human
Search an element in an array where difference between adjacent elements is 1 | x is the element to be searched in arr [ 0. . n - 1 ].
Traverse the given array starting from leftmost element.
If x is found at index i.
Jump the difference between current array element and x.
Driver program to test above function"
How can the above be solved in Python?
 ... Give me a code to search an element in an array where difference between adjacent elements is 1 and the other steps required?
assistant
#Here is code to search an element in an array where difference between adjacent elements is 1 and the other steps required.
 def search ( arr , n , x ) :
	i = 0
	while ( i < n ) :
		if ( arr [ i ] == x ) :
			return i
		i = i + abs ( arr [ i ] - x )
	print ( " number ▁ is ▁ not ▁ present ! " )
	return - 1
arr = [ 8 , 7 , 6 , 7 , 6 , 5 , 4 , 3 , 2 , 3 , 4 , 3 ]
n = len ( arr )
x = 3
print ( " Element " , x , " ▁ is ▁ present ▁ at ▁ index ▁ " , search ( arr , n , 3 ) )