Count distinct possible Bitwise XOR values of subsets of an array
Last Updated :
28 Feb, 2022
Given an array arr[] consisting of N integers, the task is to find the size of the set S such that Bitwise XOR of any subset of the array arr[] exists in the set S.
Examples:
Input: arr[] = {1, 2, 3, 4, 5}
Output: 8
Explanation:
All possible Bitwise XOR values of subsets of the array arr[] are {0, 1, 2, 3, 4, 5, 6, 7}.
Therefore, the size of the required set is 8.
Input: arr[] = {6}
Output: 1
Naive Approach: The simplest approach is to generate all possible non-empty subsets of the given array arr[] and store the Bitwise XOR of all subsets in a set. After generating all the subsets, print the size of the set obtained as the result.
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Stores the Bitwise XOR
// of every possible subset
unordered_set<int> s;
// Function to generate all
// combinations of subsets and
// store their Bitwise XOR in set S
void countXOR(int arr[], int comb[],
int start, int end,
int index, int r)
{
// If the end of the
// subset is reached
if (index == r) {
// Stores the Bitwise XOR
// of the current subset
int new_xor = 0;
// Iterate comb[] to find XOR
for (int j = 0; j < r; j++) {
new_xor ^= comb[j];
}
// Insert the Bitwise
// XOR of R elements
s.insert(new_xor);
return;
}
// Otherwise, iterate to
// generate all possible subsets
for (int i = start;
i <= end && end - i + 1 >= r - index; i++) {
comb[index] = arr[i];
// Recursive call for next index
countXOR(arr, comb, i + 1, end,
index + 1, r);
}
}
// Function to find the size of the
// set having Bitwise XOR of all the
// subsets of the given array
void maxSizeSet(int arr[], int N)
{
// Iterate over the given array
for (int r = 2; r <= N; r++) {
int comb[r + 1];
// Generate all possible subsets
countXOR(arr, comb, 0, N - 1,
0, r);
}
// Print the size of the set
cout << s.size() << endl;
}
// Driver Code
int main()
{
int arr[] = { 1, 2, 3, 4, 5 };
int N = sizeof(arr) / sizeof(arr[0]);
// Function Call
maxSizeSet(arr, N);
return 0;
}
// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG{
// Stores the Bitwise XOR
// of every possible subset
static HashSet<Integer> s;
// Function to generate all
// combinations of subsets and
// store their Bitwise XOR in set S
static void countXOR(int arr[], int comb[],
int start, int end,
int index, int r)
{
// If the end of the
// subset is reached
if (index == r)
{
// Stores the Bitwise XOR
// of the current subset
int new_xor = 0;
// Iterate comb[] to find XOR
for(int j = 0; j < r; j++)
{
new_xor ^= comb[j];
}
// Insert the Bitwise
// XOR of R elements
s.add(new_xor);
return;
}
// Otherwise, iterate to
// generate all possible subsets
for(int i = start;
i <= end && end - i + 1 >= r - index;
i++)
{
comb[index] = arr[i];
// Recursive call for next index
countXOR(arr, comb, i + 1, end, index + 1, r);
}
}
// Function to find the size of the
// set having Bitwise XOR of all the
// subsets of the given array
static void maxSizeSet(int arr[], int N)
{
// Iterate over the given array
for(int r = 1; r <= N; r++)
{
int comb[] = new int[r + 1];
// Generate all possible subsets
countXOR(arr, comb, 0, N - 1, 0, r);
}
// Print the size of the set
System.out.println(s.size());
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 1, 2, 3, 4, 5 };
int N = arr.length;
// Initialize set
s = new HashSet<>();
// Function Call
maxSizeSet(arr, N);
}
}
// This code is contributed by Kingash
# Python3 program for the above approach
# Stores the Bitwise XOR
# of every possible subset
s = set([])
# Function to generate all
# combinations of subsets and
# store their Bitwise XOR in set S
def countXOR(arr, comb, start, end, index, r):
# If the end of the
# subset is reached
if (index == r) :
# Stores the Bitwise XOR
# of the current subset
new_xor = 0
# Iterate comb[] to find XOR
for j in range(r):
new_xor ^= comb[j]
# Insert the Bitwise
# XOR of R elements
s.add(new_xor)
return
# Otherwise, iterate to
# generate all possible subsets
i = start
while i <= end and (end - i + 1) >= (r - index):
comb[index] = arr[i]
# Recursive call for next index
countXOR(arr, comb, i + 1, end, index + 1, r)
i += 1
# Function to find the size of the
# set having Bitwise XOR of all the
# subsets of the given array
def maxSizeSet(arr, N):
# Iterate over the given array
for r in range(2, N + 1):
comb = [0]*(r + 1)
# Generate all possible subsets
countXOR(arr, comb, 0, N - 1, 0, r)
# Print the size of the set
print(len(s))
arr = [ 1, 2, 3, 4, 5 ]
N = len(arr)
# Function Call
maxSizeSet(arr, N)
