human
Count positions such that all elements before it are greater | Function to count positions such that all elements before it are greater.
Count is initially 1 for the first element.
Initial Minimum.
Traverse the array.
If current element is new minimum.
Update minimum.
Increment count.
Driver Code"
Solution in Python:
 What is a code to count positions such that all elements before it are greater and the other steps required?
assistant
#Here is code to count positions such that all elements before it are greater and the other steps required.
 def getPositionCount ( a , n ) :
	count = 1 ;
	min = a [ 0 ] ;
	for i in range ( 1 , n ) :
		if ( a [ i ] <= min ) :
			min = a [ i ] ;
			count += 1 ;
	return count ;
if __name__ == " _ _ main _ _ " :
	a = [ 5 , 4 , 6 , 1 , 3 , 1 ] ;
	n = len ( a ) ;
	print ( getPositionCount ( a , n ) ) ;