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

01 Dec, 2021
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 ).
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 + 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++
Java
Python3
C#
Javascript
// 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// countsvoid 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 ind2bool 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 codeint main(){    int arr[] = {4, 4, 1, 1, 1, 3 } ;    int n = sizeof(arr)/sizeof(arr[0]);    largest3Multiple(arr, n);    return 0;}
// Java program to find the largest number// that can be mode from elements of the// array and is divisible by 3import 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 program to find the largest number# that can be mode from elements of the# array and is divisible by 3  # Number of digitsMAX_SIZE = 10  # function to sort array of digits using# countsdef 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 ind2def 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 codeif __name__ == "__main__":      arr = [4, 4, 1, 1, 1, 3]    n = len(arr)    largest3Multiple(arr, n)  # This code is contributed by# sanjeev2552
// C# program to find the largest number// that can be mode from elements of the// array and is divisible by 3using 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
<script>// JavaScript program to find the largest number// that can be mode from elements of the// array and is divisible by 3  // Number of digitsconst MAX_SIZE = 10;  // function to sort array of digits using// countsfunction sortArrayUsingCounts(arr, n){    // Store count of all elements    let count = new Uint8Array(MAX_SIZE);    for (let i = 0; i < n; i++)        count[arr[i]]++;      // Store    let index = 0;    for (let i = 0; i < MAX_SIZE; i++)        while (count[i] > 0)            arr[index++] = i, count[i]--;}  // Remove elements from arr[] at indexes ind1 and ind2function removeAndPrintResult(arr, n, ind1, ind2 = -1){    for (let i = n-1; i >=0; i--)        if (i != ind1 && i != ind2)            document.write(arr[i]) ;}  // Returns largest multiple of 3 that can be formed// using arr[] elements.function largest3Multiple(arr, n){    // Sum of all array element    let sum = arr.reduce((a, b) => a + b, 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);        return true ;    }          // Find reminder    let 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)    {        let rem_2 = new Array(2);        rem_2[0] = -1, rem_2[1] = -1;          // Traverse array elements        for (let 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)    {        let rem_1 = new Array(2);        rem_1[0] = -1, rem_1[1] = -1;          // traverse array elements        for (let 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;        }    }      document.write("Not possible");    return false;}  // Driver code    let arr = [4 , 4 , 1 , 1 , 1 , 3];    let n = arr.length;    largest3Multiple(arr, n);      // This code is contributed by Surbhi Tyagi.</script>
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 write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
sanjeev2552
princi singh
29AjayKumar
2435shivkant
surbhityagi15
divisibility
Mathematical
Sorting
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