Find the largest multiple of 3 from array of digits | Set 2 (In O(n) time and O(1) space)

Given an array of digits (contain elements from 0 to 9). Find the largest number that can be made from some or all digits of array and is divisible by 3. The same element may appear multiple times in the array, but each element in the array may only be used once.

Examples:

Input : arr[] = {5, 4, 3, 1, 1} 
Output : 4311

Input : Arr[] = {5, 5, 5, 7} 
Output : 555 

Asked In : Google Interview

We have discussed a queue based solution. Both solutions (discussed in previous and this posts) are based on the fact that a number is divisible by 3 if and only if sum of digits of the number is divisible by 3.
For example, let us consider 555, it is divisible by 3 because sum of digits is 5 + 5 + 5 = 15, which is divisible by 3. If a sum of digits is not divisible by 3 then the remainder should be either 1 or 2.
If we get remainder either ‘1’ or ‘2’, we have to remove maximum two digits to make a number that is divisible by 3:

  1. If remainder is ‘1’ : We have to remove single digit that have remainder ‘1’ or we have to remove two digit that have remainder ‘2’ ( 2 + 2 => 4 % 3 => ‘1’)
  2. If remainder is ‘2’ : .We have to remove single digit that have remainder ‘2’ or we have to remove two digit that have remainder ‘1’ ( 1 + 1 => 2 % 3 => 2 ).

Examples :

Input : arr[] = 5, 5, 5, 7 
Sum of digits = 5 + 5 + 7 = 22
Remainder = 22 % 3 = 1 
We remove smallest single digit that 
has remainder '1'. We remove 7 % 3 = 1 
So largest number divisible by 3 is : 555

Let's take an another example :
Input : arr[]  = 4 , 4 , 1 , 1 , 1 , 3
Sum of digits  = 4 + 4 + 1 + 1 + 1 + 3 = 14
Reminder = 14 % 3 = 2 
We have to remove the smallest digit that 
has remainder ' 2 ' or two digits that have
remainder '1'. Here there is no digit with 
reminder '2', so we have to remove two smallest 
digits that have remainder '1'. The digits are : 
1, 1. So largest number divisible by 3 is 4 4 3 1 

Below are implementation of above idea.

C++

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// C++ program to find the largest number
// that can be mode from elements of the
// array and is divisible by 3
#include<bits/stdc++.h>
using namespace std;
  
// Number of digits
#define MAX_SIZE 10
  
// function to sort array of digits using
// counts
void sortArrayUsingCounts(int arr[], int n)
{
    // Store count of all elements
    int count[MAX_SIZE] = {0};
    for (int i = 0; i < n; i++)
        count[arr[i]]++;
  
    // Store
    int index = 0;
    for (int i = 0; i < MAX_SIZE; i++)
        while (count[i] > 0)
            arr[index++] = i, count[i]--;
}
  
// Remove elements from arr[] at indexes ind1 and ind2
bool removeAndPrintResult(int arr[], int n, int ind1,
                                        int ind2 = -1)
{
    for (int i = n-1; i >=0; i--)
        if (i != ind1 && i != ind2)
            cout << arr[i] ;
}
  
// Returns largest multiple of 3 that can be formed
// using arr[] elements.
bool largest3Multiple(int arr[], int n)
{
    // Sum of all array element
    int sum = accumulate(arr, arr+n, 0);
  
    // Sum is divisible by 3 , no need to
    // delete an element
    if (sum%3 == 0)
        return true ;
  
    // Sort array element in increasing order
    sortArrayUsingCounts(arr, n);
  
    // Find reminder
    int remainder = sum % 3;
  
    // If remainder is '1', we have to delete either
    // one element of remainder '1' or two elements
    // of remainder '2'
    if (remainder == 1)
    {
        int rem_2[2];
        rem_2[0] = -1, rem_2[1] = -1;
  
        // Traverse array elements
        for (int i = 0 ; i < n ; i++)
        {
            // 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
                else if (rem_2[1] == -1)
                    rem_2[1] = i;
            }
        }
  
        if (rem_2[0] != -1 && 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'
    else if (remainder == 2)
    {
        int rem_1[2];
        rem_1[0] = -1, rem_1[1] = -1;
  
