human
Ceiling in a sorted array | Function to get index of ceiling of x in arr [ low . . high ].
If x is smaller than or equal to first element , then return the first element.
Otherwise , linearly search for ceil value.
if x lies between arr [ i ] and arr [ i + 1 ] including arr [ i + 1 ] , then return arr [ i + 1 ].
If we reach here then x is greater than the last element of the array , return - 1 in this case.
Driver program to check above functions"
Solution in Python:
 ... Write me a code to ceiling in a sorted array and the other steps required?
assistant
#Here is code to ceiling in a sorted array and the other steps required.
 def ceilSearch ( arr , low , high , x ) :
	if x <= arr [ low ] :
		return low
	i = low
	for i in range ( high ) :
		if arr [ i ] == x :
			return i
		if arr [ i ] < x and arr [ i + 1 ] >= x :
			return i + 1
	return - 1
arr = [ 1 , 2 , 8 , 10 , 10 , 12 , 19 ]
n = len ( arr )
x = 3
index = ceilSearch ( arr , 0 , n - 1 , x ) ;
if index == - 1 :
	print ( " Ceiling ▁ of ▁ % d ▁ doesn ' t ▁ exist ▁ in ▁ array ▁ " % x )
else :
	print ( " ceiling ▁ of ▁ % d ▁ is ▁ % d " % ( x , arr [ index ] ) )