
$arr [ $i - 1]];
// The answer is the number
// of subset from set arr[0..n-1]
// having XOR of elements as k
return $dp [ $n ][ $k ];
} // Driver Code $arr = array (1, 2, 3, 4, 5);
$k = 4;
$n = sizeof( $arr );
echo "Count of subsets is " ,
subsetXOR( $arr , $n , $k );
// This code is contributed by ajit ?> |
Count of subsets is 4
Time Complexity: O(n * m)
Auxiliary Space: O(n * m)
Efficient approach : Space optimization
In previous approach the current value dp[i][j] is only depend upon the current and previous row values of DP. So to optimize the space complexity we use a single 1D array to store the computations.
Implementation steps:
- Create a 1D vector dp of size m+1.
- Set a base case by initializing the values of DP .
- Now iterate over subproblems by the help of nested loop and get the current value from previous computations.
- Now Create a temporary vector temp used to store the current values from previous computations.
- At last return and print the final answer stored in dp[k].
Implementation:
#include <bits/stdc++.h> using namespace std;
// Function to count the number of subsets with XOR equal to k int subsetXOR( int arr[], int n, int k)
{ // Find the maximum element in the array
int max_ele = arr[0];
for ( int i = 1; i < n; i++)
if (arr[i] > max_ele)
max_ele = arr[i];
// Calculate the maximum possible XOR value
int m = (1 << ( int )(log2(max_ele) + 1)) - 1;
// If k is greater than the maximum possible XOR value, return 0
if (k > m)
return 0;
// Create a vector to store the count of subsets with XOR equal to each possible value
vector< int > dp(m + 1);
dp[0] = 1; // There is one subset with XOR equal to 0 (empty subset)
// Iterate over the array elements
for ( int i = 1; i <= n; i++)
{
// Create a temporary vector to store the previous row values
vector< int > temp = dp;
// Update the dp vector based on the previous row values
for ( int j = 0; j <= m; j++)
{
// Calculate the count of subsets with XOR equal to j using the previous row values
dp[j] = temp[j] + temp[j ^ arr[i - 1]];
}
}
// Return the count of subsets with XOR equal to k
return dp[k];
} int main()
{ int arr[] = {1, 2, 3, 4, 5};
int k = 4;
int n = sizeof (arr) / sizeof (arr[0]);
// Call the subsetXOR function and print the result
cout << "Count of subsets is " << subsetXOR(arr, n, k);
return 0;
} |
import java.util.Arrays;
public class SubsetXOR {
// Function to count the number of subsets with XOR equal to k
public static int subsetXOR( int [] arr, int n, int k) {
// Find the maximum element in the array
int maxEle = arr[ 0 ];
for ( int i = 1 ; i < n; i++) {
if (arr[i] > maxEle) {
maxEle = arr[i];
}
}
// Calculate the maximum possible XOR value
int m = ( 1 << ( int ) (Math.log(maxEle) / Math.log( 2 ) + 1 )) - 1 ;
// If k is greater than the maximum possible XOR value, return 0
if (k > m) {
return 0 ;
}
// Create an array to store the count of subsets with XOR equal to each possible value
int [] dp = new int [m + 1 ];
dp[ 0 ] = 1 ; // There is one subset with XOR equal to 0 (empty subset)
// Iterate over the array elements
for ( int i = 0 ; i < n; i++) {
// Create a temporary array to store the previous row values
int [] temp = Arrays.copyOf(dp, dp.length);
// Update the dp array based on the previous row values
for ( int j = 0 ; j <= m; j++) {
// Calculate the count of subsets with XOR equal to j using the previous row values
dp[j] = temp[j] + temp[j ^ arr[i]];
}
}
// Return the count of subsets with XOR equal to k
return dp[k];
}
public static void main(String[] args) {
int [] arr = { 1 , 2 , 3 , 4 , 5 };
int k = 4 ;
int n = arr.length;
// Call the subsetXOR function and print the result
System.out.println( "Count of subsets is " + subsetXOR(arr, n, k));
}
} |
# Function to count the number of subsets with XOR equal to k def subsetXOR(arr, k):
n = len (arr)
