human
Count the number of non | Python3 program to count number of non increasing subarrays.
Initialize result.
Initialize length of current non - increasing subarray.
Traverse through the array.
If arr [ i + 1 ] is less than or equal to arr [ i ] , then increment length.
Else Update count and reset length.
If last length is more than 1.
Driver code"
Solution in Python:

=====
Give me a code to count the number of non and the other steps required?
assistant
#Here is code to count the number of non and the other steps required.
 def countNonIncreasing ( arr , n ) :
	cnt = 0 ;
	len = 1 ;
	for i in range ( 0 , n - 1 ) :
		if ( arr [ i + 1 ] <= arr [ i ] ) :
			len += 1 ;
		else :
			cnt += ( ( ( len + 1 ) * len ) / 2 ) ;
			len = 1 ;
	if ( len > 1 ) :
		cnt += ( ( ( len + 1 ) * len ) / 2 ) ;
	return int ( cnt ) ;
if __name__ == ' _ _ main _ _ ' :
	arr = [ 5 , 2 , 3 , 7 , 1 , 1 ] ;
	n = len ( arr ) ;
	print ( countNonIncreasing ( arr , n ) ) ;