Check divisibility in a binary stream

Stream of binary number is coming, the task is to tell the number formed so far is divisible by a given number n. At any given time, you will get 0 or 1 and tell whether the number formed with these bits is divisible by n or not. Generally, e-commerce companies ask this type of questions. It was asked me in Microsoft interview. Actually that question was a bit simple, interviewer fixed the n to 3.

Method 1 (Simple but causes overflow):Keep on calculating the number formed and just check divisibility by n. 

C

/* Simple implementation of the logic,
without error handling compiled with
Microsoft visual studio 2015 */
void CheckDivisibility2(int n)
{
    int num = 0;
    std::cout << "press any key other than"
                " 0 and 1 to terminate \n";
    while (true)
    {
        int incomingBit;

        // read next incoming bit via standard
        // input. 0, 00, 000.. are same as int 0
        // ans 1, 01, 001, 00..001 is same as 1.
        std::cin >> incomingBit;

        // Update value of num formed so far
        if (incomingBit == 1)
            num = (num * 2 + 1);
        else if (incomingBit == 0)
            num = (num * 2);

        else
            break;
        if (num % n == 0)
            std::cout << "yes \n";
        else
            std::cout << "no \n";
    }
}

Python3

def CheckDivisibility2(n):
    num = 0
    print("press any key other than 0 and 1 to terminate")
    while True:
        incomingBit = input()

        # convert incoming bit to integer
        incomingBit = int(incomingBit)

        # Update value of num formed so far
        if incomingBit == 1:
            num = (num * 2 + 1)
        elif incomingBit == 0:
            num = (num * 2)
        else:
            break

        if num % n == 0:
            print("yes")
        else:
            print("no")

C#

static void CheckDivisibility2(int n)
{
  int num = 0;
  Console.WriteLine("press any key other than" +
                    " 0 and 1 to terminate");
  while (true)
  {
    int incomingBit;
    if (Int32.TryParse(Console.ReadLine(), out incomingBit))
    {

      // read next incoming bit via standard
      // input. 0, 00, 000.. are same as int 0
      // ans 1, 01, 001, 00..001 is same as 1.
      if (incomingBit == 1)
        num = (num * 2 + 1);
      else if (incomingBit == 0)
        num = (num * 2);

      else
        break;
      if (num % n == 0)
        Console.WriteLine("yes");
      else
        Console.WriteLine("no");
    }
    else
    {
      Console.WriteLine("Invalid input. Program terminated.");
      break;
    }
  }
}

Javascript

// JavaScript
function checkDivisibility2(n) {
    let num = 0;
    console.log("Press any key other than 0 and 1 to terminate");
    while (true) {
        let incomingBit = prompt();

        // convert incoming bit to integer
        incomingBit = parseInt(incomingBit);

        // Update value of num formed so far
        if (incomingBit === 1) {
            num = num * 2 + 1;
        } else if (incomingBit === 0) {
            num = num * 2;
        } else {
            break;
        }

        if (num % n === 0) {
            console.log("yes");
        } else {
            console.log("no");
        }
    }
}

Java

import java.util.Scanner;

public class CheckDivisibility2 {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = 0;
        int num = 0;

        System.out.println("Enter a value for n:");
        n = scanner.nextInt();

        System.out.println("Press any key other than 0 and 1 to terminate");
        while (true) {
            int incomingBit = scanner.nextInt();

            if (incomingBit == 1) {
                num = (num * 2 + 1);
            } else if (incomingBit == 0) {
                num = (num * 2);
            } else {
                break;
            }

            if (num % n == 0) {
                System.out.println("yes");
            } else {
                System.out.println("no");
            }
        }

        scanner.close();
    }
}

Problem in this solution: What about the overflow. Since 0 and 1 will keep on coming and the number formed will go out of range of integer.  

Method 2 (Doesn’t cause overflow) :In this solution, we just maintain the remainder if remainder is 0, the formed number is divisible by n otherwise not. This is the same technique that is used in Automata to remember the state. Here also we are remembering the state of divisibility of input number. In order to implement this technique, we need to observe how the value of a binary number changes, when it is appended by 0 or 1. Let’s take an example. Suppose you have binary number 1. If it is appended by 0 it will become 10 (2 in decimal) means 2 times of the previous value. If it is appended by 1 it will become 11(3 in decimal), 2 times of previous value +1. 
How does it help in getting the remainder?
Any number (n) can be written in the form m = an + r where a, n and r are integers and r is the remainder. So when m is multiplied by any number so the remainder. Suppose m is multiplied by x so m will be mx = xan + xr. so (mx)%n = (xan)%n + (xr)%n = 0 + (xr)%n = (xr)%n; We need to just do the above calculation (calculation of value of number when it is appended by 0 or 1 ) only over remainder.

When a binary number is appended by 0 (means 
multiplied by 2), the new remainder can be 
calculated based on current remainder only.
r = 2*r % n;

And when a binary number is appended by 1.
r = (2*r + 1) % n; 

CPP

// C++ program to check divisibility in a stream
#include <iostream>
using namespace std;

/*  A very simple implementation of the logic,
    without error handling. just to demonstrate
    the above theory. This simple version not
    restricting user from typing 000, 00 , 000.. ,
    because this all will be same as 0 for integer
    same is true for 1, 01, 001, 000...001 is same
    as 1, so ignore this type of error handling
    while reading just see the main logic is correct. */
void CheckDivisibility(int n)
{
    int remainder = 0;
    std::cout << "press any key other than 0"
                 " and 1 to terminate \n";
    while (true)
    {
        // Read next incoming bit via standard
        // input. 0, 00, 000.. are same as int 0
        // ans 1, 01, 001, 00..001 is same as 1.
        int incomingBit;
        cin >> incomingBit;

