human
Interpolation Search | If x is present in arr [ 0. . n - 1 ] , then returns index of it , else returns - 1..
Since array is sorted , an element present in array must be in range defined by corner.
Probing the position with keeping uniform distribution in mind ..
Condition of target found.
If x is larger , x is in right subarray.
If x is smaller , x is in left subarray.
Array of items in which search will be conducted.
Element to be searched.
If element was found"
Solution in Python:

What is a code to interpolation Search and the other steps required?
assistant
#Here is code to interpolation Search and the other steps required.
 def interpolationSearch ( arr , lo , hi , x ) :
	if ( lo <= hi and x >= arr [ lo ] and x <= arr [ hi ] ) :
		pos = lo + ( ( hi - lo ) // ( arr [ hi ] - arr [ lo ] ) * ( x - arr [ lo ] ) )
		if arr [ pos ] == x :
			return pos
		if arr [ pos ] < x :
			return interpolationSearch ( arr , pos + 1 , hi , x )
		if arr [ pos ] > x :
			return interpolationSearch ( arr , lo , pos - 1 , x )
	return - 1
arr = [ 10 , 12 , 13 , 16 , 18 , 19 , 20 , 21 , 22 , 23 , 24 , 33 , 35 , 42 , 47 ]
n = len ( arr )
x = 18
index = interpolationSearch ( arr , 0 , n - 1 , x )
if index != - 1 :
	print ( " Element ▁ found ▁ at ▁ index " , index )
else :
	print ( " Element ▁ not ▁ found " )