Find the largest multiple of 3 | Set 1 (Using Queue)

08 Jul, 2022
Given an array of non-negative integers. Find the largest multiple of 3 that can be formed from array elements. For example, if the input array is {8, 1, 9}, the output should be “9 8 1”, and if the input array is {8, 1, 7, 6, 0}, output should be “8 7 6 0”.
Method 1 (Brute Force) :
The simple & straight forward approach is to generate all the combinations of the elements and keep track of the largest number formed which is divisible by 3. Time Complexity: O(n x 2^n). There will be 2^n combinations of array elements. To compare each combination with the largest number so far may take O(n) time. Auxiliary Space: O(n) // to avoid integer overflow, the largest number is assumed to be stored in the form of array.
Method 2 (Tricky) :
This problem can be solved efficiently with the help of O(n) extra space. This method is based on the following facts about numbers which are multiple of 3.
A number is multiple of 3 if and only if the sum of digits of number is multiple of 3. For example, let us consider 8760, it is a multiple of 3 because sum of digits is 8 + 7+ 6+ 0 = 21, which is a multiple of 3. If a number is multiple of 3, then all permutations of it are also multiple of 3. For example, since 6078 is a multiple of 3, the numbers 8760, 7608, 7068, ..... are also multiples of 3. We get the same remainder when we divide the number and sum of digits of the number. For example, if divide number 151 and sum of it digits 7, by 3, we get the same remainder 1.
A number is multiple of 3 if and only if the sum of digits of number is multiple of 3. For example, let us consider 8760, it is a multiple of 3 because sum of digits is 8 + 7+ 6+ 0 = 21, which is a multiple of 3. 
If a number is multiple of 3, then all permutations of it are also multiple of 3. For example, since 6078 is a multiple of 3, the numbers 8760, 7608, 7068, ..... are also multiples of 3. 
We get the same remainder when we divide the number and sum of digits of the number. For example, if divide number 151 and sum of it digits 7, by 3, we get the same remainder 1.
What is the idea behind above facts? 
The value of 10%3 and 100%3 is 1. The same is true for all the higher powers of 10, because 3 divides 9, 99, 999, ... etc. Let us consider a 3 digit number n to prove above facts. Let the first, second and third digits of n be ‘a’, ‘b’ and ‘c’ respectively. n can be written as 
n = 100.a + 10.b + c 
Since (10^x)%3 is 1 for any x, the above expression gives the same remainder as following expression
 1.a + 1.b + c 
So the remainder obtained by sum of digits and ‘n’ is same.
Following is a solution based on the above observation.
Sort the array in non-decreasing order.Take three queues. One for storing elements which on dividing by 3 gives remainder as 0.The second queue stores digits which on dividing by 3 gives remainder as 1. The third queue stores digits which on dividing by 3 gives remainder as 2. Call them as queue0, queue1 and queue2Find the sum of all the digits.Three cases arise: ......4.1 The sum of digits is divisible by 3. Dequeue all the digits from the three queues. Sort them in non-increasing order. Output the array.......4.2 The sum of digits produces remainder 1 when divided by 3. Remove one item from queue1. If queue1 is empty, remove two items from queue2. If queue2 contains less than two items, the number is not possible.......4.3 The sum of digits produces remainder 2 when divided by 3. Remove one item from queue2. If queue2 is empty, remove two items from queue1. If queue1 contains less than two items, the number is not possible.Finally empty all the queues into an auxiliary array. Sort the auxiliary array in non-increasing order. Output the auxiliary array.
Sort the array in non-decreasing order.
Take three queues. One for storing elements which on dividing by 3 gives remainder as 0.The second queue stores digits which on dividing by 3 gives remainder as 1. The third queue stores digits which on dividing by 3 gives remainder as 2. Call them as queue0, queue1 and queue2
Find the sum of all the digits.
