Given an array **A[]** of size **N**, the task is to count the number of subsequences from the given array whose Bitwise XOR value is odd.

**Examples:**

Input:A[] = {1, 3, 4}Output:4Explanation:Subsequences with odd Bitwise XOR are {1}, {3}, {1, 4}, {3, 4}.

Input:A[] = {2, 8, 6}Output:0Explanation:No such subsequences are present in the array.

**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 XOR** value is odd or not. If found to be odd, then increase the count by one. After checking for all subsequences, print the count obtained.

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

**Efficient Approach:** To optimize the above approach, the idea is based on the observation that a subsequence will have odd Bitwise XOR only if the number of odd elements present in it is odd. This is because odd elements have the least significant bit equal to **1**. Therefore, XOR of the least significant bits (**1^1^1….**) up to an odd number of times sets the least significant bit of the new number formed. Therefore, the new number formed is odd.

Follow the steps below to solve the problem:

- Store the count of even and odd elements present in the array
**A[]**in**even**and**odd**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**.

- If the value of
- If the value of
**odd**is equal to**0**, then print**0**and return. - Otherwise,
- Find the total combinations with an odd number of odd elements.
- This can be calculated by
**(**^{X}C_{1}+^{X}C_{3}+…+^{X}C_{(x-(x+1)%2)}) * (^{M}C_{0}+^{M}C_{1}+…+^{M}C_{M}) = 2^{X-1}* 2^{M }= 2^{X+M-1}=**2**^{N-1}. - Therefore, Print
**2**^{N-1}

Below is the implementation of the above approach:

## C++

`// C++ program for the above approach` `#include<bits/stdc++.h>` `using` `namespace` `std;` `// Function to count the subsequences` `// having odd bitwise XOR value` `void` `countSubsequences(vector<` `int` `> A)` `{` ` ` ` ` `// Stores count of odd elements` ` ` `int` `odd = 0;` ` ` `// Stores count of even elements` ` ` `int` `even = 0;` ` ` `// Traverse the array A[]` ` ` `for` `(` `int` `el : A)` ` ` `{` ` ` ` ` `// If el is odd` ` ` `if` `(el % 2 == 1)` ` ` `odd++;` ` ` `else` ` ` `even++;` ` ` `}` ` ` `// If count of odd elements is 0` ` ` `if` `(odd == 0)` ` ` `cout << (0);` ` ` `else` ` ` `cout << (1 << (A.size() - 1));` `}` `// Driver Code` `int` `main()` `{` ` ` ` ` `// Given array A[]` ` ` `vector<` `int` `> A = { 1, 3, 4 };` ` ` ` ` `// Function call to count subsequences` ` ` `// having odd bitwise XOR value` ` ` `countSubsequences(A);` `}` `// This code is contributed by mohit kumar 29` |

## Java

`// Java program for` `// the above approach` `import` `java.io.*;` `class` `GFG {` ` ` `// Function to count the subsequences` ` ` `// having odd bitwise XOR value` ` ` `public` `static` `void` `countSubsequences(` `int` `[] A)` ` ` `{` ` ` `// Stores count of odd elements` ` ` `int` `odd = ` `0` `;` ` ` `// Stores count of even elements` ` ` `int` `even = ` `0` `;` ` ` `// Traverse the array A[]` ` ` `for` `(` `int` `el : A) {` ` ` `// If el is odd` ` ` `if` `(el % ` `2` `== ` `1` `)` ` ` `odd++;` ` ` `else` ` ` `even++;` ` ` `}` ` ` `// If count of odd elements is 0` ` ` `if` `(odd == ` `0` `)` ` ` `System.out.println(` `0` `);` ` ` `else` ` ` `System.out.println(` `1` `<< (A.length - ` `1` `));` ` ` `}` ` ` `// Driver Code` ` ` `public` `static` `void` `main(String[] args)` ` ` `{` ` ` `// Given array A[]` ` ` `int` `[] A = { ` `1` `, ` `3` `, ` `4` `};` ` ` `// Function call to count subsequences` ` ` `// having odd bitwise XOR value` ` ` `countSubsequences(A);` ` ` `}` `}` |

## Python3

`# Python3 program for the above approach` `# Function to count the subsequences` `# having odd bitwise XOR value` `def` `countSubsequences(A):` ` ` ` ` `# Stores count of odd elements` ` ` `odd ` `=` `0` ` ` `# Stores count of even elements` ` ` `even ` `=` `0` ` ` `# Traverse the array A[]` ` ` `for` `el ` `in` `A:` ` ` `# If el is odd` ` ` `if` `(el ` `%` `2` `=` `=` `1` `):` ` ` `odd ` `+` `=` `1` ` ` `else` `:` ` ` `even ` `+` `=` `1` ` ` `# If count of odd elements is 0` ` ` `if` `(odd ` `=` `=` `0` `):` ` ` `print` `(` `0` `)` ` ` `else` `:` ` ` `print` `(` `1` `<< ` `len` `(A) ` `-` `1` `)` `# Driver Code` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` `# Given array A[]` ` ` `A ` `=` `[` `1` `, ` `3` `, ` `4` `]` ` ` `# Function call to count subsequences` ` ` `# having odd bitwise XOR value` ` ` `countSubsequences(A)` `# This code is contributed by ukasp` |

## C#

`// C# program for above approach` `using` `System;` `public` `class` `GFG` `{` ` ` ` ` `// Function to count the subsequences` ` ` `// having odd bitwise XOR value` ` ` `public` `static` `void` `countSubsequences(` `int` `[] A)` ` ` `{` ` ` `// Stores count of odd elements` ` ` `int` `odd = 0;` ` ` `// Stores count of even elements` ` ` `int` `even = 0;` ` ` `// Traverse the array A[]` ` ` `foreach` `(` `int` `el ` `in` `A) {` ` ` `// If el is odd` ` ` `if` `(el % 2 == 1)` ` ` `odd++;` ` ` `else` ` ` `even++;` ` ` `}` ` ` `// If count of odd elements is 0` ` ` `if` `(odd == 0)` ` ` `Console.WriteLine(0);` ` ` `else` ` ` `Console.WriteLine(1 << (A.Length - 1));` ` ` `}` ` ` `// Driver code` ` ` `public` `static` `void` `Main(String[] args)` ` ` `{` ` ` `// Given array A[]` ` ` `int` `[] A = { 1, 3, 4 };` ` ` `// Function call to count subsequences` ` ` `// having odd bitwise XOR value` ` ` `countSubsequences(A);` ` ` `}` `}` `// This code is contributed by splevel62.` |

## Javascript

`<script>` `// Javascript program for the above approach` `// Function to count the subsequences` `// having odd bitwise XOR value` `function` `countSubsequences( A)` `{` ` ` ` ` `// Stores count of odd elements` ` ` `var` `odd = 0;` ` ` `// Stores count of even elements` ` ` `var` `even = 0;` ` ` `// Traverse the array A[]` ` ` `for` `(` `var` `e1 = 0; e1<A.length; e1++)` ` ` `{` ` ` `// If el is odd` ` ` `if` `(A[e1] % 2 == 1)` ` ` `odd++;` ` ` `else` ` ` `even++;` ` ` `}` ` ` `// If count of odd elements is 0` ` ` `if` `(odd == 0)` ` ` `document.write(0);` ` ` `else` ` ` `document.write((1 << (A.length - 1)));` `}` `// Driver Code` `// Given array A[]` `var` `A = [ 1, 3, 4 ];` `// Function call to count subsequences` `// having odd bitwise XOR value` `countSubsequences(A);` `</script>` |

**Output:**

4

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

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