        // Update remainder
        if (incomingBit == 1)
            remainder = (remainder * 2 + 1) % n;
        else if (incomingBit == 0)
            remainder = (remainder * 2) % n;
        else
            break;

        // If remainder is 0.
        if (remainder % n == 0)
            cout << "yes \n";
        else
            cout << "no \n";
    }
}

// Driver code
int main()
{
    CheckDivisibility(3);
    return 0;
}

Java

// Java program to check divisibility in a stream
import java.util.Scanner;

/*  A very simple implementation of the logic,
    without error handling. just to demonstrate
    the above theory. This simple version not
    restricting user from typing 000, 00 , 000.. ,
    because this all will be same as 0 for integer
    same is true for 1, 01, 001, 000...001 is same
    as 1, so ignore this type of error handling
    while reading just see the main logic is correct. */
class GFG {
    static void CheckDivisibility(int n)
    {
        Scanner console = new Scanner(System.in);
        int remainder = 0;
        System.out.print("press any key other than 0 and 1 to terminate \n");
        while (true)
        {
            // Read next incoming bit via standard
            // input. 0, 00, 000.. are same as int 0
            // ans 1, 01, 001, 00..001 is same as 1.
            int incomingBit = console.nextInt();

            // Update remainder
            if (incomingBit == 1)
                remainder = (remainder * 2 + 1) % n;
            else if (incomingBit == 0)
                remainder = (remainder * 2) % n;
            else
                break;

            // If remainder is 0.
            if (remainder % n == 0)
                System.out.print("yes \n");
            else
                System.out.print("no \n");
        }
}

    //Driver code
    public static void main(String[] args) {
        CheckDivisibility(3);
    }
}

//This code is contributed by phasing17

Python3

#Python3 program to check divisibility in a stream

'''  A very simple implementation of the logic,
    without error handling. just to demonstrate
    the above theory. This simple version not
    restricting user from typing 000, 00 , 000.. ,
    because this all will be same as 0 for integer
    same is true for 1, 01, 001, 000...001 is same
    as 1, so ignore this type of error handling
    while reading just see the main logic is correct. '''

def CheckDivisibility(n):

    remainder = 0
    print("press any key other than 0"
                 " and 1 to terminate")
    while (True):

        # Read next incoming bit via standard
        # input. 0, 00, 000.. are same as int 0
        # ans 1, 01, 001, 00..001 is same as 1.
        incomingBit = int(input())

        # Update remainder
        if (incomingBit == 1):
            remainder = (remainder * 2 + 1) % n
        elif (incomingBit == 0):
            remainder = (remainder * 2) % n
        else:
            break

        # If remainder is 0.
        if (remainder % n == 0):
            print("yes")
        else:
            print("no")



# Driver code
CheckDivisibility(3)

#this code is contributed by phasing17

C#

// C# program to check divisibility in a stream
using System;

/*  A very simple implementation of the logic,
    without error handling. just to demonstrate
    the above theory. This simple version not
    restricting user from typing 000, 00 , 000.. ,
    because this all will be same as 0 for integer
    same is true for 1, 01, 001, 000...001 is same
    as 1, so ignore this type of error handling
    while reading just see the main logic is correct. */
public class GFG
{
    static void CheckDivisibility(int n)
    {
        int remainder = 0;
        Console.Write("press any key other than 0 and 1 to terminate \n");
        while (true)
        {
            // Read next incoming bit via standard
            // input. 0, 00, 000.. are same as int 0
            // ans 1, 01, 001, 00..001 is same as 1.
            int incomingBit = Convert.ToInt32(Console.ReadLine());

            // Update remainder
            if (incomingBit == 1)
                remainder = (remainder * 2 + 1) % n;
            else if (incomingBit == 0)
                remainder = (remainder * 2) % n;
            else
                break;

            // If remainder is 0.
            if (remainder % n == 0)
                Console.Write("yes \n");
            else
                Console.Write("no \n");
        }
}

    //Driver Code
    public static void Main(string[] args)
    {
          //Function call
        CheckDivisibility(3);
    }
}

// This code is contributed by phasing17

Javascript

// JavaScript program to check divisibility in a stream
// A very simple implementation of the logic,
// without error handling. just to demonstrate
// the above theory. This simple version not
// restricting user from typing 000, 00 , 000.. ,
// because this all will be same as 0 for integer
// same is true for 1, 01, 001, 000...001 is same
// as 1, so ignore this type of error handling
// while reading just see the main logic is correct.
function CheckDivisibility(n) {
    const readline = require('readline').createInterface({
        input: process.stdin,
        output: process.stdout
    });
    let remainder = 0;
    console.log("press any key other than 0 and 1 to terminate");
    readline.on('line', (input) => {
        // Read next incoming bit via standard
        // input. 0, 00, 000.. are same as int 0
        // ans 1, 01, 001, 00..001 is same as 1.
        const incomingBit = parseInt(input);
        // Update remainder
        if (incomingBit === 1)
            remainder = (remainder * 2 + 1) % n;
        else if (incomingBit === 0)
            remainder = (remainder * 2) % n;
        else {
            readline.close();
            return;
        }
        // If remainder is 0.
        if (remainder % n === 0)
            console.log("yes");
        else
            console.log("no");
    });
}

// Driver code
CheckDivisibility(3);

// This code is contributed by shivamsharma215

Input:

1
0
1
0
1
-1

Output:

Press any key other than 0 and 1 to terminate 
no 
no 
no 
no 
yes 

Time Complexity: O(N), where N is the number of inputs
Auxiliary Space: O(1)

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