Three cases arise: ......4.1 The sum of digits is divisible by 3. Dequeue all the digits from the three queues. Sort them in non-increasing order. Output the array.......4.2 The sum of digits produces remainder 1 when divided by 3. Remove one item from queue1. If queue1 is empty, remove two items from queue2. If queue2 contains less than two items, the number is not possible.......4.3 The sum of digits produces remainder 2 when divided by 3. Remove one item from queue2. If queue2 is empty, remove two items from queue1. If queue1 contains less than two items, the number is not possible.
......4.1 The sum of digits is divisible by 3. Dequeue all the digits from the three queues. Sort them in non-increasing order. Output the array.
......4.2 The sum of digits produces remainder 1 when divided by 3. Remove one item from queue1. If queue1 is empty, remove two items from queue2. If queue2 contains less than two items, the number is not possible.
......4.3 The sum of digits produces remainder 2 when divided by 3. Remove one item from queue2. If queue2 is empty, remove two items from queue1. If queue1 contains less than two items, the number is not possible.
Finally empty all the queues into an auxiliary array. Sort the auxiliary array in non-increasing order. Output the auxiliary array.
The below code works only if the input arrays has numbers from 0 to 9. It can be easily extended for any positive integer array. We just have to modify the part where we sort the array in decreasing order, at the end of code.  
C++
C
Java
Python3
C#
Javascript
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std; // This function puts all elements of 3 queues in the auxiliary arrayvoid populateAux(int aux[], queue<int> queue0, queue<int> queue1,                 queue<int> queue2, int* top){    // Put all items of first queue in aux[]    while (!queue0.empty()) {        aux[(*top)++] = queue0.front();        queue0.pop();    }     // Put all items of second queue in aux[]    while (!queue1.empty()) {        aux[(*top)++] = queue1.front();        queue1.pop();    }     // Put all items of third queue in aux[]    while (!queue2.empty()) {        aux[(*top)++] = queue2.front();        queue2.pop();    }} // The main function that finds the largest possible multiple of// 3 that can be formed by arr[] elementsint findMaxMultupleOf3(int arr[], int size){    // Step 1: sort the array in non-decreasing order    sort(arr, arr + size);     // Create 3 queues to store numbers with remainder 0, 1    // and 2 respectively    queue<int> queue0, queue1, queue2;     // Step 2 and 3 get the sum of numbers and place them in    // corresponding queues    int i, sum;    for (i = 0, sum = 0; i < size; ++i) {        sum += arr[i];        if ((arr[i] % 3) == 0)            queue0.push(arr[i]);        else if ((arr[i] % 3) == 1)            queue1.push(arr[i]);        else            queue2.push(arr[i]);    }     // Step 4.2: The sum produces remainder 1    if ((sum % 3) == 1) {        // either remove one item from queue1        if (!queue1.empty())            queue1.pop();         // or remove two items from queue2        else {            if (!queue2.empty())                queue2.pop();            else                return 0;             if (!queue2.empty())                queue2.pop();            else                return 0;        }    }     // Step 4.3: The sum produces remainder 2    else if ((sum % 3) == 2) {        // either remove one item from queue2        if (!queue2.empty())            queue2.pop();         // or remove two items from queue1        else {            if (!queue1.empty())                queue1.pop();            else                return 0;             if (!queue1.empty())                queue1.pop();            else                return 0;        }    }     int aux[size], top = 0;     // Empty all the queues into an auxiliary array.    populateAux(aux, queue0, queue1, queue2, &top);     // sort the array in non-increasing order    sort(aux, aux + top, greater<int>());     // print the result    for (int i = 0; i < top; ++i)        cout << aux[i] << " ";     return top;}int main(){     int arr[] = { 8, 1, 7, 6, 0 };    int size = sizeof(arr) / sizeof(arr[0]);     if (findMaxMultupleOf3(arr, size) == 0)        cout << "Not Possible";     return 0;}
