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Find largest multiple of 3 in array of digits

Here given code implementation process.

/*
    C Program 
    Find largest multiple of 3 in array of digits
*/
#include <stdio.h>

// Print result
void printResult(int result[])
{
    int i = 0;
    int j = 0;

    for (j = 9; j >= 0; --j)
    {
        i = result[j];

        while(i > 0)
        {
            printf("%d",j);
            i--;
        }
    }
    printf("\n");
}
// Handles the request of find the largest multiple of 3
void multipleOfThree(int data[],int n)
{

    int result[10];
    int sum = 0;
    int i = 0;

    // Set initial count digit
    for (i = 0; i < 10; ++i)
    {
        result[i] = 0;
    }

    for (i = 0; i < n; ++i)
    {
        if(data[i] > 9 || data[i] < 0)
        {
            // When array not contain valid digit
            return ;
        }
        sum += data[i];
        result[data[i]]++;
    }

   
    if(sum % 3 == 0)
    {
        // All elements of array is part of result
        printResult(result);
        return;
    }
    else if(sum % 3 == 1)
    {
        // When need to remove single digit which is divisible by 3 and remainder is 1
        
        for (i = 1; i < 10; ++i)
        {
            if(result[i] > 0 && (result[i] % 3) == 1 )
            {
                // Remove smallest
                result[i]--;
                printResult(result);
                return;
            }
        }

    }
    else
    {
        // When remender is 2

        int p1 = -1;
        int p2 = -1;

        for (i = 1; i < 10; ++i)
        {
            if( result[i] > 0 && i % 3 == 1 )
            {
               
                if(p1 == -1)
                {
                    p1 = i;
                }
                else if(p2==-1)
                {
                    p2 = i;
                }
            }
            else if(result[i] > 0 && i % 3 == 2)
            {
                // We get a single digit which reminder is 2 when it's divided by 3
                result[i]--;
                printResult(result);
                return;
            }
        }

        if(p1 != -1 && p2 != -1)
        {
            // Remove 2 smallest
            result[p1]--;

            if(result[p1] > 0)
            {
                // When p1 are exist
               result[p1]--; 
            }
            else
            {
                result[p2]--; 
            }
           
            printResult(result);
            return;
        }
    }
    // When array sum is not multiple of 3
    printf("None\n");
}

int main(int argc, char const *argv[])
{ 
    // Define arrays of positive digits
    int data1[] = {7,0,1,2,8,1,1,6} ; 
    int data2[] = {1,1,0,0,0} ; 
    int data3[] = {9, 2, 4, 8, 3, 0, 0, 0, 0, 1, 4,3,2,3,2,3,6,2,4,6,5,2,3,5,2,4,1,0,0,0,0,0} ; 

    // Get the size of data1
    int n = sizeof(data1)/sizeof(data1[0]); 

    multipleOfThree(data1,n);

    // Get the size of data2
    n = sizeof(data2)/sizeof(data2[0]); 
    
    multipleOfThree(data2,n);

    // Get the size of data3
    n = sizeof(data3)/sizeof(data3[0]); 
    
    multipleOfThree(data3,n);