# Find the maximum element in the array
max_ele = max (arr)
# Calculate the maximum possible XOR value
m = ( 1 << ( int (max_ele).bit_length())) - 1
# If k is greater than the maximum possible XOR value, return 0
if k > m:
return 0
# Create a list to store the count of subsets with XOR equal to each possible value
dp = [ 0 ] * (m + 1 )
dp[ 0 ] = 1 # There is one subset with XOR equal to 0 (empty subset)
# Iterate over the array elements
for i in range ( 1 , n + 1 ):
# Create a temporary list to store the previous row values
temp = dp[:]
# Update the dp list based on the previous row values
for j in range (m + 1 ):
# Calculate the count of subsets with XOR equal to j using the previous row values
dp[j] = temp[j] + temp[j ^ arr[i - 1 ]]
# Return the count of subsets with XOR equal to k
return dp[k]
if __name__ = = "__main__" :
arr = [ 1 , 2 , 3 , 4 , 5 ]
k = 4
# Call the subsetXOR function and print the result
print ( "Count of subsets is" , subsetXOR(arr, k))
|
using System;
using System.Linq;
using System.Collections.Generic;
class SubsetXORProgram {
// Function to count the number of subsets with XOR
// equal to k
static int SubsetXOR( int [] arr, int n, int k)
{
// Find the maximum element in the array
int maxEle = arr.Max();
// Calculate the maximum possible XOR value
int m = (1 << ( int )(Math.Ceiling(
Math.Log(maxEle + 1, 2))))
- 1;
// If k is greater than the maximum possible XOR
// value, return 0
if (k > m)
return 0;
// Create a list to store the count of subsets with
// XOR equal to each possible value
List< int > dp = new List< int >( new int [m + 1]);
dp[0] = 1; // There is one subset with XOR equal to
// 0 (empty subset)
// Iterate over the array elements
for ( int i = 1; i <= n; i++) {
// Create a temporary list to store the previous
// row values
List< int > temp = new List< int >(dp);
// Update the dp list based on the previous row
// values
for ( int j = 0; j <= m; j++) {
// Calculate the count of subsets with XOR
// equal to j using the previous row values
dp[j] = temp[j] + temp[j ^ arr[i - 1]];
}
}
// Return the count of subsets with XOR equal to k
return dp[k];
}
static void Main()
{
int [] arr = { 1, 2, 3, 4, 5 };
int k = 4;
int n = arr.Length;
// Call the SubsetXOR function and print the result
Console.WriteLine( "Count of subsets is "
+ SubsetXOR(arr, n, k));
}
} |
// Function to count the number of subsets with XOR equal to k function subsetXOR(arr, n, k) {
// Find the maximum element in the array
let maxEle = arr[0];
for (let i = 1; i < n; i++) {
if (arr[i] > maxEle) {
maxEle = arr[i];
}
}
// Calculate the maximum possible XOR value
let m = (1 << (Math.log2(maxEle) + 1)) - 1;
// If k is greater than the maximum possible XOR value, return 0
if (k > m) {
return 0;
}
// Create an array to store the count of subsets with XOR equal to each possible value
let dp = new Array(m + 1).fill(0);
dp[0] = 1; // There is one subset with XOR equal to 0 (empty subset)
// Iterate over the array elements
for (let i = 0; i < n; i++) {
// Create a temporary array to store the previous row values
let temp = [...dp];
// Update the dp array based on the previous row values
for (let j = 0; j <= m; j++) {
// Calculate the count of subsets with XOR equal to j using the previous row values
dp[j] = temp[j] + temp[j ^ arr[i]];
}
}
// Return the count of subsets with XOR equal to k
return dp[k];
} // Main function function main() {
let arr = [1, 2, 3, 4, 5];
let k = 4;
let n = arr.length;
// Call the subsetXOR function and print the result
console.log( "Count of subsets is " + subsetXOR(arr, n, k));
} // Call the main function main(); |
Output
Count of subsets is 4
Time Complexity: O(n * m)
Auxiliary Space: O(m)
This article is contributed by Pranay Pandey.