Given an array **arr[]** of **N **positive integers, the task is to find the length of the longest subsequence such that Bitwise XOR of all integers in the subsequence is odd.

**Examples:**

Input:N = 7, arr[] ={2, 3, 4, 1, 5, 6, 7}Output:6Explanation:The subsequence of maximum length is {2, 3, 4, 5, 6, 7}with XOR of all elements as 1.Other subsequences also exists.

Input:N = 4, arr[] = {2, 4, 6, 8}Output:0Explanation:No possible subsequence exits.

**Naive Approach:** The naive idea is to generate all possible subsequence of the given array and check if Bitwise XOR of any subsequence is odd or not. If there exist subsequences whose Bitwise XOR is odd then print the maximum length of that among those subsequences.

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

**Auxiliary Space:**

*O(1)*

**Efficient Approach:** To optimize the above naive approach the idea is to count the number of odd and even element in the given array and find the length of the longest subsequence as below:

- Count the number of even and odd elements in
**arr[]**. - If the count of odd values is equal to array size i.e.,
**N**, then we have two possible cases:- If the size of the array is odd then the max length will be equal to
**N** - Else max length will be equal to
**N – 1**.

- If the size of the array is odd then the max length will be equal to
- If the count of even values is equal to array size then the max length will be zero.
- Now if both types of elements are present in the given array, then the max length will include all the even elements and for odd elements, we include all of them if the count of odd values is odd else we include
**odd – 1**elements. - Print the maximum length of the longest subsequence after the above steps.

Below is the implementation of the above approach:

## C++

`// C++ program for the above approach` `#include <iostream>` `using` `namespace` `std;` `// Function for find max XOR subsequence` `// having odd value` `int` `maxXORSubsequence(` `int` `arr[], ` `int` `n)` `{` ` ` `// Initialize odd and even count` ` ` `int` `odd = 0, even = 0;` ` ` `// Count the number of odd and even` ` ` `// numbers in given array` ` ` `for` `(` `int` `i = 0; i < n; i++) {` ` ` `if` `(arr[i] & 1)` ` ` `odd++;` ` ` `else` ` ` `even++;` ` ` `}` ` ` `int` `maxlen;` ` ` `if` `(odd == n) {` ` ` `// if all values are odd` ` ` `// in given array` ` ` `if` `(odd % 2 == 0)` ` ` `maxlen = n - 1;` ` ` `else` ` ` `maxlen = n;` ` ` `}` ` ` `else` `if` `(even == n) {` ` ` `// if all values are even` ` ` `// in given array` ` ` `maxlen = 0;` ` ` `}` ` ` `else` `{` ` ` `// if both odd and even are` ` ` `// present in given array` ` ` `if` `(odd % 2 == 0)` ` ` `maxlen = even + odd - 1;` ` ` `else` ` ` `maxlen = even + odd;` ` ` `}` `}` `// Driver Code` `int` `main()` `{` ` ` `// Given array arr[]` ` ` `int` `arr[] = { 2, 3, 4, 5, 6, 7 };` ` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]);` ` ` `// Function Call` ` ` `cout << maxXORSubsequence(arr, n);` `}` |

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## Java

`// Java program for the above approach` `class` `GFG{` ` ` `// Function for find max XOR subsequence` `// having odd value` `static` `int` `maxXORSubsequence(` `int` `arr[], ` `int` `n)` `{` ` ` ` ` `// Initialize odd and even count` ` ` `int` `i, odd = ` `0` `, even = ` `0` `;` ` ` `// Count the number of odd and even` ` ` `// numbers in given array` ` ` `for` `(i = ` `0` `; i < n; i++)` ` ` `{` ` ` `if` `((arr[i] & ` `1` `) != ` `0` `)` ` ` `odd++;` ` ` `else` ` ` `even++;` ` ` `}` ` ` `int` `maxlen;` ` ` `if` `(odd == n)` ` ` `{` ` ` ` ` `// If all values are odd` ` ` `// in given array` ` ` `if` `(odd % ` `2` `== ` `0` `)` ` ` `maxlen = n - ` `1` `;` ` ` `else` ` ` `maxlen = n;` ` ` `}` ` ` `else` `if` `(even == n)` ` ` `{` ` ` `// If all values are even` ` ` `// in given array` ` ` `maxlen = ` `0` `;` ` ` `}` ` ` `else` ` ` `{` ` ` `// If both odd and even are` ` ` `// present in given array` ` ` `if` `(odd % ` `2` `== ` `0` `)` ` ` `maxlen = even + odd - ` `1` `;` ` ` `else` ` ` `maxlen = even + odd;` ` ` `}` ` ` `return` `maxlen;` `}` `// Driver Code` `public` `static` `void` `main (String []args)` `{` ` ` ` ` `// Given array arr[]` ` ` `int` `arr[] = { ` `2` `, ` `3` `, ` `4` `, ` `5` `, ` `6` `, ` `7` `};` ` ` `int` `n = arr.length;` ` ` `// Function Call` ` ` `System.out.print( maxXORSubsequence(arr, n));` `}` `}` `// This code is contributed by chitranayal` |