    return 0;
}

Output

8761110
None
98665544443333322222211000000000
/*
  Java Program for
  Find largest multiple of 3 in array of digits
*/
class Multiplier
{
    // Print result
    public void printResult(int[] result)
    {
        int i = 0;
        int j = 0;
        for (j = 9; j >= 0; --j)
        {
            i = result[j];
            while (i > 0)
            {
                System.out.print(j);
                i--;
            }
        }
        System.out.print("\n");
    }
    // Handles the request of find the largest multiple of 3
    public void multipleOfThree(int[] data, int n)
    {
        int[] result = new int[10];
        int sum = 0;
        int i = 0;
        // Set initial count digit
        for (i = 0; i < 10; ++i)
        {
            result[i] = 0;
        }
        for (i = 0; i < n; ++i)
        {
            if (data[i] > 9 || data[i] < 0)
            {
                // When array not contain valid digit
                return;
            }
            sum += data[i];
            result[data[i]]++;
        }
        if (sum % 3 == 0)
        {
            // All elements of array is part of result
            printResult(result);
            return;
        }
        else if (sum % 3 == 1)
        {
            // When need to remove single digit which is divisible by 3 and remainder is 1
            for (i = 1; i < 10; ++i)
            {
                if (result[i] > 0 && (result[i] % 3) == 1)
                {
                    // Remove smallest
                    result[i]--;
                    printResult(result);
                    return;
                }
            }
        }
        else
        {
            // When remender is 2
            int p1 = -1;
            int p2 = -1;
            for (i = 1; i < 10; ++i)
            {
                if (result[i] > 0 && i % 3 == 1)
                {
                    if (p1 == -1)
                    {
                        p1 = i;
                    }
                    else if (p2 == -1)
                    {
                        p2 = i;
                    }
                }
                else if (result[i] > 0 && i % 3 == 2)
                {
                    // We get a single digit which reminder is 2 when it's divided by 3
                    result[i]--;
                    printResult(result);
                    return;
                }
            }
            if (p1 != -1 && p2 != -1)
            {
                // Remove 2 smallest
                result[p1]--;
                if (result[p1] > 0)
                {
                    // When p1 are exist
                    result[p1]--;
                }
                else
                {
                    result[p2]--;
                }
                printResult(result);
                return;
            }
        }
        // When array sum is not multiple of 3
        System.out.print("None\n");
    }
    public static void main(String[] args)
    {
        Multiplier task = new Multiplier();
        // Define arrays of positive digits
        int[] data1 = {
            7 , 0 , 1 , 2 , 8 , 1 , 1 , 6
        };
        int[] data2 = {
            1 , 1 , 0 , 0 , 0
        };
        int[] data3 = {
            9 , 2 , 4 , 8 , 3 , 0 , 0 , 0 , 0 , 1 , 4 , 3 , 2 , 3 , 2 , 
            3 , 6 , 2 , 4 , 6 , 5 , 2 , 3 , 5 , 2 , 4 , 1 , 0 , 0 , 0 , 0 , 0
        };
        // Get the size of data1
        int n = data1.length;
        task.multipleOfThree(data1, n);
        // Get the size of data2
        n = data2.length;
        task.multipleOfThree(data2, n);
        // Get the size of data3
        n = data3.length;
        task.multipleOfThree(data3, n);
    }
}

Output

8761110
None
98665544443333322222211000000000
// Include header file
#include <iostream>

using namespace std;
/*
  C++ Program for
  Find largest multiple of 3 in array of digits
*/
class Multiplier
{
	public:
		// Print result
		void printResult(int result[])
		{
			int i = 0;
			int j = 0;
			for (j = 9; j >= 0; --j)
			{
				i = result[j];
				while (i > 0)
				{
					cout << j;
					i--;
				}
			}
			cout << "\n";
		}
	// Handles the request of find the largest multiple of 3
	void multipleOfThree(int data[], int n)
	{
		int result[10];
		int sum = 0;
		int i = 0;
		// Set initial count digit
		for (i = 0; i < 10; ++i)
		{
			result[i] = 0;
		}
		for (i = 0; i < n; ++i)
		{
			// When array not contain valid digit
			if (data[i] > 9 || data[i] < 0)
			{
				return;
			}
			sum += data[i];
			result[data[i]]++;
		}
		if (sum % 3 == 0)
		{
			// All elements of array is part of result
			this->printResult(result);
			return;
		}
		else if (sum % 3 == 1)
		{
			// When need to remove single digit which is divisible by 3 and remainder is 1
			for (i = 1; i < 10; ++i)
			{
				if (result[i] > 0 && (result[i] % 3) == 1)
				{
					// Remove smallest
					result[i]--;
					this->printResult(result);
					return;
				}
			}
		}
		else
		{
			// When remender is 2
			int p1 = -1;
			int p2 = -1;
			for (i = 1; i < 10; ++i)
			{
				if (result[i] > 0 && i % 3 == 1)
				{
					if (p1 == -1)
					{
						p1 = i;
					}
					else if (p2 == -1)
					{
						p2 = i;
					}
				}
				else if (result[i] > 0 && i % 3 == 2)
				{
					// We get a single digit which reminder is 2 when it's divided by 3
					result[i]--;
					this->printResult(result);
					return;
				}
			}
			if (p1 != -1 && p2 != -1)
			{
				// Remove 2 smallest
				result[p1]--;
				if (result[p1] > 0)
				{
					// When p1 are exist
					result[p1]--;
				}
				else
				{
					result[p2]--;
				}
				this->printResult(result);
				return;
			}
		}
		// When array sum is not multiple of 3
		cout << "None\n";
	}
};
int main()
{
	Multiplier task = Multiplier();
	// Define arrays of positive digits
	int data1[] = {
		7 , 0 , 1 , 2 , 8 , 1 , 1 , 6
	};
	int data2[] = {
		1 , 1 , 0 , 0 , 0
	};
	int data3[] = {
		9 , 2 , 4 , 8 , 3 , 0 , 0 , 0 , 0 , 1 , 4 , 3 , 2 , 3 , 2 , 3 , 
        6 , 2 , 4 , 6 , 5 , 2 , 3 , 5 , 2 , 4 , 1 , 0 , 0 , 0 , 0 , 0
	};
	// Get the size of data1
	int n = sizeof(data1) / sizeof(data1[0]);
	task.multipleOfThree(data1, n);
	// Get the size of data2
	n = sizeof(data2) / sizeof(data2[0]);
	task.multipleOfThree(data2, n);
	// Get the size of data3
	n = sizeof(data3) / sizeof(data3[0]);
	task.multipleOfThree(data3, n);
	return 0;
}