/* A program to find the largest multiple of 3 from an array of elements */#include <stdio.h>#include <stdlib.h> // A queue nodetypedef struct Queue {    int front;    int rear;    int capacity;    int* array;} Queue; // A utility function to create a queue with given capacityQueue* createQueue(int capacity){    Queue* queue = (Queue*)malloc(sizeof(Queue));    queue->capacity = capacity;    queue->front = queue->rear = -1;    queue->array = (int*)malloc(queue->capacity * sizeof(int));    return queue;} // A utility function to check if queue is emptyint isEmpty(Queue* queue){    return queue->front == -1;} // A function to add an item to queuevoid Enqueue(Queue* queue, int item){    queue->array[++queue->rear] = item;    if (isEmpty(queue))        ++queue->front;} // A function to remove an item from queueint Dequeue(Queue* queue){    int item = queue->array[queue->front];    if (queue->front == queue->rear)        queue->front = queue->rear = -1;    else        queue->front++;     return item;} // A utility function to print array contentsvoid printArr(int* arr, int size){    int i;    for (i = 0; i < size; ++i)        printf("%d ", arr[i]);} /* Following two functions are needed for library function qsort().   Refer following link for help of qsort()   http:// www.cplusplus.com/reference/clibrary/cstdlib/qsort/ */int compareAsc(const void* a, const void* b){    return *(int*)a > *(int*)b;}int compareDesc(const void* a, const void* b){    return *(int*)a < *(int*)b;} // This function puts all elements of 3 queues in the auxiliary arrayvoid populateAux(int* aux, Queue* queue0, Queue* queue1,                 Queue* queue2, int* top){    // Put all items of first queue in aux[]    while (!isEmpty(queue0))        aux[(*top)++] = Dequeue(queue0);     // Put all items of second queue in aux[]    while (!isEmpty(queue1))        aux[(*top)++] = Dequeue(queue1);     // Put all items of third queue in aux[]    while (!isEmpty(queue2))        aux[(*top)++] = Dequeue(queue2);} // The main function that finds the largest possible multiple of// 3 that can be formed by arr[] elementsint findMaxMultupleOf3(int* arr, int size){    // Step 1: sort the array in non-decreasing order    qsort(arr, size, sizeof(int), compareAsc);     // Create 3 queues to store numbers with remainder 0, 1    // and 2 respectively    Queue* queue0 = createQueue(size);    Queue* queue1 = createQueue(size);    Queue* queue2 = createQueue(size);     // Step 2 and 3 get the sum of numbers and place them in    // corresponding queues    int i, sum;    for (i = 0, sum = 0; i < size; ++i) {        sum += arr[i];        if ((arr[i] % 3) == 0)            Enqueue(queue0, arr[i]);        else if ((arr[i] % 3) == 1)            Enqueue(queue1, arr[i]);        else            Enqueue(queue2, arr[i]);    }     // Step 4.2: The sum produces remainder 1    if ((sum % 3) == 1) {        // either remove one item from queue1        if (!isEmpty(queue1))            Dequeue(queue1);         // or remove two items from queue2        else {            if (!isEmpty(queue2))                Dequeue(queue2);            else                return 0;             if (!isEmpty(queue2))                Dequeue(queue2);            else                return 0;        }    }     // Step 4.3: The sum produces remainder 2    else if ((sum % 3) == 2) {        // either remove one item from queue2        if (!isEmpty(queue2))            Dequeue(queue2);         // or remove two items from queue1        else {            if (!isEmpty(queue1))                Dequeue(queue1);            else                return 0;             if (!isEmpty(queue1))                Dequeue(queue1);            else                return 0;        }    }     int aux[size], top = 0;     // Empty all the queues into an auxiliary array.    populateAux(aux, queue0, queue1, queue2, &top);     // sort the array in non-increasing order    qsort(aux, top, sizeof(int), compareDesc);     // print the result    printArr(aux, top);     return top;} // Driver program to test above functionsint main(){    int arr[] = { 8, 1, 7, 6, 0 };    int size = sizeof(arr) / sizeof(arr[0]);     if (findMaxMultupleOf3(arr, size) == 0)        printf("Not Possible");     return 0;}
