What least number should be replaced for * so that number 372 * 0 is exactly divisible by 3?

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Given a very large number num (1 <= num <= 10^1000), print the number of digits that needs to be removed to make the number exactly divisible by 3. If it is not possible then print -1.

Examples : 

The idea is based on the fact that a number is multiple of 3 if and only if sum of its digits is multiple of 3 (See this for details).
One important observation used here is that the answer is at most 2 if an answer exists. So here are the only options for the function: 

1. Sum of digits is already equal to 0 modulo 3. Thus, we don’t have to erase any digits.
2. There exists such a digit that equals sum modulo 3. Then we just have to erase a single digit
3. All the digits are neither divisible by 3 nor equal to sum modulo 3. So two of such digits will sum up to number, which equals sum modulo 3, (2+2) mod 3=1, (1+1) mod 3=2

C++

#include

using namespace std;

int divisible(string num)

{

    int n = num.length();

    int sum = accumulate(begin(num),

                         end(num), 0) - '0' * 1;

    if (sum % 3 == 0)

        return 0;

    if (n == 1)

        return -1;

    for (int i = 0; i < n; i++)

        if (sum % 3 == (num[i] - '0') % 3)

            return 1;

    if (n == 2)

        return -1;

    return 2;

}

int main()

{

    string num = "1234";

    cout << divisible(num);

    return 0;

}

Java

import java.io.*;

class GFG {

    static int divisible(String num)

    {

        int n = num.length();

        int sum = 0;

        for (int i = 0; i < n; i++)

            sum += (int)(num.charAt(i));

        if (sum % 3 == 0)

            return 0;

        if (n == 1)

            return -1;

        for (int i = 0; i < n; i++)

            if (sum % 3 == (num.charAt(i) - '0') % 3)

                return 1;

        if (n == 2)

            return -1;

        return 2;

    }

    public static void main(String[] args)

    {

        String num = "1234";

        System.out.println(divisible(num));

    }

}

Python3

def divisible(num):

    n = len(num)

    sum_ = 0

    for i in range(n):

        sum_ += int(num[i])

    if (sum_ % 3 == 0):

        return 0

    if (n == 1):

        return -1

    for i in range(n):

        if (sum_ % 3 == int(num[i]) % 3):

            return 1

    if (n == 2):

        return -1

    return 2

if __name__ == '__main__':

    num = "1234"

    print(divisible(num))

C#

using System;

class GFG {

    static int divisible(String num)

    {

        int n = num.Length;

        int sum = 0;

        for (int i = 0; i < n; i++)

            sum += (int)(num[i]);

        if (sum % 3 == 0)

            return 0;

        if (n == 1)

            return -1;

        for (int i = 0; i < n; i++)

            if (sum % 3 == (num[i] - '0') % 3)

                return 1;

        if (n == 2)

            return -1;

        return 2;

    }

    public static void Main()

    {

        string num = "1234";

        Console.WriteLine(divisible(num));

    }

}

PHP

function divisible($num)

{

    $n = strlen($num);

    $sum = ($num); ($num); 0 - '0';

    if ($sum % 3 == 0)

        return 0;

    if ($n == 1)

        return -1;

    for ($i = 0; $i < $n; $i++)

        if ($sum % 3 == ($num[$i] - '0') % 3)

            return 1;        

    if ($n == 2)

        return -1;

    return 2;

}

$num = "1234";

echo divisible($num);

?>

Javascript

Output : 

Time Complexity: O(N), as we are using a loop to traverse N times.

Auxiliary Space: O(1), as we are not using any extra space.

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