Calculate largest multiple of 3 in Python

By Vineet Raj

In this article, you will learn to Calculate the largest multiple of 3 in Python.

Problem: largest multiple of 3 in Python from the given digits

We have to find the largest multiple of 3 that can be formed from array elements(we have an array of positive integers).

For example, if we have an input list of {5,6,4}, the output will be “54”, and if the input list is {5,6,7,4,1}, the output will be “7641”.

Basic Approach-

Generate all the combinations of given digits and the maximum number that is divisible by 3 is the result.
But, it will have a time complexity of O(2^n).

Better Approach-

The sum of the digits of the multiple is divisible by 3.

453 is divisible by 3 and (4+5+3=12) is also divisible by 3.

#Python program
MAX_SIZE = 10
def sortArray(arr, n): 

  count = [0]*MAX_SIZE 
  for i in range(n): 
    count[arr[i]] += 1
 
  index = 0
  for i in range(MAX_SIZE): 
    while count[i] > 0: 
      arr[index] = i 
      index += 1
      count[i] -= 1
def removePrintResult(arr, n, ind1, ind2=-1): 
  for i in range(n-1, -1, -1): 
    if i != ind1 and i != ind2: 
      print(arr[i], end="") 


 
def largest3Multiple(arr, n): 

  # Sum of all array element 
  s = sum(arr) 

  # Sum is divisible by 3, no need to 
  # delete an element 
  if s % 3 == 0: 
    return True

  # Sort array element in increasing order 
  sortArray(arr, n) 

  # Find reminder 
  remainder = s % 3

  # If remainder is '1', we have to delete either 
  # one element of remainder '1' or two elements 
  # of remainder '2' 
  if remainder == 1: 
    rem_2 = [0]*2
    rem_2[0] = -1; rem_2[1] = -1

    # Traverse array elements 
    for i in range(n): 

      # Store first element of remainder '1' 
      if arr[i] % 3 == 1: 
        removeAndPrintResult(arr, n, i) 
        return True

      if arr[i] % 3 == 2: 

        # If this is first occurrence of remainder 2 
        if rem_2[0] == -1: 
          rem_2[0] = i 

        # If second occurrence 
        elif rem_2[1] == -1: 
          rem_2[1] = i 

    if rem_2[0] != -1 and rem_2[1] != -1: 
      removeAndPrintResult(arr, n, rem_2[0], rem_2[1]) 
      return True

  # If remainder is '2', we have to delete either 
  # one element of remainder '2' or two elements 
  # of remainder '1' 
  elif remainder == 2: 
    rem_1 = [0]*2
    rem_1[0] = -1; rem_1[1] = -1

    # traverse array elements 
    for i in range(n): 

      # store first element of remainder '2' 
      if arr[i] % 3 == 2: 
        removeAndPrintResult(arr, n, i) 
        return True

      if arr[i] % 3 == 1: 

        # If this is first occurrence of remainder 1 
        if rem_1[0] == -1: 
          rem_1[0] = i 
        
        # If second occurrence 
        elif rem_1[1] == -1: 
          rem_1[1] = i 
          
    if rem_1[0] != -1 and rem_1[1] != -1: 
      removeAndPrintResult(arr, n, rem_1[0], rem_1[1]) 
      return True

  print("Not possible") 
  return False

# Driver code 
if __name__ == "__main__": 

  arr = [5,1,2,1] 
  n = len(arr) 
  largest3Multiple(arr, n) 

Output:

663

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