# This code is contributed by decode2207.
// C# program for the above approach
using System;
using System.Collections.Generic;
public class GFG
{
// Stores the Bitwise XOR
// of every possible subset
static HashSet<int> s;
// Function to generate all
// combinations of subsets and
// store their Bitwise XOR in set S
static void countXOR(int []arr, int []comb,
int start, int end,
int index, int r)
{
// If the end of the
// subset is reached
if (index == r)
{
// Stores the Bitwise XOR
// of the current subset
int new_xor = 0;
// Iterate comb[] to find XOR
for(int j = 0; j < r; j++)
{
new_xor ^= comb[j];
}
// Insert the Bitwise
// XOR of R elements
s.Add(new_xor);
return;
}
// Otherwise, iterate to
// generate all possible subsets
for(int i = start;
i <= end && end - i + 1 >= r - index;
i++)
{
comb[index] = arr[i];
// Recursive call for next index
countXOR(arr, comb, i + 1, end, index + 1, r);
}
}
// Function to find the size of the
// set having Bitwise XOR of all the
// subsets of the given array
static void maxSizeSet(int []arr, int N)
{
// Iterate over the given array
for(int r = 1; r <= N; r++)
{
int []comb = new int[r + 1];
// Generate all possible subsets
countXOR(arr, comb, 0, N - 1, 0, r);
}
// Print the size of the set
Console.WriteLine(s.Count);
}
// Driver Code
public static void Main(String[] args)
{
int []arr = { 1, 2, 3, 4, 5 };
int N = arr.Length;
// Initialize set
s = new HashSet<int>();
// Function Call
maxSizeSet(arr, N);
}
}
// This code is contributed by Amit Katiyar
<script>
// JavaScript program for the above approach
// Stores the Bitwise XOR
// of every possible subset
let s;
// Function to generate all
// combinations of subsets and
// store their Bitwise XOR in set S
function countXOR(arr,comb,start,end,index,r)
{
// If the end of the
// subset is reached
if (index == r)
{
// Stores the Bitwise XOR
// of the current subset
let new_xor = 0;
// Iterate comb[] to find XOR
for(let j = 0; j < r; j++)
{
new_xor ^= comb[j];
}
// Insert the Bitwise
// XOR of R elements
s.add(new_xor);
return;
}
// Otherwise, iterate to
// generate all possible subsets
for(let i = start;
i <= end && end - i + 1 >= r - index;
i++)
{
comb[index] = arr[i];
// Recursive call for next index
countXOR(arr, comb, i + 1, end, index + 1, r);
}
}
// Function to find the size of the
// set having Bitwise XOR of all the
// subsets of the given array
function maxSizeSet(arr,N)
{
// Iterate over the given array
for(let r = 1; r <= N; r++)
{
let comb = new Array(r + 1);
// Generate all possible subsets
countXOR(arr, comb, 0, N - 1, 0, r);
}
// Print the size of the set
document.write(s.size);
}
// Driver Code
let arr=[1, 2, 3, 4, 5 ];
let N = arr.length;
// Initialize set
s = new Set();
// Function Call
maxSizeSet(arr, N);
// This code is contributed by avanitrachhadiya2155
</script>
Time Complexity: O(NN)
Auxiliary Space: O(N)
Efficient approach: The above approach can be optimized by using the Greedy Approach and Gaussian Elimination. Follow the steps below to solve the problem:
- Initialize an auxiliary array, say dp[] of size 20, that stores the mask of each array element and initializes it with 0.
- Initialize a variable, say ans as 0, to store the size of the array dp[].
- Traverse the given array arr[] and for each array element arr[], perform the following steps:
- Initialize a variable, say mask as arr[i], to check the position of the most significant set bit.
- If the ith bit from the right of arr[i] is set and dp[i] is 0, then update the array dp[i] as 2i and increment the value of ans by 1 and break out of the loop.
- Otherwise, update the value of the mask as the Bitwise XOR of the mask and dp[i].
- After completing the above steps, print the value of 2ans as the resultant size of the required set of elements.
