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Maximize sum of array by repeatedly removing an element from pairs whose concatenation is a multiple of 3 | Function to calculate sum of digits of an integer.
Function to calculate maximum sum of array after removing pairs whose concatenation is divisible by 3.
Stores the sum of digits of array element.
Find the sum of digits.
If i is divisible by 3.
Otherwise , if i modulo 3 is 1.
Otherwise , if i modulo 3 is 2.
Return the resultant sum of array elements.
Driver Code"
Solution in Python:
 Give me a code to maximize sum of array by repeatedly removing an element from pairs whose concatenation is a multiple of 3 and the other steps required?
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#Here is code to maximize sum of array by repeatedly removing an element from pairs whose concatenation is a multiple of 3 and the other steps required.
 def getSum ( n ) :
	ans = 0
	for i in str ( n ) :
		ans += int ( i )
	return ans
def getMax ( arr ) :
	maxRem0 = 0
	rem1 = 0
	rem2 = 0
	for i in arr :
		digitSum = getSum ( i )
		if digitSum % 3 == 0 :
			maxRem0 = max ( maxRem0 , i )
		elif digitSum % 3 == 1 :
			rem1 += i
		else :
			rem2 += i
	print ( maxRem0 + max ( rem1 , rem2 ) )
arr = [ 23 , 12 , 43 , 3 , 56 ]
getMax ( arr )