Output

8761110
None
98665544443333322222211000000000
// Include namespace system
using System;
/*
  C# Program for
  Find largest multiple of 3 in array of digits
*/
public class Multiplier
{
	// Print result
	public void printResult(int[] result)
	{
		int i = 0;
		int j = 0;
		for (j = 9; j >= 0; --j)
		{
			i = result[j];
			while (i > 0)
			{
				Console.Write(j);
				i--;
			}
		}
		Console.Write("\n");
	}
	// Handles the request of find the largest multiple of 3
	public void multipleOfThree(int[] data, int n)
	{
		int[] result = new int[10];
		int sum = 0;
		int i = 0;
		// Set initial count digit
		for (i = 0; i < 10; ++i)
		{
			result[i] = 0;
		}
		for (i = 0; i < n; ++i)
		{
			// When array not contain valid digit
			if (data[i] > 9 || data[i] < 0)
			{
				return;
			}
			sum += data[i];
			result[data[i]]++;
		}
		if (sum % 3 == 0)
		{
			// All elements of array is part of result
			printResult(result);
			return;
		}
		else if (sum % 3 == 1)
		{
			// When need to remove single digit which is divisible by 3 and remainder is 1
			for (i = 1; i < 10; ++i)
			{
				if (result[i] > 0 && (result[i] % 3) == 1)
				{
					// Remove smallest
					result[i]--;
					printResult(result);
					return;
				}
			}
		}
		else
		{
			// When remender is 2
			int p1 = -1;
			int p2 = -1;
			for (i = 1; i < 10; ++i)
			{
				if (result[i] > 0 && i % 3 == 1)
				{
					if (p1 == -1)
					{
						p1 = i;
					}
					else if (p2 == -1)
					{
						p2 = i;
					}
				}
				else if (result[i] > 0 && i % 3 == 2)
				{
					// We get a single digit which reminder is 2 when it's divided by 3
					result[i]--;
					printResult(result);
					return;
				}
			}
			if (p1 != -1 && p2 != -1)
			{
				// Remove 2 smallest
				result[p1]--;
				if (result[p1] > 0)
				{
					// When p1 are exist
					result[p1]--;
				}
				else
				{
					result[p2]--;
				}
				printResult(result);
				return;
			}
		}
		// When array sum is not multiple of 3
		Console.Write("None\n");
	}
	public static void Main(String[] args)
	{
		Multiplier task = new Multiplier();
		// Define arrays of positive digits
		int[] data1 = {
			7 , 0 , 1 , 2 , 8 , 1 , 1 , 6
		};
		int[] data2 = {
			1 , 1 , 0 , 0 , 0
		};
		int[] data3 = {
			9 , 2 , 4 , 8 , 3 , 0 , 0 , 0 , 0 , 1 , 4 , 3 , 2 , 3 , 2 , 3 , 6 , 
            2 , 4 , 6 , 5 , 2 , 3 , 5 , 2 , 4 , 1 , 0 , 0 , 0 , 0 , 0
		};
		// Get the size of data1
		int n = data1.Length;
		task.multipleOfThree(data1, n);
		// Get the size of data2
		n = data2.Length;
		task.multipleOfThree(data2, n);
		// Get the size of data3
		n = data3.Length;
		task.multipleOfThree(data3, n);
	}
}