/* Java program to find the largest multipleof 3 that can be formed from an arrayof elements */import java.util.Arrays;import java.util.Queue;import java.util.LinkedList;import java.util.Collections; public class Geeks {     // This function puts all elements of 3 queues in the    // array auxiliary    public static int populateAux(int aux[], Queue<Integer> queue0,                        Queue<Integer> queue1, Queue<Integer> queue2)    {        int top=0;        // Put all items of first queue in aux[]        while(!queue0.isEmpty())        {            aux[top++]=queue0.remove();        }         // Put all items of second queue in aux[]        while(!queue1.isEmpty())        {            aux[top++]=queue1.remove();        }         // Put all items of third queue in aux[]        while(!queue2.isEmpty())        {            aux[top++]=queue2.remove();        }         //Return number of integer added to aux[]        return top;    }     // The main function that finds the largest possible multiple of    // 3 that can be formed by arr[] elements    public static boolean findMaxMultupleOf3(int arr[])    {        // Step 1: sort the array in non-decreasing order        Arrays.sort(arr);         // Create 3 queues to store numbers with remainder 0, 1        // and 2 respectively        Queue<Integer> queue0=new LinkedList<>();        Queue<Integer> queue1=new LinkedList<>();        Queue<Integer> queue2=new LinkedList<>();         // Step 2 and 3 get the sum of numbers and place them in        // corresponding queues        int sum=0;        for (int i = 0; i < arr.length; ++i)        {            sum += arr[i];            if ((arr[i] % 3) == 0)                queue0.add(arr[i]);            else if ((arr[i] % 3) == 1)                queue1.add(arr[i]);            else                queue2.add(arr[i]);        }         // Step 4.2: The sum produces remainder 1        if ((sum % 3) == 1)        {            // either remove one item from queue1            if (!queue1.isEmpty())                queue1.remove();             // or remove two items from queue2            else            {                if (!queue2.isEmpty())                    queue2.remove();                else                    return false;                 if (!queue2.isEmpty())                    queue2.remove();                else                    return false;            }        }        // Step 4.3: The sum produces remainder 2        else if ((sum % 3) == 2)        {            // either remove one item from queue2            if (!queue2.isEmpty())                queue2.remove();            // or remove two items from queue1            else            {                if (!queue1.isEmpty())                    queue1.remove();                else                    return false;                 if (!queue1.isEmpty())                    queue1.remove();                else                    return false;            }        }                 int aux[]=new int[arr.length];        // Empty all the queues into an auxiliary array        // and get the number of integers added to aux[]        int top=populateAux(aux,queue0,queue1,queue2);         // sort the array in non-increasing order        Arrays.sort(aux,0,top);         // print the result        for (int i = top-1; i>=0; i--)            System.out.print(aux[i]+" ") ;         return true;    }     public static void main(String args[])    {        int arr[] = { 8, 1, 7, 6, 0 };        if (!findMaxMultupleOf3(arr))        System.out.println("Not possible") ;    }} // This code is contributed by Gaurav Tiwari
# Python3 program to find the largest multiple# of 3 that can be formed from an array# of elements aux = [] # This function puts all elements of 3 queues in the# array auxiliarydef populateAux(queue0, queue1, queue2):    global aux    top = 0     # Put all items of first queue in aux[]    while(len(queue0) > 0):        aux[top] = queue0.pop(0)        top += 1     # Put all items of second queue in aux[]    while(len(queue1) != 0):        aux[top] = queue1.pop(0)        top += 1     # Put all items of third queue in aux[]    while(len(queue2) != 0):        aux[top] = queue2.pop(0)        top += 1     # Return number of integer added to aux[]    return top  # The main function that finds the largest possible multiple of# 3 that can be formed by arr[] elementsdef findMaxMultupleOf3(arr):    global aux    # Step 1: sort the array in non-decreasing order    arr.sort()     # Create 3 queues to store numbers with remainder 0, 1    # and 2 respectively    queue0 = []    queue1 = []    queue2 = []     # Step 2 and 3 get the sum of numbers and place them in    # corresponding queues    sum = 0    for i in range(len(arr)):         sum += arr[i]        if ((arr[i] % 3) == 0):            queue0.append(arr[i])        elif ((arr[i] % 3) == 1):            queue1.append(arr[i])        else:            queue2.append(arr[i])     # Step 4.2: The sum produces remainder 1    if ((sum % 3) == 1):         # either remove one item from queue1        if (len(queue1) != 0):            queue1.pop(0)         # or remove two items from queue2        else:             if (len(queue2) != 0):                