        // traverse array elements
        for (int i = 0; i < n; i++)
        {
            // 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
                else if (rem_1[1] == -1)
                    rem_1[1] = i;
            }
        }
  
        if (rem_1[0] != -1 && rem_1[1] != -1)
        {
            removeAndPrintResult(arr, n, rem_1[0], rem_1[1]);
            return true;
        }
    }
  
    cout << "Not possible";
    return false;
}
  
// Driver code
int main()
{
    int arr[] = {4 , 4 , 1 , 1 , 1 , 3} ;
    int n = sizeof(arr)/sizeof(arr[0]);
    largest3Multiple(arr, n);
    return 0;
}

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Java

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// Java program to find the largest number
// that can be mode from elements of the
// array and is divisible by 3
import java.util.*;
  
class GFG 
{
  
    // Number of digits
    static int MAX_SIZE = 10;
  
    // function to sort array of digits using
    // counts
    static void sortArrayUsingCounts(int arr[], 
                                     int n)
    {
        // Store count of all elements
        int[] count = new int[MAX_SIZE];
        for (int i = 0; i < n; i++) 
        {
            count[arr[i]]++;
        }
  
        // Store
        int index = 0;
        for (int i = 0; i < MAX_SIZE; i++) 
        {
            while (count[i] > 0
            {
                arr[index++] = i;
                count[i]--;
            }
        }
    }
  
    // Remove elements from arr[]
    // at indexes ind1 and ind2
    static void removeAndPrintResult(int arr[], int n, 
                                     int ind1, int ind2)
    {
        for (int i = n - 1; i >= 0; i--)
        {
            if (i != ind1 && i != ind2) 
            {
                System.out.print(arr[i]);
            }
        }
    }
  
    // Returns largest multiple of 3 
    // that can be formed using
    // arr[] elements.
    static boolean largest3Multiple(int arr[], 
                                    int n)
    {
        // Sum of all array element
        int sum = accumulate(arr, 0, n);
  
        // If sum is divisible by 3, 
        // no need to delete an element
        if (sum % 3 == 0
        {
            return true;
        }
  
        // Sort array element in increasing order
        sortArrayUsingCounts(arr, n);
  
        // Find reminder
        int remainder = sum % 3;
  
        // If remainder is '1', we have to 
        // delete either one element of 
        // remainder '1' or two elements of 
        // remainder '2'
        if (remainder == 1)
        {
            int[] rem_2 = new int[2];
            rem_2[0] = -1;
            rem_2[1] = -1;
  
            // Traverse array elements
            for (int i = 0; i < n; i++) 
            {
                  
                // Store first element of remainder '1'
                if (arr[i] % 3 == 1)
                {
                    removeAndPrintResult(arr, n, i, -1);
                    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
                    else if (rem_2[1] == -1
                    {
                        rem_2[1] = i;
                    }
                }
            }
  
            if (rem_2[0] != -1 && 
                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'
        else if (remainder == 2
        {
            int[] rem_1 = new int[2];
            rem_1[0] = -1;
            rem_1[1] = -1;
  
            // traverse array elements
            for (int i = 0; i < n; i++) 
            {
                  
                // store first element of remainder '2'
                if (arr[i] % 3 == 2
                {
                    removeAndPrintResult(arr, n, i, -1);
                    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
                    else if (rem_1[1] == -1
                    {
                        rem_1[1] = i;
                    }
                }
            }
  
            if (rem_1[0] != -1 && 
                rem_1[1] != -1)
            {
                removeAndPrintResult(arr, n, rem_1[0], 
                                             rem_1[1]);
                return true;
            }
        }
        System.out.print("Not possible");
        return false;
    }
  
    static int accumulate(int[] arr,
                          int start, 
                          int end) 
    {
        int sum = 0;
        for (int i = 0; i < arr.length; i++) 
        {
            sum += arr[i];
        }
        return sum;
    }
      
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = {4, 4, 1, 1, 1, 3};
        int n = arr.length;
        largest3Multiple(arr, n);
    }
}
  
// This code is contributed
// by Princi Singh

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Python3

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# Python program to find the largest number
# that can be mode from elements of the
# array and is divisible by 3
  
# Number of digits
MAX_SIZE = 10
  
# function to sort array of digits using
# counts
def sortArrayUsingCounts(arr, n):
  
    # Store count of all elements
    count = [0]*MAX_SIZE
    for i in range(n):
        count[arr[i]] += 1
  
    # Store
    index = 0
    for i in range(MAX_SIZE):
        while count[i] > 0:
            arr[index] = i
            index += 1
            count[i] -= 1
  
