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Length of longest subsequence whose XOR value is odd
• Difficulty Level : Easy
• Last Updated : 30 Mar, 2021

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: 6
Explanation: 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: 0
Explanation: 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*2N)
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:

1. Count the number of even and odd elements in arr[].
2. If the count of odd values is equal to array size i.e., N, then we have two possible cases:
1. If the size of the array is odd then the max length will be equal to N
2. Else max length will be equal to N – 1.
3. If the count of even values is equal to array size then the max length will be zero.
4. 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.
5. 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 using namespace std; // Function for find max XOR subsequence// having odd valueint 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 Codeint 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);}

## Java

 // Java program for the above approachclass GFG{     // Function for find max XOR subsequence// having odd valuestatic 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 Codepublic 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

## Python3

 # Python3 program for the above approach # Function for find max XOR subsequence# having odd valuedef 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 codeif __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_

## C#

 // C# program for the above approachusing System;class GFG{      // Function for find max XOR subsequence// having odd valuestatic 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 Codepublic 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

## Javascript


Output:
6

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

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