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
```

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 is C++ implementation of above idea.

```// 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;
}
```

Output:

``` 555
```

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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