Output

8761110
None
98665544443333322222211000000000
<?php
/*
  Php Program for
  Find largest multiple of 3 in array of digits
*/
class Multiplier
{
	// Print result
	public	function printResult( & $result)
	{
		$i = 0;
		$j = 0;
		for ($j = 9; $j >= 0; --$j)
		{
			$i = $result[$j];
			while ($i > 0)
			{
				echo $j;
				$i--;
			}
		}
		echo "\n";
	}
	// Handles the request of find the largest multiple of 3
	public	function multipleOfThree( & $data, $n)
	{
		$result = array_fill(0, 10, 0);
		$sum = 0;
		$i = 0;
		for ($i = 0; $i < $n; ++$i)
		{
			// When array not contain valid digit
			if ($data[$i] > 9 || $data[$i] < 0)
			{
				return;
			}
			$sum += $data[$i];
			$result[$data[$i]]++;
		}
		if ($sum % 3 == 0)
		{
			// All elements of array is part of result
			$this->printResult($result);
			return;
		}
		else if ($sum % 3 == 1)
		{
			// When need to remove single digit which is divisible by 3 and remainder is 1
			for ($i = 1; $i < 10; ++$i)
			{
				if ($result[$i] > 0 && ($result[$i] % 3) == 1)
				{
					// Remove smallest
					$result[$i]--;
					$this->printResult($result);
					return;
				}
			}
		}
		else
		{
			// When remender is 2
			$p1 = -1;
			$p2 = -1;
			for ($i = 1; $i < 10; ++$i)
			{
				if ($result[$i] > 0 && $i % 3 == 1)
				{
					if ($p1 == -1)
					{
						$p1 = $i;
					}
					else if ($p2 == -1)
					{
						$p2 = $i;
					}
				}
				else if ($result[$i] > 0 && $i % 3 == 2)
				{
					// We get a single digit which reminder is 2 when it's divided by 3
					$result[$i]--;
					$this->printResult($result);
					return;
				}
			}
			if ($p1 != -1 && $p2 != -1)
			{
				// Remove 2 smallest
				$result[$p1]--;
				if ($result[$p1] > 0)
				{
					// When p1 are exist
					$result[$p1]--;
				}
				else
				{
					$result[$p2]--;
				}
				$this->printResult($result);
				return;
			}
		}
		// When array sum is not multiple of 3
		echo "None\n";
	}
}

function main()
{
	$task = new Multiplier();
	// Define arrays of positive digits
	$data1 = array(7, 0, 1, 2, 8, 1, 1, 6);
	$data2 = array(1, 1, 0, 0, 0);
	$data3 = array(9, 2, 4, 8, 3, 0, 0, 0, 0, 1, 4, 3, 2, 3, 2, 3, 
                   6, 2, 4, 6, 5, 2, 3, 5, 2, 4, 1, 0, 0, 0, 0, 0);
	// Get the size of data1
	$n = count($data1);
	$task->multipleOfThree($data1, $n);
	// Get the size of data2
	$n = count($data2);
	$task->multipleOfThree($data2, $n);
	// Get the size of data3
	$n = count($data3);
	$task->multipleOfThree($data3, $n);
}
main();

