human
Check if an element is present in an array using at most floor ( N / 2 ) + 2 comparisons | Function to check whether X is present in the array A [ ].
Initialise a pointer.
Store the number of comparisons.
Variable to store product.
Check is N is odd.
Update i and T.
Traverse the array.
Check if i < N.
Update T.
Check if T is equal to 0.
Driver Code.
Given Input.
Function Call"
How can the above be solved in Python?

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Give me a code to check if an element is present in an array using at most floor ( N / 2 ) + 2 comparisons and the other steps required?
assistant
#Here is code to check if an element is present in an array using at most floor ( N / 2 ) + 2 comparisons and the other steps required.
 def findElement ( A , N , X ) :
	i = 0
	Comparisons = 0
	T = 1
	Found = " No "
	Comparisons += 1
	if ( N % 2 == 1 ) :
		i = 1
		T *= ( A [ 0 ] - X )
	while ( i < N ) :
		Comparisons += 1
		T *= ( A [ i ] - X )
		T *= ( A [ i + 1 ] - X )
		i += 2
	Comparisons += 1
	if ( T == 0 ) :
		print ( " Yes " , Comparisons )
	else :
		print ( " No " )
if __name__ == ' _ _ main _ _ ' :
	A = [ - 3 , 5 , 11 , 3 , 100 , 2 , 88 , 22 , 7 , 900 , 23 , 4 , 1 ]
	N = len ( A )
	X = 1
	findElement ( A , N , X )