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Minimum removals required to make a given array Bitonic | Function to coutnt minimum array elements required to be removed to make an array bitonic.
left [ i ] : Stores the length of LIS up to i - th index.
right [ i ] : Stores the length of decreasing subsequence over the range [ i , N ].
Calculate the length of LIS up to i - th index.
Traverse the array upto i - th index.
If arr [ j ] is less than arr [ i ].
Update left [ i ].
Calculate the length of decreasing subsequence over the range [ i , N ].
Traverse right [ ] array.
If arr [ i ] is greater than arr [ j ].
Update right [ i ].
Stores length of the longest bitonic array.
Traverse left [ ] and right [ ] array.
Update maxLen.
Function to prminimum removals required to make given array bitonic.
Driver Code"
How can the above be solved in Python?

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Can you provide a code to minimum removals required to make a given array Bitonic and the other steps required?
assistant
#Here is code to minimum removals required to make a given array Bitonic and the other steps required.
 def min_element_removal ( arr , N ) :
	left = [ 1 ] * N
	right = [ 1 ] * ( N )
	for i in range ( 1 , N ) :
		for j in range ( i ) :
			if ( arr [ j ] < arr [ i ] ) :
				left [ i ] = max ( left [ i ] , left [ j ] + 1 )
	for i in range ( N - 2 , - 1 , - 1 ) :
		for j in range ( N - 1 , i , - 1 ) :
			if ( arr [ i ] > arr [ j ] ) :
				right [ i ] = max ( right [ i ] , right [ j ] + 1 )
	maxLen = 0
	for i in range ( 1 , N - 1 ) :
		maxLen = max ( maxLen , left [ i ] + right [ i ] - 1 )
	print ( ( N - maxLen ) )
def makeBitonic ( arr , N ) :
	if ( N == 1 ) :
		print ( "0" )
		return
	if ( N == 2 ) :
		if ( arr [ 0 ] != arr [ 1 ] ) :
			print ( "0" )
		else :
			print ( "1" )
		return
	min_element_removal ( arr , N )
if __name__ == ' _ _ main _ _ ' :
	arr = [ 2 , 1 , 1 , 5 , 6 , 2 , 3 , 1 ]
	N = len ( arr )
	makeBitonic ( arr , N )