human
Write me a code to k and the other steps required given the following:
K | Python3 program to find kth element from two sorted arrays.
In case we have reached end of array 1.
In case we have reached end of array 2.
k should never reach 0 or exceed sizes of arrays.
Compare first elements of arrays and return.
Size of array 1 is less than k / 2.
Last element of array 1 is not kth We can directly return the ( k - m ) th element in array 2.
Size of array 2 is less than k / 2.
Normal comparison , move starting index of one array k / 2 to the right.
Driver code"
How can the above be solved in Python?
assistant
#Here is code to k and the other steps required.
 def kth ( arr1 , arr2 , m , n , k , st1 = 0 , st2 = 0 ) :
	if ( st1 == m ) :
		return arr2 [ st2 + k - 1 ]
	if ( st2 == n ) :
		return arr1 [ st1 + k - 1 ]
	if ( k == 0 or k > ( m - st1 ) + ( n - st2 ) ) :
		return - 1
	if ( k == 1 ) :
		if ( arr1 [ st1 ] < arr2 [ st2 ] ) :
			return arr1 [ st1 ]
		else :
			return arr2 [ st2 ]
	curr = int ( k / 2 )
	if ( curr - 1 >= m - st1 ) :
		if ( arr1 [ m - 1 ] < arr2 [ st2 + curr - 1 ] ) :
			return arr2 [ st2 + ( k - ( m - st1 ) - 1 ) ]
		else :
			return kth ( arr1 , arr2 , m , n , k - curr , st1 , st2 + curr )
	if ( curr - 1 >= n - st2 ) :
		if ( arr2 [ n - 1 ] < arr1 [ st1 + curr - 1 ] ) :
			return arr1 [ st1 + ( k - ( n - st2 ) - 1 ) ]
		else :
			return kth ( arr1 , arr2 , m , n , k - curr , st1 + curr , st2 )
	else :
		if ( arr1 [ curr + st1 - 1 ] < arr2 [ curr + st2 - 1 ] ) :
			return kth ( arr1 , arr2 , m , n , k - curr , st1 + curr , st2 )
		else :
			return kth ( arr1 , arr2 , m , n , k - curr , st1 , st2 + curr )
arr1 = [ 2 , 3 , 6 , 7 , 9 ]
arr2 = [ 1 , 4 , 8 , 10 ]
k = 5
print ( kth ( arr1 , arr2 , 5 , 4 , k ) )