human
Count number of rotated strings which have more number of vowels in the first half than second half | Function to return the count of rotated strings which have more number of vowels in the first half than the second half.
Create a new string.
Pre array to store count of all vowels.
Compute the prefix array.
To store the required answer.
Find all rotated strings.
Right and left index of the string.
x1 stores the number of vowels in the rotated string.
Left stores the number of vowels in the first half of rotated string.
Right stores the number of vowels in the second half of rotated string.
If the count of vowels in the first half is greater than the count in the second half.
Return the required answer.
Driver code"
How can the above be solved in Python?
assistant
#Here is code to count number of rotated strings which have more number of vowels in the first half than second half and the other steps required.
 def cntRotations ( s , n ) :
	str = s + s ;
	pre = [ 0 ] * ( 2 * n ) ;
	for i in range ( 2 * n ) :
		if ( i != 0 ) :
			pre [ i ] += pre [ i - 1 ] ;
		if ( str [ i ] == ' a ' or str [ i ] == ' e ' or str [ i ] == ' i ' or str [ i ] == ' o ' or str [ i ] == ' u ' ) :
			pre [ i ] += 1 ;
	ans = 0 ;
	for i in range ( n - 1 , 2 * n - 1 , 1 ) :
		r = i ; l = i - n ;
		x1 = pre [ r ] ;
		if ( l >= 0 ) :
			x1 -= pre [ l ] ;
		r = ( int ) ( i - n / 2 ) ;
		left = pre [ r ] ;
		if ( l >= 0 ) :
			left -= pre [ l ] ;
		right = x1 - left ;
		if ( left > right ) :
			ans += 1 ;
	return ans ;
s = " abecidft " ;
n = len ( s ) ;
print ( cntRotations ( s , n ) ) ;