# 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

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 + 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++

 // C++ program to find the largest number // that can be mode from elements of the // array and is divisible by 3 #include 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);             // Sort array element in increasing order     sortArrayUsingCounts(arr, n);         // Sum is divisible by 3 , no need to     // delete an element     if (sum%3 == 0)     {             removeAndPrintResult(arr,n,-1,-1);           return true ;     }       // 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; }

## Java

 // 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);                   // Sort array element in increasing order         sortArrayUsingCounts(arr, n);                   // If sum is divisible by 3,          // no need to delete an element         if (sum % 3 == 0)          {             removeAndPrintResult(arr, n, -1, -1);             return true;         }           // 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);     } }

## Python3

 # Python3 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)       # Sort array element in increasing order     sortArrayUsingCounts(arr, n)       # Sum is divisible by 3, no need to     # delete an element     if s % 3 == 0:         removeAndPrintResult(arr, n, -1)         return True          # 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

## C#

 // 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);           // Sort array element in increasing order         sortArrayUsingCounts(arr, n);           // If sum is divisible by 3,          // no need to delete an element         if (sum % 3 == 0)          {             removeAndPrintResult(arr, n, -1, -1);             return true;         }               // 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

## Javascript



Output:

4431

Time Complexity : O(n)

Previous
Next