human
k | Function to find kth missing number.
If the total missing number count is equal to k we can iterate backwards for the first missing number and that will be the answer ..
To further optimize we check if the previous element ' s ▁ ▁ missing ▁ number ▁ count ▁ is ▁ equal ▁ ▁ to ▁ k . ▁ Eg : ▁ arr ▁ = ▁ [ 4,5,6,7,8 ] ▁ ▁ If ▁ you ▁ observe ▁ in ▁ the ▁ example ▁ array , ▁ ▁ the ▁ total ▁ count ▁ of ▁ missing ▁ numbers ▁ for ▁ all ▁ ▁ the ▁ indices ▁ are ▁ same , ▁ and ▁ we ▁ are ▁ ▁ aiming ▁ to ▁ narrow ▁ down ▁ the ▁ ▁ search ▁ window ▁ and ▁ achieve ▁ O ( logn ) ▁ ▁ time ▁ complexity ▁ which ▁ ▁ otherwise ▁ would ' ve been O ( n ) ..
Else we return arr [ mid ] - 1..
Here we appropriately narrow down the search window ..
In case the upper limit is - ve it means the missing number set is 1 , 2 , . . , k and hence we directly return k ..
Else we find the residual count of numbers which we 'd then add to  arr[u] and get the missing kth number..
Return arr [ u ] + k.
Driver Code.
Function Call"
How can the above be solved in Python?

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Give me a code to k and the other steps required?
assistant
#Here is code to k and the other steps required.
 def missingK ( arr , k ) :
	n = len ( arr )
	l = 0
	u = n - 1
	mid = 0
	while ( l <= u ) :
		mid = ( l + u ) // 2 ;
		numbers_less_than_mid = arr [ mid ] - ( mid + 1 ) ;
		if ( numbers_less_than_mid == k ) :
		if ( mid > 0 and ( arr [ mid - 1 ] - ( mid ) ) == k ) :
			u = mid - 1 ;
			continue ;
		return arr [ mid ] - 1 ;
		if ( numbers_less_than_mid < k ) :
		l = mid + 1 ;
		elif ( k < numbers_less_than_mid ) :
		u = mid - 1 ;
	if ( u < 0 ) :
		return k ;
	less = arr [ u ] - ( u + 1 ) ;
	k -= less ;
	return arr [ u ] + k ;
if __name__ == ' _ _ main _ _ ' :
	arr = [ 2 , 3 , 4 , 7 , 11 ] ;
	k = 5 ;
	print ( " Missing ▁ kth ▁ number ▁ = ▁ " + str ( missingK ( arr , k ) ) )