# Number of subsets with same AND, OR and XOR values in an Array

Given an array arr[] of size N consisting of non-negative integers, the task is to find the number of non-empty subsets of the array such that the bitwise AND, bitwise OR and bitwise XOR values of the subsequence are equal to each other.

Note: Since the answer can be large, mod it with 1000000007.

Examples:

Input: arr[] = [1, 3, 2, 1, 2, 1]
Output: 7
Explanation:
One of the subsequences with equal bitwise Xor, bitwise or and bitwise AND is {1, 1, 1}.

Input: arr = [2, 3, 4, 5]
Output: 4

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive Approach The naive approach for this problem is to traverse through all the subsets of the array in an iterative manner, and for each subset find the bitwise AND, OR and XOR value and check whether they are equal or not. Finally, return the count of such equal subsets.

Below is the implementation of the above approach:

## C++

 // C++ implementation to find the number // of subsets with equal bitwise AND, // OR and XOR values    #include using namespace std; const int mod = 1000000007;    // Function to find the number of // subsets with equal bitwise AND, // OR and XOR values int countSubsets(int a[], int n) {     int answer = 0;        // Traverse through all the subsets     for (int i = 0; i < (1 << n); i++) {            int bitwiseAND = -1;         int bitwiseOR = 0;         int bitwiseXOR = 0;            // Finding the subsets with the bits         // of 'i' which are set         for (int j = 0; j < n; j++) {                // Computing the bitwise AND             if (i & (1 << j)) {                 if (bitwiseAND == -1)                     bitwiseAND = a[j];                 else                     bitwiseAND &= a[j];                    // Computing the bitwise OR                 bitwiseOR |= a[j];                    // Computing the bitwise XOR                 bitwiseXOR ^= a[j];             }         }            // Comparing all the three values         if (bitwiseAND == bitwiseOR             && bitwiseOR == bitwiseXOR)             answer = (answer + 1) % mod;     }     return answer; }    // Driver code int main() {     int N = 6;     int A[N] = { 1, 3, 2, 1, 2, 1 };        cout << countSubsets(A, N);        return 0; }

## Java

 // Java implementation to find the number // of subsets with equal bitwise AND, // OR and XOR values import java.io.*;    class GFG { static int mod = 1000000007;    // Function to find the number of // subsets with equal bitwise AND, // OR and XOR values static int countSubsets(int a[], int n) {     int answer = 0;        // Traverse through all the subsets     for (int i = 0; i < (1 << n); i++) {            int bitwiseAND = -1;         int bitwiseOR = 0;         int bitwiseXOR = 0;            // Finding the subsets with the bits         // of 'i' which are set         for (int j = 0; j < n; j++) {                // Computing the bitwise AND             if ((i & (1 << j)) == 0) {                 if (bitwiseAND == -1)                     bitwiseAND = a[j];                 else                     bitwiseAND &= a[j];                    // Computing the bitwise OR                 bitwiseOR |= a[j];                    // Computing the bitwise XOR                 bitwiseXOR ^= a[j];             }         }            // Comparing all the three values         if (bitwiseAND == bitwiseOR             && bitwiseOR == bitwiseXOR)             answer = (answer + 1) % mod;     }     return answer; }    // Driver Code public static void main (String[] args) {     int N = 6;     int A[] = { 1, 3, 2, 1, 2, 1 };        System.out.print(countSubsets(A, N)); } }    // This code is contributed by shivanisinghss2110

## Python3

 # Python3 implementation to find the number  # of subsets with equal bitwise AND,  # OR and XOR values     mod = 1000000007;     # Function to find the number of  # subsets with equal bitwise AND,  # OR and XOR values  def countSubsets(a, n) :        answer = 0;         # Traverse through all the subsets      for i in range(1 << n) :            bitwiseAND = -1;          bitwiseOR = 0;          bitwiseXOR = 0;             # Finding the subsets with the bits          # of 'i' which are set          for j in range(n) :                # Computing the bitwise AND              if (i & (1 << j)) :                 if (bitwiseAND == -1) :                     bitwiseAND = a[j];                  else :                     bitwiseAND &= a[j];                     # Computing the bitwise OR                  bitwiseOR |= a[j];                     # Computing the bitwise XOR                  bitwiseXOR ^= a[j];             # Comparing all the three values          if (bitwiseAND == bitwiseOR and bitwiseOR == bitwiseXOR) :             answer = (answer + 1) % mod;             return answer;     # Driver code  if __name__ == "__main__" :             N = 6;      A = [ 1, 3, 2, 1, 2, 1 ];         print(countSubsets(A, N));     # This code is contributed by AnkitRai01