queue2.pop(0)            else:                return False             if (len(queue2) != 0):                queue2.pop(0)            else:                return False     # Step 4.3: The sum produces remainder 2    elif ((sum % 3) == 2):         # either remove one item from queue2        if (len(queue2) != 0):            queue2.pop(0)        # or remove two items from queue1        else:             if (len(queue1) != 0):                queue1.pop(0)            else:                return False             if (len(queue1) != 0):                queue1.pop(0)            else:                return False     aux = [0 for i in range(len(arr))]    # Empty all the queues into an auxiliary array    # and get the number of integers added to aux[]    top = populateAux(queue0, queue1, queue2)     # sort the array in non-increasing order    aux.sort(reverse=True)     # print the result    for i in range(0, top):        print(aux[i], end=" ")     return True # Driver codearr = [8, 1, 7, 6, 0]if (not findMaxMultupleOf3(arr)):    print("Not possible")  # This code is contributed by phasing17
/* C# program to find the largest multipleof 3 that can be formed from an arrayof elements */using System;using System.Collections.Generic; class Geeks{     // This function puts all elements of 3 queues in the    // array auxiliary    public static int populateAux(int []aux, Queue<int> queue0,                        Queue<int> queue1, Queue<int> queue2)    {        int top = 0;        // Put all items of first queue in aux[]        while(queue0.Count != 0)        {            aux[top++] = queue0.Dequeue();        }         // Put all items of second queue in aux[]        while(queue1.Count != 0)        {            aux[top++] = queue1.Dequeue();        }         // Put all items of third queue in aux[]        while(queue2.Count != 0)        {            aux[top++] = queue2.Dequeue();        }         //Return number of integer added to aux[]        return top;    }     // The main function that finds the largest possible multiple of    // 3 that can be formed by arr[] elements    public static bool findMaxMultupleOf3(int []arr)    {        // Step 1: sort the array in non-decreasing order        Array.Sort(arr);         // Create 3 queues to store numbers with remainder 0, 1        // and 2 respectively        Queue<int> queue0 = new Queue<int>();        Queue<int> queue1 = new Queue<int>();        Queue<int> queue2 = new Queue<int>();         // Step 2 and 3 get the sum of numbers and place them in        // corresponding queues        int sum=0;        for (int i = 0; i < arr.Length; ++i)        {            sum += arr[i];            if ((arr[i] % 3) == 0)                queue0.Enqueue(arr[i]);            else if ((arr[i] % 3) == 1)                queue1.Enqueue(arr[i]);            else                queue2.Enqueue(arr[i]);        }         // Step 4.2: The sum produces remainder 1        if ((sum % 3) == 1)        {            // either remove one item from queue1            if (queue1.Count != 0)                queue1.Dequeue();             // or remove two items from queue2            else            {                if (queue2.Count != 0)                    queue2.Dequeue();                else                    return false;                 if (queue2.Count != 0)                    queue2.Dequeue();                else                    return false;            }        }        // Step 4.3: The sum produces remainder 2        else if ((sum % 3) == 2)        {            // either remove one item from queue2            if (queue2.Count != 0)                queue2.Dequeue();            // or remove two items from queue1            else            {                if (queue1.Count != 0)                    queue1.Dequeue();                else                    return false;                 if (queue1.Count != 0)                    queue1.Dequeue();                else                    return false;            }        }                 int []aux = new int[arr.Length];                 // Empty all the queues into an auxiliary array        // and get the number of integers added to aux[]        int top = populateAux(aux,queue0,queue1,queue2);         // sort the array in non-increasing order        Array.Sort(aux,0,top);         // print the result        for (int i = top-1; i >= 0; i--)            Console.Write(aux[i]+" ") ;         return true;    }     // Driver code    public static void Main()    {        int []arr = { 8, 1, 7, 6, 0 };        if (!findMaxMultupleOf3(arr))        Console.WriteLine("Not possible") ;    }} /* This code contributed by PrinciRaj1992 */