  
# Remove elements from arr[] at indexes ind1 and ind2
def removeAndPrintResult(arr, n, ind1, ind2=-1):
    for i in range(n-1, -1, -1):
        if i != ind1 and i != ind2:
            print(arr[i], end="")
  
  
# Returns largest multiple of 3 that can be formed
# using arr[] elements.
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
    sortArrayUsingCounts(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 = [4, 4, 1, 1, 1, 3]
    n = len(arr)
    largest3Multiple(arr, n)
  
# This code is contributed by
# sanjeev2552

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C#

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// C# program to find the largest number
// that can be mode from elements of the
// array and is divisible by 3
using System;
      
class GFG 
{
  
    // Number of digits
    static int MAX_SIZE = 10;
  
    // function to sort array of digits using
    // counts
    static void sortArrayUsingCounts(int []arr, 
                                     int n)
    {
        // Store count of all elements
        int[] count = new int[MAX_SIZE];
        for (int i = 0; i < n; i++) 
        {
            count[arr[i]]++;
        }
  
        // Store
        int index = 0;
        for (int i = 0; i < MAX_SIZE; i++) 
        {
            while (count[i] > 0) 
            {
                arr[index++] = i;
                count[i]--;
            }
        }
    }
  
    // Remove elements from arr[]
    // at indexes ind1 and ind2
    static void removeAndPrintResult(int []arr, int n, 
                                     int ind1, int ind2)
    {
        for (int i = n - 1; i >= 0; i--)
        {
            if (i != ind1 && i != ind2) 
            {
                Console.Write(arr[i]);
            }
        }
    }
  
    // Returns largest multiple of 3 
    // that can be formed using
    // arr[] elements.
    static Boolean largest3Multiple(int []arr, 
                                    int n)
    {
        // Sum of all array element
        int sum = accumulate(arr, 0, n);
  
        // If sum is divisible by 3, 
        // no need to delete an element
        if (sum % 3 == 0) 
        {
            return true;
        }
  
        // Sort array element in increasing order
        sortArrayUsingCounts(arr, n);
  
        // Find reminder
        int remainder = sum % 3;
  
        // If remainder is '1', we have to 
        // delete either one element of 
        // remainder '1' or two elements of 
        // remainder '2'
        if (remainder == 1)
        {
            int[] rem_2 = new int[2];
            rem_2[0] = -1;
            rem_2[1] = -1;
  
            // Traverse array elements
            for (int i = 0; i < n; i++) 
            {
                  
                // Store first element of remainder '1'
                if (arr[i] % 3 == 1)
                {
                    removeAndPrintResult(arr, n, i, -1);
                    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
                    else if (rem_2[1] == -1) 
                    {
                        rem_2[1] = i;
                    }
                }
            }
  
            if (rem_2[0] != -1 && 
                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'
        else if (remainder == 2) 
        {
            int[] rem_1 = new int[2];
            rem_1[0] = -1;
            rem_1[1] = -1;
  
            // traverse array elements
            for (int i = 0; i < n; i++) 
            {
                  
                // store first element of remainder '2'
                if (arr[i] % 3 == 2) 
                {
                    removeAndPrintResult(arr, n, i, -1);
                    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
                    else if (rem_1[1] == -1) 
                    {
                        rem_1[1] = i;
                    }
                }
            }
  
            if (rem_1[0] != -1 && 
                rem_1[1] != -1)
            {
                removeAndPrintResult(arr, n, rem_1[0], 
                                             rem_1[1]);
                return true;
            }
        }
        Console.Write("Not possible");
        return false;
    }
  
    static int accumulate(int[] arr,
                          int start, 
                          int end) 
    {
        int sum = 0;
        for (int i = 0; i < arr.Length; i++) 
        {
            sum += arr[i];
        }
        return sum;
    }
      
    // Driver code
    public static void Main(String[] args)
    {
        int []arr = {4, 4, 1, 1, 1, 3};
        int n = arr.Length;
        largest3Multiple(arr, n);
    }
}
  
// This code is contributed by 29AjayKumar

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Output:

4431

Time Complexity : O(n)
This article is contributed by Nishant Singh. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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