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:
- 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 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++ 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 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 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);
} |
// Java program for the above approach import java.io.*;
public 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 |
# 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_ |
// 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 |
<script> // Javascript program for the above approach // Function for find max XOR subsequence // having odd value function maxXORSubsequence(arr, n)
{ // Initialize odd and even count
let odd = 0, even = 0;
// Count the number of odd and even
// numbers in given array
for (let i = 0; i < n; i++)
{
if ((arr[i] & 1) != 0)
odd++;
else
even++;
}
let 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 // Given array arr[] let arr = [ 2, 3, 4, 5, 6, 7 ]; let n = arr.length; // Function Call document.write(maxXORSubsequence(arr, n)); // This code is contributed by divyesh072019 </script> |
6
Time Complexity: O(N)
Auxiliary Space: O(1)