Count subsequences having odd Bitwise OR values in an array

Given an array arr[] consisting of N positive integers, the task is to find the number of subsequences from the given array whose Bitwise OR value is odd.

Examples:

Input: arr = [2, 4, 1]
Output: 4
Explanation: Subsequences with odd Bitwise OR values are {1}, {2, 1}, {4, 1}, {2, 4, 1}

Input: arr = [1, 3, 4]
Output: 6

Naive Approach: The simplest approach to solve the problem is to generate all the subsequences of the given array and for each subsequence, check if its Bitwise OR value is odd or not. If it is odd, then increase the count by one. After checking for all subsequences, print the count obtained.
Time Complexity:
Auxiliary Space:

Efficient Approach: Given problem can be solved by observing that for a subsequence to have an odd Bitwise OR value at least one element of the subsequence should be odd. Therefore at least one element in the subsequence should have the least significant digit equal to 1. Follow the steps below to solve this problem:

• Store the count of even and odd elements present in the array arr[] in even and odd variables respectively.
• Traverse the array A[] using the variable i
  • If the value of A[i] is odd, increase the value of odd by 1.
  • Otherwise, increase the value of even by 1.
• Total combinations with at least one odd element and any number of even elements can be given by: [2^(odd elements) – 1] * 2^(even elements). Since at least one odd element is needed so empty set of combinations of 1 is excluded

Below is the implementation of the approach:

C++

// C++ implementation for the above approach
#include <bits/stdc++.h>

using namespace std;

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

    // Stores count of even elements
    int even = 0;

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

        // If element is odd
        if (x & 1)
            odd++;
        else
            even++;
    }

    // Return the final answer
    return ((1 << odd) - 1) *
              (1 << even);
}

// Driver Code
int main()
{
    // Given array arr[]
    vector<int> arr = {2, 4, 1};

    cout << countSubsequences(arr);
}

Java

// Java implementation for the above approach
import java.io.*;

class GFG {

      // Function to count the subsequences
    // having odd bitwise OR value
    static int countSubsequences(int arr[])
    {

        // Stores count of odd elements
        int odd = 0;

        // Stores count of even elements
        int even = 0;

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

            // If element is odd
            if ((arr[i] & 1) != 0)
                odd++;
            else
                even++;
        }

        // Return the final answer
        return ((1 << odd) - 1) *
                  (1 << even);
    }

    // Driver Code
    public static void main (String[] args) {
        // Given array arr[]
        int arr[] = {2, 4, 1};

        System.out.println(countSubsequences(arr));
    }
}

// This code is contributed by Dharanendra L V.

Python3

# Python3 implementation for the above approach

# Function to count the subsequences
# having odd bitwise OR value
def countSubsequences(arr) :

    # Stores count of odd elements
    odd = 0;

    # Stores count of even elements
    even = 0;

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

        # If element is odd
        if (x & 1) :
            odd += 1;
        else :
            even += 1;

    # Return the final answer
    return ((1 << odd) - 1) * (1 << even);


# Driver Code
if __name__ == "__main__" :

    # Given array arr[]
    arr = [2, 4, 1];

    print(countSubsequences(arr));

    # This code is contributed by AnkThon

C#

// Java implementation for the above approach
using System;

class GFG {

      // Function to count the subsequences
    // having odd bitwise OR value
    static int countSubsequences(int []arr)
    {

        // Stores count of odd elements
        int odd = 0;

        // Stores count of even elements
        int even = 0;

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

            // If element is odd
            if ((arr[i] & 1) != 0)
                odd++;
            else
                even++;
        }

        // Return the final answer
        return ((1 << odd) - 1) *
                  (1 << even);
    }

    // Driver Code
    public static void Main (String[] args) {
        // Given array arr[]
        int []arr = {2, 4, 1};

        Console.Write(countSubsequences(arr));
    }
}

// This code is contributed by shivanisinghss2110

Javascript

    <script>
        // JavaScript Program to implement
        // the above approach

// Function to count the subsequences
// having odd bitwise OR value
function countSubsequences( arr)
{

    // Stores count of odd elements
    let odd = 0;

    // Stores count of even elements
    let even = 0;

    // Traverse the array arr[]
    for (let x of arr) {

        // If element is odd
        if (x & 1)
            odd++;
        else
            even++;
    }

    // Return the final answer
    return ((1 << odd) - 1) *
              (1 << even);
}

// Driver Code

    // Given array arr[]
    let arr = [2, 4, 1];

    document.write(countSubsequences(arr));

     // This code is contributed by Potta Lokesh

    </script>

Output

4

Time Complexity: 
Auxiliary Space: