human
Count ways to make Bitwise XOR of odd and even indexed elements equal by removing an array element | Function to count ways to make Bitwise XOR of odd and even indexed elements equal by removing an array element.
Stores xor of odd and even indexed elements from the end.
Stores xor of odd and even indexed elements from the start.
Stores the required count.
Traverse the array in reverse.
If i is odd.
If i is even.
Traverse the array.
If i is odd.
If i is even.
Removing arr [ i ] , post_even stores XOR of odd indexed elements.
Removing arr [ i ] , post_odd stores XOR of even indexed elements.
Check if they are equal.
If i is odd , xor it with curr_odd.
If i is even , xor it with curr_even.
Finally print res.
Drivers Code.
Given array.
Given size.
Function call"
How can the above be solved in Python?

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Give me a code to count ways to make Bitwise XOR of odd and even indexed elements equal by removing an array element and the other steps required?
assistant
#Here is code to count ways to make Bitwise XOR of odd and even indexed elements equal by removing an array element and the other steps required.
 def Remove_one_element ( arr , n ) :
	post_odd = 0
	post_even = 0
	curr_odd = 0
	curr_even = 0
	res = 0
	for i in range ( n - 1 , - 1 , - 1 ) :
		if ( i % 2 ) :
			post_odd ^= arr [ i ]
		else :
			post_even ^= arr [ i ]
	for i in range ( n ) :
		if ( i % 2 ) :
			post_odd ^= arr [ i ]
		else :
			post_even ^= arr [ i ]
		X = curr_odd ^ post_even
		Y = curr_even ^ post_odd
		if ( X == Y ) :
			res += 1
		if ( i % 2 ) :
			curr_odd ^= arr [ i ]
		else :
			curr_even ^= arr [ i ]
	print ( res )
if __name__ == " _ _ main _ _ " :
	arr = [ 1 , 0 , 1 , 0 , 1 ]
	N = len ( arr )
	Remove_one_element ( arr , N )