Count of subsequences having odd Bitwise AND values in the given array

Given an array arr[] of N integers, the task is to find the number of subsequences of the given array such that their Bitwise AND value is Odd.

Examples:

Input: arr[] = {2, 3, 1}
Output: 3
Explanation: The subsequences of the given array having odd Bitwise AND values are {3} = 3, {1} = 1, {3, 1} = 3 & 1 = 1.

Input: arr[] = {1, 3, 3}
Output: 7

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

Naive Approach: The given problem can be solved by generating all the subsequences of the given array arr[] and keep track of the count of subsequences such that their Bitwise AND value is odd.

Time Complexity: O(N * 2^N)
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized by the observation that for the Bitwise AND value of a set of integers to be odd, the least significant bit of each and every element of the set must be a set bit. Therefore, for a subsequence to have an odd Bitwise AND value, all the elements of the subsequence must be odd. Using this observation, the given problem can be solved using the steps below:

• Create a variable odd, which stores the total number of odd integers in the given array arr[]. Initialize it with 0.
• Iterate through the array and increment the value of odd by 1 if the current integer is an odd integer.
• The count of non-empty subsequences of an array having X elements is 2^X - 1. Therefore, the number of non-empty subsequences with all odd elements is 2^odd - 1,  which is the required answer.

Below is the implementation of the approach:

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find count of subsequences
// having odd bitwise AND value
int countSubsequences(vector<int> arr)
{
    // Stores count of odd elements
    int odd = 0;

    // Traverse the array arr[]
    for (int x : arr) {

        // If x is odd increment count
        if (x & 1)
            odd++;
    }

    // Return Answer
    return (1 << odd) - 1;
}

// Driver Code
int main()
{
    vector<int> arr = { 1, 3, 3 };

    // Function Call
    cout << countSubsequences(arr);

    return 0;
}

// Java program for the above approach
import java.io.*;
class GFG {

    // Function to find count of subsequences
    // having odd bitwise AND value
    static int countSubsequences(int arr[], int N)
    {
        // Stores count of odd elements
        int odd = 0;

        // Traverse the array arr[]
        for (int i = 0; i < N; i++) {

            // If x is odd increment count
            if ((arr[i] & 1) % 2 == 1)
                odd++;
        }

        // Return Answer
        return (1 << odd) - 1;
    }

    // Driver Code
    public static void main(String[] args)
    {
        int N = 3;

        int arr[] = { 1, 3, 3 };
        // Function Call
        System.out.println(countSubsequences(arr, N));
    }
}

// This code is contributed by dwivediyash

# python program for the above approach

# Function to find count of subsequences
# having odd bitwise AND value
def countSubsequences(arr):

    # Stores count of odd elements
    odd = 0

    # Traverse the array arr[]
    for x in arr:

        # If x is odd increment count
        if (x & 1):
            odd = odd + 1

    # Return Answer
    return (1 << odd) - 1

# Driver Code
if __name__ == "__main__":

    arr = [1, 3, 3]

    # Function Call
    print(countSubsequences(arr))
    
# This code is contributed by rakeshsahni

// C# program for the above approach
using System;

public class GFG
{

    // Function to find count of subsequences
    // having odd bitwise AND value
    static int countSubsequences(int []arr, int N)
    {
      
        // Stores count of odd elements
        int odd = 0;

        // Traverse the array arr[]
        for (int i = 0; i < N; i++) {

            // If x is odd increment count
            if ((arr[i] & 1) % 2 == 1)
                odd++;
        }

        // Return Answer
        return (1 << odd) - 1;
    }

    // Driver Code
    public static void Main(string[] args)
    {
        int N = 3;

        int []arr = { 1, 3, 3 };
        
        // Function Call
        Console.WriteLine(countSubsequences(arr, N));
    }
}

// This code is contributed by AnkThon

<script>
// Javascript program for the above approach

// Function to find count of subsequences
// having odd bitwise AND value
function countSubsequences(arr)
{
    // Stores count of odd elements
    let odd = 0;

    // Traverse the array arr[]
    for (let x = 0; x < arr.length; x++) {

        // If x is odd increment count
        if (arr[x] & 1)
            odd++;
    }

    // Return Answer
    return (1 << odd) - 1;
}

// Driver Code
    let arr = [ 1, 3, 3 ];

    // Function Call
    document.write(countSubsequences(arr));

// This code is contributed by subham348.
</script>

7

Time Complexity: O(N)
Auxiliary Space: O(1)