Output

8761110
None
98665544443333322222211000000000
/*
  Node Js Program for
  Find largest multiple of 3 in array of digits
*/
class Multiplier
{
	// Print result
	printResult(result)
	{
		var i = 0;
		var j = 0;
		for (j = 9; j >= 0; --j)
		{
			i = result[j];
			while (i > 0)
			{
				process.stdout.write(""+j);
				i--;
			}
		}
		process.stdout.write("\n");
	}
	// Handles the request of find the largest multiple of 3
	multipleOfThree(data, n)
	{
		var result = Array(10).fill(0);
		var sum = 0;
		var i = 0;
		for (i = 0; i < n; ++i)
		{
			// When array not contain valid digit
			if (data[i] > 9 || data[i] < 0)
			{
				return;
			}
			sum += data[i];
			result[data[i]]++;
		}
		if (sum % 3 == 0)
		{
			// All elements of array is part of result
			this.printResult(result);
			return;
		}
		else if (sum % 3 == 1)
		{
			// When need to remove single digit which is divisible by 3 and remainder is 1
			for (i = 1; i < 10; ++i)
			{
				if (result[i] > 0 && (result[i] % 3) == 1)
				{
					// Remove smallest
					result[i]--;
					this.printResult(result);
					return;
				}
			}
		}
		else
		{
			// When remender is 2
			var p1 = -1;
			var p2 = -1;
			for (i = 1; i < 10; ++i)
			{
				if (result[i] > 0 && i % 3 == 1)
				{
					if (p1 == -1)
					{
						p1 = i;
					}
					else if (p2 == -1)
					{
						p2 = i;
					}
				}
				else if (result[i] > 0 && i % 3 == 2)
				{
					// We get a single digit which reminder is 2 when it's divided by 3
					result[i]--;
					this.printResult(result);
					return;
				}
			}
			if (p1 != -1 && p2 != -1)
			{
				// Remove 2 smallest
				result[p1]--;
				if (result[p1] > 0)
				{
					// When p1 are exist
					result[p1]--;
				}
				else
				{
					result[p2]--;
				}
				this.printResult(result);
				return;
			}
		}
		// When array sum is not multiple of 3
		process.stdout.write("None\n");
	}
}

function main()
{
	var task = new Multiplier();
	// Define arrays of positive digits
	var data1 = [7, 0, 1, 2, 8, 1, 1, 6];
	var data2 = [1, 1, 0, 0, 0];
	var data3 = [9, 2, 4, 8, 3, 0, 0, 0, 0, 1, 4, 3, 2, 3, 2, 3, 6, 2, 
                 4, 6, 5, 2, 3, 5, 2, 4, 1, 0, 0, 0, 0, 0];
	// Get the size of data1
	var n = data1.length;
	task.multipleOfThree(data1, n);
	// Get the size of data2
	n = data2.length;
	task.multipleOfThree(data2, n);
	// Get the size of data3
	n = data3.length;
	task.multipleOfThree(data3, n);
}
main();

Output

8761110
None
98665544443333322222211000000000
#   Python 3 Program for
#   Find largest multiple of 3 in array of digits

class Multiplier :
	#  Print result
	def printResult(self, result) :
		i = 0
		j = 9
		while (j >= 0) :
			i = result[j]
			while (i > 0) :
				print(j, end = "")
				i -= 1
			
			j -= 1
		
		print(end = "\n")
	
	#  Handles the request of find the largest multiple of 3
	def multipleOfThree(self, data, n) :
		result = [0] * (10)
		sum = 0
		i = 0
		while (i < n) :
			#  When array not contain valid digit
			if (data[i] > 9 or data[i] < 0) :
				return
			
			sum += data[i]
			result[data[i]] += 1
			i += 1
		
		if (sum % 3 == 0) :
			#  All elements of array is part of result
			self.printResult(result)
			return
		
		elif(sum % 3 == 1) :
			#  When need to remove single digit which is divisible by 3 and remainder is 1
			i = 1
			while (i < 10) :
				if (result[i] > 0 and(result[i] % 3) == 1) :
					#  Remove smallest
					result[i] -= 1
					self.printResult(result)
					return
				
				i += 1
			
		else :
			#  When remender is 2
			p1 = -1
			p2 = -1
			i = 1
			while (i < 10) :
				if (result[i] > 0 and i % 3 == 1) :
					if (p1 == -1) :
						p1 = i
					
					elif(p2 == -1) :
						p2 = i
					
				
				elif(result[i] > 0 and i % 3 == 2) :
					#  We get a single digit which reminder is 2 when it's divided by 3
					result[i] -= 1
					self.printResult(result)
					return
				
				i += 1
			
			if (p1 != -1 and p2 != -1) :
				#  Remove 2 smallest
				result[p1] -= 1
				if (result[p1] > 0) :
					#  When p1 are exist
					result[p1] -= 1
				else :
					result[p2] -= 1
				
				self.printResult(result)
				return
			
		
		#  When array sum is not multiple of 3
		print("None")
	

def main() :
	task = Multiplier()
	#  Define arrays of positive digits
	data1 = [7, 0, 1, 2, 8, 1, 1, 6]
	data2 = [1, 1, 0, 0, 0]
	data3 = [9, 2, 4, 8, 3, 0, 0, 0, 0, 1, 4, 3, 2, 3, 2, 3, 6, 2, 4,
             6, 5, 2, 3, 5, 2, 4, 1, 0, 0, 0, 0, 0]
	#  Get the size of data1
	n = len(data1)
	task.multipleOfThree(data1, n)
	#  Get the size of data2
	n = len(data2)
	task.multipleOfThree(data2, n)
	#  Get the size of data3
	n = len(data3)
	task.multipleOfThree(data3, n)

if __name__ == "__main__": main()