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## Python3

`# Python3 program for the above approach ` `# Function for find max XOR subsequence ` `# having odd value` `def` `maxXorSubsequence(arr, n):` ` ` ` ` `# Initialize odd and even count ` ` ` `odd ` `=` `0` ` ` `even ` `=` `0` ` ` ` ` `# Count the number of odd and even ` ` ` `# numbers in given array ` ` ` `for` `i ` `in` `range` `(` `0` `, n):` ` ` `if` `arr[i] ` `%` `2` `!` `=` `0` `:` ` ` `odd ` `+` `=` `1` ` ` `else` `:` ` ` `even ` `+` `=` `1` ` ` `if` `odd ` `=` `=` `n:` ` ` ` ` `# If all values are odd ` ` ` `# in given array ` ` ` `if` `odd ` `%` `2` `=` `=` `0` `:` ` ` `maxlen ` `=` `n ` `-` `1` ` ` `else` `:` ` ` `maxlen ` `=` `n` ` ` ` ` `elif` `even ` `=` `=` `n:` ` ` ` ` `# If all values are even ` ` ` `# in given array ` ` ` `maxlen ` `=` `0` ` ` `else` `:` ` ` ` ` `# If both odd and even are ` ` ` `# present in given array ` ` ` `if` `odd ` `%` `2` `=` `=` `0` `:` ` ` `maxlen ` `=` `even ` `+` `odd ` `-` `1` ` ` `else` `:` ` ` `maxlen ` `=` `even ` `+` `odd` ` ` ` ` `return` `maxlen` `# Driver code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` ` ` `# Given array arr[] ` ` ` `arr ` `=` `[ ` `2` `, ` `3` `, ` `4` `, ` `5` `, ` `6` `, ` `7` `]` ` ` `n ` `=` `len` `(arr)` ` ` ` ` `# Function Call ` ` ` `print` `(maxXorSubsequence(arr,n))` ` ` `# This code is contributed by virusbuddah_` |

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## C#

`// C# program for the above approach` `using` `System;` `class` `GFG{` ` ` `// Function for find max XOR subsequence` `// having odd value` `static` `int` `maxXORSubsequence(` `int` `[] arr, ` `int` `n)` `{` ` ` ` ` `// Initialize odd and even count` ` ` `int` `i, odd = 0, even = 0;` ` ` ` ` `// Count the number of odd and even` ` ` `// numbers in given array` ` ` `for` `(i = 0; i < n; i++)` ` ` `{` ` ` `if` `((arr[i] & 1) != 0)` ` ` `odd++;` ` ` `else` ` ` `even++;` ` ` `}` ` ` ` ` `int` `maxlen;` ` ` ` ` `if` `(odd == n)` ` ` `{` ` ` ` ` `// If all values are odd` ` ` `// in given array` ` ` `if` `(odd % 2 == 0)` ` ` `maxlen = n - 1;` ` ` `else` ` ` `maxlen = n;` ` ` `}` ` ` `else` `if` `(even == n)` ` ` `{` ` ` ` ` `// If all values are even` ` ` `// in given array` ` ` `maxlen = 0;` ` ` `}` ` ` `else` ` ` `{` ` ` ` ` `// If both odd and even are` ` ` `// present in given array` ` ` `if` `(odd % 2 == 0)` ` ` `maxlen = even + odd - 1;` ` ` `else` ` ` `maxlen = even + odd;` ` ` `}` ` ` `return` `maxlen;` `}` ` ` `// Driver Code` `public` `static` `void` `Main (` `string` `[]args)` `{` ` ` ` ` `// Given array arr[]` ` ` `int` `[]arr = { 2, 3, 4, 5, 6, 7 };` ` ` `int` `n = arr.Length;` ` ` ` ` `// Function Call` ` ` `Console.Write( maxXORSubsequence(arr, n));` `}` `}` ` ` `// This code is contributed by rock_cool` |

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**Output:**

6

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

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