human
Find the Number of Permutations that satisfy the given condition in an array | Function to calculate x ^ y recursively.
Function to return the number of permutations that satisfy the given condition in an array.
If there is only one element then only one permutation is available.
Sort the array for calculating the number of elements occurring twice.
If the maximum element is occurring twice , then the number of permutations satisfying the condition is 0.
This variable will store the number of element occurring twice.
Loop to check the number of elements occurring twice.
Check if this element is occurring twice.
If this element is occurring twice then check if this number is occurring more than twice.
If element occurring thrice then no permutation will satisfy the given condition.
Since we have checked the next element as well , then we can increment the loop variable.
Driver code"
How can the above be solved in Python?

Can you provide a code to find the Number of Permutations that satisfy the given condition in an array and the other steps required?
assistant
#Here is code to find the Number of Permutations that satisfy the given condition in an array and the other steps required.
 def pow ( x , y ) :
	if ( y == 1 ) :
		return x
	if ( y == 0 ) :
		return 1
	temp = pow ( x , y // 2 )
	temp *= temp
	if ( y & 1 ) :
		temp *= x
	return temp
def noOfPermutations ( a , n ) :
	if ( n == 1 ) :
		return 1
	a . sort ( )
	if ( a [ n - 1 ] == a [ n - 2 ] ) :
		return 0
	x = 0
	for i in range ( n - 2 ) :
		if ( a [ i ] == a [ i + 1 ] ) :
			if ( a [ i ] == a [ i + 2 ] ) :
				return 0
			x += 1
			i += 1
	return pow ( 2 , n - 2 * x - 1 )
if __name__ == " _ _ main _ _ " :
	a = [ 1 , 2 , 2 , 3 , 4 ]
	n = len ( a )
	num = noOfPermutations ( a , n )
	print ( num )