Output

8761110
None
98665544443333322222211000000000
#   Ruby Program for
#   Find largest multiple of 3 in array of digits

class Multiplier 
	#  Print result
	def printResult(result) 
		i = 0
		j = 9
		while (j >= 0) 
			i = result[j]
			while (i > 0) 
				print(j)
				i -= 1
			end

			j -= 1
		end

		print("\n")
	end

	#  Handles the request of find the largest multiple of 3
	def multipleOfThree(data, n) 
		result = Array.new(10) {0}
		sum = 0
		i = 0
		while (i < n) 
			#  When array not contain valid digit
			if (data[i] > 9 || data[i] < 0) 
				return
			end

			sum += data[i]
			result[data[i]] += 1
			i += 1
		end

		if (sum % 3 == 0) 
			#  All elements of array is part of result
			self.printResult(result)
			return
		elsif(sum % 3 == 1) 
			#  When need to remove single digit which is divisible by 3 and remainder is 1
			i = 1
			while (i < 10) 
				if (result[i] > 0 && (result[i] % 3) == 1) 
					#  Remove smallest
					result[i] -= 1
					self.printResult(result)
					return
				end

				i += 1
			end

		else 
			#  When remender is 2
			p1 = -1
			p2 = -1
			i = 1
			while (i < 10) 
				if (result[i] > 0 && i % 3 == 1) 
					if (p1 == -1) 
						p1 = i
					elsif(p2 == -1) 
						p2 = i
					end

				elsif(result[i] > 0 && i % 3 == 2) 
					#  We get a single digit which reminder is 2 when it's divided by 3
					result[i] -= 1
					self.printResult(result)
					return
				end

				i += 1
			end

			if (p1 != -1 && p2 != -1) 
				#  Remove 2 smallest
				result[p1] -= 1
				if (result[p1] > 0) 
					#  When p1 are exist
					result[p1] -= 1
				else 
					result[p2] -= 1
				end

				self.printResult(result)
				return
			end

		end

		#  When array sum is not multiple of 3
		print("None\n")
	end

end

def main() 
	task = Multiplier.new()
	#  Define arrays of positive digits
	data1 = [7, 0, 1, 2, 8, 1, 1, 6]
	data2 = [1, 1, 0, 0, 0]
	data3 = [9, 2, 4, 8, 3, 0, 0, 0, 0, 1, 4, 3, 2, 3, 2, 3, 6, 2, 4, 
             6, 5, 2, 3, 5, 2, 4, 1, 0, 0, 0, 0, 0]
	#  Get the size of data1
	n = data1.length
	task.multipleOfThree(data1, n)
	#  Get the size of data2
	n = data2.length
	task.multipleOfThree(data2, n)
	#  Get the size of data3
	n = data3.length
	task.multipleOfThree(data3, n)
end

main()

