human
Give me a code to maximum length of same indexed subarrays from two given arrays satisfying the given condition and the other steps required given the following:
Maximum length of same indexed subarrays from two given arrays satisfying the given condition | Function to find maximum length of subarray such that sum of maximum element in subarray in brr [ ] and sum of subarray in [ ] arr * K is at most C.
Base Case.
Let maximum length be 0.
Perform Binary search.
Find mid value.
Check if the current mid satisfy the given condition.
If yes , then store length.
Otherwise.
Return maximum length stored.
Function to check if it is possible to have such a subarray of length K.
Finds the maximum element in each window of size k.
Check for window of size K.
For all possible subarrays of length k.
Until deque is empty.
Calculate the total cost and check if less than equal to c.
Find sum of current subarray and the total cost.
Include the new element of current subarray.
Discard the element of last subarray.
Remove all the elements in the old window.
Calculate total cost and check <= c.
If current subarray length satisfies.
If it is not possible.
Driver Code"
How can the above be solved in Python?
assistant
#Here is code to maximum length of same indexed subarrays from two given arrays satisfying the given condition and the other steps required.
 def maxLength ( a , b , n , c ) :
	if ( n == 0 ) :
		return 0
	max_length = 0
	low = 0
	high = n
	while ( low <= high ) :
		mid = int ( low + ( high - low ) / 2 )
		if ( possible ( a , b , n , c , mid ) ) :
			max_length = mid
			low = mid + 1
		else :
			high = mid - 1
	return max_length
def possible ( a , b , n , c , k ) :
	dq = [ ]
	Sum = 0
	for i in range ( k ) :
		Sum += a [ i ]
		while ( len ( dq ) > 0 and b [ i ] > b [ dq [ len ( dq ) - 1 ] ] ) :
			dq . pop ( len ( dq ) - 1 )
		dq . append ( i )
	total_cost = Sum * k + b [ dq [ 0 ] ]
	if ( total_cost <= c ) :
		return True
	for i in range ( k , n ) :
		Sum += a [ i ]
		Sum -= a [ i - k ]
		while ( len ( dq ) > 0 and dq [ 0 ] <= i - k ) :
			dq . pop ( 0 )
		while ( len ( dq ) > 0 and b [ i ] > b [ dq [ len ( dq ) - 1 ] ] ) :
			dq . pop ( len ( dq ) - 1 )
		dq . append ( i )
		total_cost = Sum * k + b [ dq [ 0 ] ]
		if ( total_cost <= c ) :
			return True
	return False
A = [ 1 , 2 , 1 , 6 , 5 , 5 , 6 , 1 ]
B = [ 14 , 8 , 15 , 15 , 9 , 10 , 7 , 12 ]
N = len ( A )
C = 40
print ( maxLength ( A , B , N , C ) )