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}
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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.
C++
//Author - RainX (ABHIJIT ROY, NIT AGARTALA)#include <bits/stdc++.h>using namespace std;void generate(vector<int>& arr, int curr, int n, unordered_map<int, int>& XOR, int xorSubset) { if (curr == n) { XOR[xorSubset]++; return; } generate(arr, curr + 1, n, XOR, xorSubset ^ arr[curr]); generate(arr, curr + 1, n, XOR, xorSubset);}int subsetXOR(vector<int>& arr, int N, int K) { unordered_map<int, int> P1, P2; generate(arr, 0, N / 2, P1, 0); // first half, took O(2^(n/2)) generate(arr, N / 2, N, P2, 0); // second half, took O(2^(n/2)) int cnt = 0; for (auto x : P1) { // finding corresponding subset in P2 int find = K ^ x.first; cnt += (P2[find] * x.second); } return cnt;}int main() { vector<int> arr = { 1, 2, 3, 4, 5 }; int k = 4; int N = 5; cout << "Count of subsets is " << subsetXOR(arr, N, k); return 0;}// Code by RainX (Abhijit Roy, NIT AGARTALA) |
Java
//Author: Amar Singh IIT BHU Varanasiimport java.util.*;class Main { 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); return; } generate(arr, curr + 1, n, XOR, xorSubset ^ arr.get(curr)); generate(arr, curr + 1, n, XOR, xorSubset); } 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(arr, N / 2, N, P2, 0); int cnt = 0; for (int x : P1.keySet()) { int find = K ^ x; if (P2.containsKey(find)) { cnt += P2.get(find) * P1.get(x); } } return cnt; } public static void main(String[] args) { List<Integer> arr = Arrays.asList(1, 2, 3, 4, 5); int K = 4; int N = arr.size(); System.out.println("Count of subsets is " + subsetXOR(arr, N, K)); }}//Author: Amar Singh IIT BHU Varanasi |
Output
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]
C++
// 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 functionint 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) |
C
// 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 functionint 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
// Java dynamic programming solution to finding the number// of subsets having xor of their elements as kimport 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) |
Python3
# Python 3 arr dynamic programming solution # to finding the number of subsets having # xor of their elements as kimport 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 Codearr = [1, 2, 3, 4, 5]k = 4n = len(arr)print("Count of subsets is", subsetXOR(arr, n, k))# This code is contributed# by sahishelangia |
C#
// C# dynamic programming solution to finding the number// of subsets having xor of their elements as kusing 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. |
PHP
<?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?> |
Javascript
<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> |
Output
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:
C++
#include <bits/stdc++.h>using namespace std;// Function to count the number of subsets with XOR equal to kint 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;} |
Output
Count of subsets is 4
Time Complexity: O(n * m)
Auxiliary Space: O(m)
This article is contributed by Pranay Pandey. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
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