## C#

 // C# implementation to find the number // of subsets with equal bitwise AND, // OR and XOR values  using System;    class GFG { static int mod = 1000000007;     // Function to find the number of // subsets with equal bitwise AND, // OR and XOR values static int countSubsets(int []a, int n) {     int answer = 0;         // Traverse through all the subsets     for (int i = 0; i < (1 << n); i++) {             int bitwiseAND = -1;         int bitwiseOR = 0;         int bitwiseXOR = 0;             // Finding the subsets with the bits         // of 'i' which are set         for (int j = 0; j < n; j++) {                 // Computing the bitwise AND             if ((i & (1 << j)) == 0) {                 if (bitwiseAND == -1)                     bitwiseAND = a[j];                 else                     bitwiseAND &= a[j];                     // Computing the bitwise OR                 bitwiseOR |= a[j];                     // Computing the bitwise XOR                 bitwiseXOR ^= a[j];             }         }             // Comparing all the three values         if (bitwiseAND == bitwiseOR             && bitwiseOR == bitwiseXOR)             answer = (answer + 1) % mod;     }     return answer; }     // Driver Code public static void Main(String[] args) {     int N = 6;     int []A = { 1, 3, 2, 1, 2, 1 };         Console.Write(countSubsets(A, N)); } }    // This code is contributed by 29AjayKumar

Output:

7

Time Complexity: O(N * 2N) where N is the size of the array.

Efficient Approach: The efficient approach lies behind the property of bitwise operations.

• Using the property of bitwise AND, and bitwise OR we can say that if a & b == a | b, then a is equal to b. So if the AND and OR values of the subset are equal then all the elements of the subset are identical (say x). So the AND and OR values are equal to x.
• Since all the values of subsequence are equal to each other, two case arise for XOR value:
1. Subset size is odd: The XOR value equals to x.
2. Subset size is even: The XOR values equals to 0.
• Therefore, from the above observation, we can come to the conclusion that all odd-sized subsequences/subsets with equal elements follow the property.
• In addition to this if all the elements of the subset are 0, then then the subset will follow the property (irrespective of subset size). So all the subsets which have only 0 as their element will be added to the answer.
• If frequency of some element is K, the then number of odd sized subsets it can form is 2K – 1, and the total non-empty subsets it can form is 2K – 1.

Below is the implementation of the above approach:

## C++

 // C++ program to find the number // of subsets with equal bitwise // AND, OR and XOR values    #include using namespace std; const int mod = 1000000007;    // Function to find the number of // subsets with equal bitwise AND, // OR and XOR values int countSubsets(int a[], int n) {     int answer = 0;        // Precompute the modded powers     // of two for subset counting     int powerOfTwo[100005];     powerOfTwo[0] = 1;        // Loop to iterate and find the modded     // powers of two for subset counting     for (int i = 1; i < 100005; i++)         powerOfTwo[i]             = (powerOfTwo[i - 1] * 2)               % mod;        // Map to store the frequency of     // each element     unordered_map frequency;        // Loop to compute the frequency     for (int i = 0; i < n; i++)         frequency[a[i]]++;        // For every element > 0, the number of     // subsets formed using this element only     // is equal to 2 ^ (frequency[element]-1).     // And for 0, we have to find all     // the subsets, so 2^(frequency[element]) -1     for (auto el : frequency) {            // If element is greater than 0         if (el.first != 0)             answer                 = (answer % mod                    + powerOfTwo[el.second - 1])                   % mod;            else             answer                 = (answer % mod                    + powerOfTwo[el.second]                    - 1 + mod)                   % mod;     }     return answer; }    // Driver code int main() {     int N = 6;     int A[N] = { 1, 3, 2, 1, 2, 1 };        cout << countSubsets(A, N);        return 0; }

## Java

 // Java program to find the number // of subsets with equal bitwise // AND, OR and XOR values        import java.util.*;    class GFG{ static int mod = 1000000007;     // Function to find the number of // subsets with equal bitwise AND, // OR and XOR values static int countSubsets(int a[], int n) {     int answer = 0;         // Precompute the modded powers     // of two for subset counting     int []powerOfTwo = new int[100005];     powerOfTwo[0] = 1;         // Loop to iterate and find the modded     // powers of two for subset counting     for (int i = 1; i < 100005; i++)         powerOfTwo[i]             = (powerOfTwo[i - 1] * 2)               % mod;         // Map to store the frequency of     // each element     HashMap frequency = new HashMap();         // Loop to compute the frequency     for (int i = 0; i < n; i++)         if(frequency.containsKey(a[i])){             frequency.put(a[i], frequency.get(a[i])+1);         }else{             frequency.put(a[i], 1);     }         // For every element > 0, the number of     // subsets formed using this element only     // is equal to 2 ^ (frequency[element]-1).     // And for 0, we have to find all     // the subsets, so 2^(frequency[element]) -1     for (Map.Entry el : frequency.entrySet()) {             // If element is greater than 0         if (el.getKey() != 0)             answer                 = (answer % mod                    + powerOfTwo[el.getValue() - 1])                   % mod;             else             answer                 = (answer % mod                    + powerOfTwo[el.getValue()]                    - 1 + mod)                   % mod;     }     return answer; }     // Driver code public static void main(String[] args) {     int N = 6;     int A[] = { 1, 3, 2, 1, 2, 1 };         System.out.print(countSubsets(A, N));     } }    // This code is contributed by 29AjayKumar

Output:

7

Time Complexity: O(N), where N is the size of the array.

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