human
Count of all possible reverse bitonic subarrays | Function that counts all the reverse bitonic subarray in arr [ ].
To store the count of reverse bitonic subarray.
Iterate the array and select the starting element.
Iterate for selecting the ending element for subarray.
Subarray arr [ i to j ].
For 1 length , increment the count and continue.
For Decreasing Subarray.
Check if only Decreasing.
For Increasing Subarray.
Print the total count of subarrays.
Given array arr [ ].
Function Call"
How can the above be solved in Python?

Write me a code to count of all possible reverse bitonic subarrays and the other steps required?
assistant
#Here is code to count of all possible reverse bitonic subarrays and the other steps required.
 def countReversebitonic ( arr , n ) :
	c = 0 ;
	for i in range ( n ) :
		for j in range ( i , n ) :
			temp = arr [ i ]
			f = 0 ;
			if ( j == i ) :
				c += 1 ;
				continue ;
			k = i + 1 ;
			while ( k <= j and temp > arr [ k ] ) :
				temp = arr [ k ] ;
				k += 1 ;
			if ( k > j ) :
				c += 1 ;
				f = 2 ;
			while ( k <= j and temp < arr [ k ] and f != 2 ) :
				temp = arr [ k ] ;
				k += 1 ;
			if ( k > j and f != 2 ) :
				c += 1 ;
				f = 0 ;
	print ( c )
arr = [ 2 , 3 , 1 , 4 ] ;
countReversebitonic ( arr , len ( arr ) ) ;