Output

8761110
None
98665544443333322222211000000000

/*
  Scala Program for
  Find largest multiple of 3 in array of digits
*/
class Multiplier
{
	// Print result
	def printResult(result: Array[Int]): Unit = {
		var i: Int = 0;
		var j: Int = 9;
		while (j >= 0)
		{
			i = result(j);
			while (i > 0)
			{
				print(j);
				i -= 1;
			}
			j -= 1;
		}
		print("\n");
	}
	// Handles the request of find the largest multiple of 3
	def multipleOfThree(data: Array[Int], n: Int): Unit = {
		var result: Array[Int] = Array.fill[Int](10)(0);
		var sum: Int = 0;
		var i: Int = 0;
		while (i < n)
		{
			// When array not contain valid digit
			if (data(i) > 9 || data(i) < 0)
			{
				return;
			}
			sum += data(i);
			result(data(i)) += 1;
			i += 1;
		}
		if (sum % 3 == 0)
		{
			// All elements of array is part of result
			this.printResult(result);
			return;
		}
		else if (sum % 3 == 1)
		{
			// When need to remove single digit which is divisible by 3 and remainder is 1
			i = 1;
			while (i < 10)
			{
				if (result(i) > 0 && (result(i) % 3) == 1)
				{
					// Remove smallest
					result(i) -= 1;
					this.printResult(result);
					return;
				}
				i += 1;
			}
		}
		else
		{
			// When remender is 2
			var p1: Int = -1;
			var p2: Int = -1;
			i = 1;
			while (i < 10)
			{
				if (result(i) > 0 && i % 3 == 1)
				{
					if (p1 == -1)
					{
						p1 = i;
					}
					else if (p2 == -1)
					{
						p2 = i;
					}
				}
				else if (result(i) > 0 && i % 3 == 2)
				{
					// We get a single digit which reminder is 2 when it's divided by 3
					result(i) -= 1;
					this.printResult(result);
					return;
				}
				i += 1;
			}
			if (p1 != -1 && p2 != -1)
			{
				// Remove 2 smallest
				result(p1) -= 1;
				if (result(p1) > 0)
				{
					// When p1 are exist
					result(p1) -= 1;
				}
				else
				{
					result(p2) -= 1;
				}
				this.printResult(result);
				return;
			}
		}
		// When array sum is not multiple of 3
		print("None\n");
	}
}
object Main
{
	def main(args: Array[String]): Unit = {
		var task: Multiplier = new Multiplier();
		// Define arrays of positive digits
		var data1: Array[Int] = Array(7, 0, 1, 2, 8, 1, 1, 6);
		var data2: Array[Int] = Array(1, 1, 0, 0, 0);
		var data3: Array[Int] = Array(9, 2, 4, 8, 3, 0, 0, 0, 0, 1, 4, 3, 2, 3, 2, 
                                      3, 6, 2, 4, 6, 5, 2, 3, 5, 2, 4, 1, 0, 0, 0, 0, 0);
		// Get the size of data1
		var n: Int = data1.length;
		task.multipleOfThree(data1, n);
		// Get the size of data2
		n = data2.length;
		task.multipleOfThree(data2, n);
		// Get the size of data3
		n = data3.length;
		task.multipleOfThree(data3, n);
	}
}

Output

8761110
None
98665544443333322222211000000000
/*
  Swift 4 Program for
  Find largest multiple of 3 in array of digits
*/
class Multiplier
{
	// Print result
	func printResult(_ result: [Int])
	{
		var i: Int = 0;
		var j: Int = 9;
		while (j >= 0)
		{
			i = result[j];
			while (i > 0)
			{
				print(j, terminator: "");
				i -= 1;
			}
			j -= 1;
		}
		print(terminator: "\n");
	}
	// Handles the request of find the largest multiple of 3
	func multipleOfThree(_ data: [Int], _ n: Int)
	{
		var result: [Int] = Array(repeating: 0, count: 10);
		var sum: Int = 0;
		var i: Int = 0;
		while (i < n)
		{
			// When array not contain valid digit
			if (data[i] > 9 || data[i] < 0)
			{
				return;
			}
			sum += data[i];
			result[data[i]] +=   1;
			i += 1;
		}
		if (sum % 3 == 0)
		{
			// All elements of array is part of result
			self.printResult(result);
			return;
		}
		else if (sum % 3 == 1)
		{
			// When need to remove single digit which is divisible by 3 and remainder is 1
			i = 1;
			while (i < 10)
			{
				if (result[i] > 0 && (result[i] % 3) == 1)
				{
					// Remove smallest
					result[i] -= 1;
					self.printResult(result);
					return;
				}
				i += 1;
			}
		}
		else
		{
			// When remender is 2
			var p1: Int = -1;
			var p2: Int = -1;
			i = 1;
			while (i < 10)
			{
				if (result[i] > 0 && i % 3 == 1)
				{
					if (p1 == -1)
					{
						p1 = i;
					}
					else if (p2 == -1)
					{
						p2 = i;
					}
				}
				else if (result[i] > 0 && i % 3 == 2)
				{
					// We get a single digit which reminder is 2 when it"s divided by 3
					result[i] -= 1;
					self.printResult(result);
					return;
				}
				i += 1;
			}
			if (p1  != -1 && p2  != -1)
			{
				// Remove 2 smallest
				result[p1] -= 1;
				if (result[p1] > 0)
				{
					// When p1 are exist
					result[p1] -= 1;
				}
				else
				{
					result[p2] -= 1;
				}
				self.printResult(result);
				return;
			}
		}
		// When array sum is not multiple of 3
		print("None");
	}
}
func main()
{
	let task: Multiplier = Multiplier();
	// Define arrays of positive digits
	let data1: [Int] = [7, 0, 1, 2, 8, 1, 1, 6];
	let data2: [Int] = [1, 1, 0, 0, 0];
	let data3: [Int] = [9, 2, 4, 8, 3, 0, 0, 0, 0, 1, 4, 3, 2, 3, 2, 3, 6, 2, 
                        4, 6, 5, 2, 3, 5, 2, 4, 1, 0, 0, 0, 0, 0];
	// Get the size of data1
	var n: Int = data1.count;
	task.multipleOfThree(data1, n);
	// Get the size of data2
	n = data2.count;
	task.multipleOfThree(data2, n);
	// Get the size of data3
	n = data3.count;
	task.multipleOfThree(data3, n);
}
main();

