human
Count of possible arrays from prefix | Function to find power of a number ..
Function to find factorial of a number ..
Function to print no of arrays.
c variable counts the no of pairs.
Map to store the frequency of each element.
Sum of all elements of the array.
Variable to check if it is possible to make any array.
First element of suffix array and the last element of prefix array.
Check if the element exists in the map.
If elements of any pair are equal and their frequency is not divisible by 2 update the isArrayPossible variable to false and break through the loop.
If elements of any pair are not equal and their frequency is not same update the isArrayPossible variable to false and break through the loop.
Check if frequency is greater than zero.
update the count of pairs.
Multiply the answer by 2 ^ ( frequency of pairs ) since the elements of the pair are not the same in this condition.
Divide the answer by the factorial of no of similar pairs.
Make frequency of both these elements 0.
Update the count of pairs.
Divide the answer by the factorial of no . of similar pairs.
Make frequency of this element 0.
Check if it is possible to make the array and there are n - 1 pairs whose sum will be equal to s1.
Driver code.
Function calling"
Solution in Python:

Can you provide a code to count of possible arrays from prefix and the other steps required?
assistant
#Here is code to count of possible arrays from prefix and the other steps required.
 def power ( a , b ) :
	result = 1 ;
	while ( b > 0 ) :
		if ( b % 2 == 1 ) :
			result = result * a ;
		a = a * a ;
		b = b // 2 ;
	return result ;
def factorial ( n ) :
	fact = 1 ;
	for i in range ( 1 , n + 1 ) :
		fact = fact * i ;
	return fact ;
def findNoOfArrays ( a , n ) :
	sum = 0 ; c = 0 ;
	mp = dict . fromkeys ( a , 0 ) ;
	for i in range ( 2 * n ) :
		mp [ a [ i ] ] += 1 ;
		sum = sum + a [ i ] ;
	isArrayPossible = True ;
	ans = factorial ( n - 1 ) ;
	s1 = sum // ( n + 1 ) ;
	if ( mp [ s1 ] >= 2 ) :
		mp [ s1 ] = mp [ s1 ] - 2 ;
	else :
		isArrayPossible = False ;
	if ( isArrayPossible ) :
		for first , second in mp . items ( ) :
			if ( first == s1 - first ) :
				if ( mp [ first ] % 2 != 0 ) :
					isArrayPossible = False ;
					break ;
			if ( first != s1 - first ) :
				if s1 - first in mp :
					if ( mp [ first ] != mp [ s1 - first ] ) :
						isArrayPossible = False ;
						break ;
			if ( second > 0 ) :
				if ( first != s1 - first ) :
					c = c + second ;
					ans = ans * power ( 2 , second ) ;
					ans = ans / factorial ( second ) ;
					mp [ first ] = 0 ;
					mp [ s1 - first ] = 0 ;
				if ( first == s1 - first ) :
					c = c + second // 2 ;
					ans = ans // factorial ( second // 2 ) ;
					mp [ first ] = 0 ;
	if ( c < n - 1 or isArrayPossible == False ) :
		print ( "0" ) ;
	else :
		print ( ans ) ;
if __name__ == " _ _ main _ _ " :
	arr1 = [ 5 , 2 , 3 , 5 ] ;
	n1 = len ( arr1 ) ;
	findNoOfArrays ( arr1 , n1 // 2 ) ;
	arr2 = [ - 1 , - 1 , - 1 , 0 , 1 , 0 , 1 , 0 , 1 , 0 , 0 , 0 ] ;
	n2 = len ( arr2 ) ;
	findNoOfArrays ( arr2 , n2 // 2 ) ;