human
Count all possible groups of size 2 or 3 that have sum as multiple of 3 | Returns count of all possible groups that can be formed from elements of a [ ] ..
Create an array C [ 3 ] to store counts of elements with remainder 0 , 1 and 2. c [ i ] would store count of elements with remainder i.
To store the result.
Count elements with remainder 0 , 1 and 2.
Case 3. a : Count groups of size 2 from 0 remainder elements.
Case 3. b : Count groups of size 2 with one element with 1 remainder and other with 2 remainder.
Case 4. a : Count groups of size 3 with all 0 remainder elements.
Case 4. b : Count groups of size 3 with all 1 remainder elements.
Case 4. c : Count groups of size 3 with all 2 remainder elements.
Case 4. c : Count groups of size 3 with different remainders.
Return total count stored in res.
Driver program"
How can the above be solved in Python?
 Can you provide a code to count all possible groups of size 2 or 3 that have sum as multiple of 3 and the other steps required?
assistant
#Here is code to count all possible groups of size 2 or 3 that have sum as multiple of 3 and the other steps required.
 def findgroups ( arr , n ) :
	c = [ 0 , 0 , 0 ]
	res = 0
	for i in range ( 0 , n ) :
		c [ arr [ i ] % 3 ] += 1
	res += ( ( c [ 0 ] * ( c [ 0 ] - 1 ) ) >> 1 )
	res += c [ 1 ] * c [ 2 ]
	res += ( c [ 0 ] * ( c [ 0 ] - 1 ) * ( c [ 0 ] - 2 ) ) / 6
	res += ( c [ 1 ] * ( c [ 1 ] - 1 ) * ( c [ 1 ] - 2 ) ) / 6
	res += ( ( c [ 2 ] * ( c [ 2 ] - 1 ) * ( c [ 2 ] - 2 ) ) / 6 )
	res += c [ 0 ] * c [ 1 ] * c [ 2 ]
	return res
arr = [ 3 , 6 , 7 , 2 , 9 ]
n = len ( arr )
print ( " Required ▁ number ▁ of ▁ groups ▁ are " , int ( findgroups ( arr , n ) ) )