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:

- 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’)
- 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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