Output

8761110
None
98665544443333322222211000000000
/*
  Kotlin Program for
  Find largest multiple of 3 in array of digits
*/
class Multiplier
{
	// Print result
	fun printResult(result: Array < Int > ): Unit
	{
		var i: Int ;
		var j: Int = 9;
		while (j >= 0)
		{
			i = result[j];
			while (i > 0)
			{
				print(j);
				i -= 1;
			}
			j -= 1;
		}
		print("\n");
	}
	// Handles the request of find the largest multiple of 3
	fun multipleOfThree(data: Array < Int > , n: Int): Unit
	{
		var result: Array < Int > = Array(10)
		{
			0
		};
		var sum: Int = 0;
		var i: Int = 0;
		while (i < n)
		{
			// When array not contain valid digit
			if (data[i] > 9 || data[i] < 0)
			{
				return;
			}
			sum += data[i];
			result[data[i]] += 1;
			i += 1;
		}
		if (sum % 3 == 0)
		{
			// All elements of array is part of result
			this.printResult(result);
			return;
		}
		else if (sum % 3 == 1)
		{
			// When need to remove single digit which is divisible by 3 and remainder is 1
			i = 1;
			while (i < 10)
			{
				if (result[i] > 0 && (result[i] % 3) == 1)
				{
					// Remove smallest
					result[i] -= 1;
					this.printResult(result);
					return;
				}
				i += 1;
			}
		}
		else
		{
			// When remender is 2
			var p1: Int = -1;
			var p2: Int = -1;
			i = 1;
			while (i < 10)
			{
				if (result[i] > 0 && i % 3 == 1)
				{
					if (p1 == -1)
					{
						p1 = i;
					}
					else if (p2 == -1)
					{
						p2 = i;
					}
				}
				else if (result[i] > 0 && i % 3 == 2)
				{
					// We get a single digit which reminder is 2 when it's divided by 3
					result[i] -= 1;
					this.printResult(result);
					return;
				}
				i += 1;
			}
			if (p1 != -1 && p2 != -1)
			{
				// Remove 2 smallest
				result[p1] -= 1;
				if (result[p1] > 0)
				{
					// When p1 are exist
					result[p1] -= 1;
				}
				else
				{
					result[p2] -= 1;
				}
				this.printResult(result);
				return;
			}
		}
		// When array sum is not multiple of 3
		print("None\n");
	}
}
fun main(args: Array <String> ): Unit
{
	var task: Multiplier = Multiplier();
	// Define arrays of positive digits
	var data1: Array <Int> = arrayOf(7, 0, 1, 2, 8, 1, 1, 6);
	var data2: Array <Int> = arrayOf(1, 1, 0, 0, 0);
	var data3: Array <Int> = arrayOf(9, 2, 4, 8, 3, 0, 0, 0, 0, 1, 4, 3, 2, 3, 2, 3, 
                                       6, 2, 4, 6, 5, 2, 3, 5, 2, 4, 1, 0, 0, 0, 0, 0);
	// Get the size of data1
	var n: Int = data1.count();
	task.multipleOfThree(data1, n);
	// Get the size of data2
	n = data2.count();
	task.multipleOfThree(data2, n);
	// Get the size of data3
	n = data3.count();
	task.multipleOfThree(data3, n);
}

Output

8761110
None
98665544443333322222211000000000
Last updated on by Kalkicode




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