human
Number of quadruples where the first three terms are in AP and last three terms are in GP | Function to return the count of quadruples.
Hash table to count the number of occurrences.
Traverse and increment the count.
Run two nested loop for second and third element.
If they are same.
Initially decrease the count.
Find the first element using common difference.
Find the fourth element using GP y ^ 2 = x * z property.
If it is an integer.
If not equal.
Same elements.
Later increase the value for future calculations.
Driver code"
Solution in Python:
assistant
#Here is code to number of quadruples where the first three terms are in AP and last three terms are in GP and the other steps required.
 def countQuadruples ( a , n ) :
	mpp = dict . fromkeys ( a , 0 ) ;
	for i in range ( n ) :
		mpp [ a [ i ] ] += 1 ;
	count = 0 ;
	for j in range ( n ) :
		for k in range ( n ) :
			if ( j == k ) :
				continue ;
			mpp [ a [ j ] ] -= 1 ;
			mpp [ a [ k ] ] -= 1 ;
			first = a [ j ] - ( a [ k ] - a [ j ] ) ;
			if first not in mpp :
				mpp [ first ] = 0 ;
			fourth = ( a [ k ] * a [ k ] ) // a [ j ] ;
			if fourth not in mpp :
				mpp [ fourth ] = 0 ;
			if ( ( a [ k ] * a [ k ] ) % a [ j ] == 0 ) :
				if ( a [ j ] != a [ k ] ) :
					count += mpp [ first ] * mpp [ fourth ] ;
				else :
					count += ( mpp [ first ] * ( mpp [ fourth ] - 1 ) ) ;
			mpp [ a [ j ] ] += 1 ;
			mpp [ a [ k ] ] += 1 ;
	return count ;
if __name__ == " _ _ main _ _ " :
	a = [ 2 , 6 , 4 , 9 , 2 ] ;
	n = len ( a ) ;
	print ( countQuadruples ( a , n ) ) ;