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
int const size = 20;
// Stores the mask of the vector
int dp[size];
// Stores the current size of dp[]
int ans;
// Function to store the
// mask of given integer
void insertVector(int mask)
{
// Iterate over the range [0, 20]
for (int i = 0; i < 20; i++) {
// If i-th bit 0
if ((mask & 1 << i) == 0)
continue;
// If dp[i] is zero
if (!dp[i]) {
// Store the position in dp
dp[i] = mask;
// Increment the answer
++ans;
// Return from the loop
return;
}
// mask = mask XOR dp[i]
mask ^= dp[i];
}
}
// Function to find the size of the
// set having Bitwise XOR of all the
// subset of the given array
void maxSizeSet(int arr[], int N)
{
// Traverse the array
for (int i = 0; i < N; i++) {
insertVector(arr[i]);
}
// Print the answer
cout << (1 << ans) << endl;
}
// Driver Code
int main()
{
int arr[] = { 1, 2, 3, 4, 5 };
int N = sizeof(arr) / sizeof(arr[0]);
// Function Call
maxSizeSet(arr, N);
return 0;
}
// Java program for the above approach
class GFG{
final static int size = 20;
// Stores the mask of the vector
static int[] dp = new int[size];
// Stores the current size of dp[]
static int ans;
// Function to store the
// mask of given integer
static void insertVector(int mask)
{
// Iterate over the range [0, 20]
for (int i = 0; i < 20; i++) {
// If i-th bit 0
if ((mask & 1 << i) == 0)
continue;
// If dp[i] is zero
if (dp[i]==0) {
// Store the position in dp
dp[i] = mask;
// Increment the answer
++ans;
// Return from the loop
return;
}
// mask = mask XOR dp[i]
mask ^= dp[i];
}
}
// Function to find the size of the
// set having Bitwise XOR of all the
// subset of the given array
static void maxSizeSet(int[] arr, int N)
{
// Traverse the array
for (int i = 0; i < N; i++) {
insertVector(arr[i]);
}
// Print the answer
System.out.println(1<<ans);
}
// Driver code
public static void main(String[] args)
{
int[] arr = { 1, 2, 3, 4, 5 };
int N = arr.length;
// Function Call
maxSizeSet(arr, N);
}
}
//This code is contributed by Hritik Dwivedi
# Python3 program for the above approach
# Stores the mask of the vector
dp = [0]*20
# Stores the current 20 of dp[]
ans = 0
# Function to store the
# mask of given integer
def insertVector(mask):
global dp, ans
# Iterate over the range [0, 20]
for i in range(20):
# If i-th bit 0
if ((mask & 1 << i) == 0):
continue
# If dp[i] is zero
if (not dp[i]):
# Store the position in dp
dp[i] = mask
# Increment the answer
ans += 1
# Return from the loop
return
# mask = mask XOR dp[i]
mask ^= dp[i]
# Function to find the 20 of the
# set having Bitwise XOR of all the
# subset of the given array
def maxSizeSet(arr, N):
# Traverse the array
for i in range(N):
insertVector(arr[i])
# Print the answer
print ((1 << ans))
# Driver Code
if __name__ == '__main__':
arr = [1, 2, 3, 4, 5]
N = len(arr)
# Function Call
maxSizeSet(arr, N)
# This code is contributed by mohit kumar 29.
// C# program for the above approach
using System;
class GFG{
static int size = 20;
// Stores the mask of the vector
static int[] dp = new int[size];
// Stores the current size of dp[]
static int ans;
// Function to store the
// mask of given integer
static void insertVector(int mask)
{
// Iterate over the range [0, 20]
for(int i = 0; i < 20; i++)
{
// If i-th bit 0
if ((mask & 1 << i) == 0)
continue;
// If dp[i] is zero
if (dp[i] == 0)
{
// Store the position in dp
dp[i] = mask;
// Increment the answer
++ans;
// Return from the loop
return;
}
// mask = mask XOR dp[i]
mask ^= dp[i];
}
}
// Function to find the size of the
// set having Bitwise XOR of all the
// subset of the given array
static void maxSizeSet(int[] arr, int N)
{
// Traverse the array
for(int i = 0; i < N; i++)
{
insertVector(arr[i]);
}
// Print the answer
Console.WriteLine(1 << ans);
}
// Driver code
public static void Main(string[] args)
{
int[] arr = { 1, 2, 3, 4, 5 };
int N = arr.Length;
// Function Call
maxSizeSet(arr, N);
}
}
// This code is contributed by ukasp
<script>
// Javascript program for the above approach
let size = 20;
// Stores the mask of the vector
var dp = new Array(size).fill(0);
// Stores the current size of dp[]
let ans = 0;
// Function to store the
// mask of given integer
function insertVector(mask)
{
// Iterate over the range [0, 20]
for(let i = 0; i < 20; i++)
{
// If i-th bit 0
if ((mask & 1 << i) == 0)
continue;
// If dp[i] is zero
if (dp[i] == 0)
{
// Store the position in dp
dp[i] = mask;
// Increment the answer
++ans;
// Return from the loop
return;
}
// mask = mask XOR dp[i]
mask ^= dp[i];
}
}
// Function to find the size of the
// set having Bitwise XOR of all the
// subset of the given array
function maxSizeSet(arr, N)
{
// Traverse the array
for(let i = 0; i < N; i++)
{
insertVector(arr[i]);
}
// Print the answer
document.write(1<<ans);
}
// Driver Code
let arr = [ 1, 2, 3, 4, 5 ];
let N = arr.length;
// Function Call
maxSizeSet(arr, N);
// This code is contributed by nitin_sharma
</script>
Time Complexity: O(M * N)
Auxiliary Space: O(M * N)
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