<script> /* Javascript program to find the largest multipleof 3 that can be formed from an arrayof elements */ // This function puts all elements of 3 queues in the    // array auxiliaryfunction populateAux(aux, queue0, queue1, queue2){    let top = 0;             // Put all items of first queue in aux[]        while(queue0.length != 0)        {            aux[top++] = queue0.shift();        }           // Put all items of second queue in aux[]        while(queue1.length != 0)        {            aux[top++] = queue1.shift();        }           // Put all items of third queue in aux[]        while(queue2.length != 0)        {            aux[top++] = queue2.shift();        }           //Return number of integer added to aux[]        return top;} // The main function that finds the largest possible multiple of    // 3 that can be formed by arr[] elementsfunction findMaxMultupleOf3(arr){    // Step 1: sort the array in non-decreasing order        arr.sort(function(a, b){return a - b;});           // Create 3 queues to store numbers with remainder 0, 1        // and 2 respectively        let queue0 = [];        let queue1 = [];        let queue2 = [];           // Step 2 and 3 get the sum of numbers and place them in        // corresponding queues        let sum = 0;        for (let i = 0; i < arr.length; ++i)        {            sum += arr[i];            if ((arr[i] % 3) == 0)                queue0.push(arr[i]);            else if ((arr[i] % 3) == 1)                queue1.push(arr[i]);            else                queue2.push(arr[i]);        }           // Step 4.2: The sum produces remainder 1        if ((sum % 3) == 1)        {            // either remove one item from queue1            if (queue1.length != 0)                queue1.shift();               // or remove two items from queue2            else            {                if (queue2.length != 0)                    queue2.shift();                else                    return false;                   if (queue2.length != 0)                    queue2.shift();                else                    return false;            }        }        // Step 4.3: The sum produces remainder 2        else if ((sum % 3) == 2)        {            // either remove one item from queue2            if (queue2.length != 0)                queue2.shift();            // or remove two items from queue1            else            {                if (queue1.length != 0)                    queue1.shift();                else                    return false;                   if (queue1.length != 0)                    queue1.shift();                else                    return false;            }        }                   let aux=new Array(arr.length);        // Empty all the queues into an auxiliary array        // and get the number of integers added to aux[]        let top=populateAux(aux, queue0, queue1, queue2);           // sort the array in non-increasing order        aux.sort(function(a, b){return a - b;});           // print the result        for (let i = top - 1; i >= 0; i--)            document.write(aux[i]+" ") ;           return true;} // Driver codelet arr = [ 8, 1, 7, 6, 0 ];if (!findMaxMultupleOf3(arr))        document.write("Not possible") ; // This code is contributed by rag2127.</script>
8 7 6 0
 
The above method can be optimized in following ways. 
We can use Heap Sort or Merge Sort to make the time complexity O(nLogn). We can avoid extra space for queues. We know at most two items will be removed from the input array. So we can keep track of two items in two variables. At the end, instead of sorting the array again in descending order, we can print the ascending sorted array in reverse order. While printing in reverse order, we can skip the two elements to be removed.
We can use Heap Sort or Merge Sort to make the time complexity O(nLogn). 
We can avoid extra space for queues. We know at most two items will be removed from the input array. So we can keep track of two items in two variables. 
At the end, instead of sorting the array again in descending order, we can print the ascending sorted array in reverse order. While printing in reverse order, we can skip the two elements to be removed.
Time Complexity: O(nLogn), assuming a O(nLogn) algorithm is used for sorting. 
Find the largest multiple of 3 | Set 2 (In O(n) time and O(1) space)
_Gaurav_Tiwari
princiraj1992
avanitrachhadiya2155
hardikkoriintern
phasing17
cpp-queue
divisibility
Queue
Queue
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