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Check if binary string multiple of 3 using DFA

  • Difficulty Level : Easy
  • Last Updated : 23 Sep, 2021

Given a string of binary characters, check if it is multiple of 3 or not.
Examples : 

Input :  1 0 1 0
Output : NO
Explanation : (1 0 1 0) is 10 and hence
not a multiple of 3

Input :  1 1 0 0
Output : YES
Explanation : (1 1 0 0) is 12 and hence 
a multiple of 3

Approach 1 : One simple method is to convert the binary number into its decimal representation and then check if it is a multiple of 3 or not. Now, when it comes to DFA (Deterministic Finite Automata), there is no concept of memory i.e. you cannot store the string when it is provided, so the above method would not be applicable. In simple terms, a DFA takes a string as input and process it. If it reaches final state, it is accepted, else rejected. As you cannot store the string, so input is taken character by character.

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The DFA for given problem is : 



As, when a number is divided by 3, there are only 3 possibilities. The remainder can be either 0, 1 or 2. Here, state 0 represents that the remainder when the number is divided by 3 is 0. State 1 represents that the remainder when the number is divided by 3 is 1 and similarly state 2 represents that the remainder when the number is divided by 3 is 2. So if a string reaches state 0 in the end, it is accepted otherwise rejected.

Below is the implementation of above approach : 

C++




// C++ Program to illustrate
// DFA for multiple of 3
#include <bits/stdc++.h>
using namespace std;
 
// checks if binary characters
// are multiple of 3
bool isMultiple3(char c[], int size)
{
    // initial state is 0th
    char state = '0';
 
    for (int i = 0; i < size; i++) {
 
        // storing binary digit
        char digit = c[i];
 
        switch (state) {
 
        // when state is 0
        case '0':
            if (digit == '1')
                state = '1';
            break;
 
        // when state is 1
        case '1':
            if (digit == '0')
                state = '2';
            else
                state = '0';
            break;
 
        // when state is 2
        case '2':
            if (digit == '0')
                state = '1';
            break;
        }
    }
 
    // if final state is 0th state
    if (state == '0')
        return true;
    return false;
}
 
// Driver's Code
int main()
{
    // size of binary array
    int size = 5;
 
    // array of binary numbers
    // Here it is 21 in decimal
    char c[] = { '1', '0', '1', '0', '1' };
 
    // if binary numbers are a multiple of 3
    if (isMultiple3(c, size))
        cout << "YES\n";
    else
        cout << "NO\n";
 
    return 0;
}

Java




// Java Program to illustrate
// DFA for multiple of 3
import java.io.*;
 
class GFG
{
    // checks if binary characters
    // are multiple of 3
    static boolean isMultiple3(char c[], int size)
    {
        // initial state is 0th
        char state = '0';
     
        for (int i = 0; i < size; i++) {
     
            // storing binary digit
            char digit = c[i];
     
            switch (state) {
     
            // when state is 0
            case '0':
                if (digit == '1')
                    state = '1';
                break;
     
            // when state is 1
            case '1':
                if (digit == '0')
                    state = '2';
                else
                    state = '0';
                break;
     
            // when state is 2
            case '2':
                if (digit == '0')
                    state = '1';
                break;
            }
        }
     
        // if final state is 0th state
        if (state == '0')
            return true;
        return false;
    }
     
 
 
 
    // Driver Code
    public static void main (String[] args)
    {
        // size of binary array
        int size = 5;
     
        // array of binary numbers
        // Here it is 21 in decimal
        char c[] = { '1', '0', '1', '0', '1' };
     
        // if binary numbers are a multiple of 3
        if (isMultiple3(c, size))
            System.out.println ("YES");
        else
            System.out.println ("NO");
             
    }
}
// This code is contributed by vt_m

Python3




# Python program to check if the binary String is divisible
# by 3.
 
 
 
 
# Function to check if the binary String is divisible by 3.
def CheckDivisibilty(A):
    oddbits = 0;
    evenbits = 0;
    for counter in range(len(A)):
 
        # checking if the bit is nonzero
        if (A[counter] == '1'):
 
            # checking if the nonzero bit is at even
            # position
            if (counter % 2 == 0):
                evenbits+=1;
            else:
                oddbits+=1;
             
         
     
 
    # Checking if the difference of non-zero oddbits and
    # evenbits is divisible by 3.
    if (abs(oddbits - evenbits) % 3 == 0):
        print("Yes" + "");
    else:
        print("No" + "");
     
 
 
# Driver Program
if __name__ == '__main__':
    A = "10101";
    CheckDivisibilty(A);
     
 
 
# This code contributed by umadevi9616 Added code in Python

C#




// C# Program to illustrate
// DFA for multiple of 3
using System;
 
class GFG {
     
    // checks if binary characters
    // are multiple of 3
    static bool isMultiple3(char []c, int size)
    {
        // initial state is 0th
        char state = '0';
     
        for (int i = 0; i < size; i++)
        {
     
            // storing binary digit
            char digit = c[i];
     
            switch (state)
            {
     
                // when state is 0
                case '0':
                    if (digit == '1')
                        state = '1';
                    break;
         
                // when state is 1
                case '1':
                    if (digit == '0')
                        state = '2';
                    else
                        state = '0';
                    break;
         
                // when state is 2
                case '2':
                    if (digit == '0')
                        state = '1';
                    break;
            }
        }
     
        // if final state is 0th state
        if (state == '0')
            return true;
             
        return false;
    }
 
    // Driver Code
    public static void Main ()
    {
         
        // size of binary array
        int size = 5;
     
        // array of binary numbers
        // Here it is 21 in decimal
        char []c = { '1', '0', '1', '0', '1' };
     
        // if binary numbers are a multiple of 3
        if (isMultiple3(c, size))
            Console.WriteLine ("YES");
        else
            Console.WriteLine ("NO");
    }
}
 
// This code is contributed by vt_m.

PHP




<?php
// PHP Program to illustrate
// DFA for multiple of 3
 
// checks if binary characters
// are multiple of 3
function isMultiple3($c,$size)
{
     
    // initial state is 0th
    $state = '0';
 
    for ($i = 0; $i < $size; $i++)
    {
 
        // storing binary digit
        $digit = $c[$i];
 
        switch ($state)
        {
 
            // when state is 0
            case '0':
                if ($digit == '1')
                    $state = '1';
                break;
     
            // when state is 1
            case '1':
                if ($digit == '0')
                    $state = '2';
                else
                    $state = '0';
                break;
     
            // when state is 2
            case '2':
                if ($digit == '0')
                    $state = '1';
                break;
        }
    }
 
    // if final state is 0th state
    if ($state == '0')
        return true;
    return false;
}
 
    // Driver Code
 
    // size of binary array
    $size = 5;
 
    // array of binary numbers
    // Here it is 21 in decimal
    $c= array('1', '0', '1', '0', '1');
 
    // if binary numbers are
    // a multiple of 3
    if (isMultiple3($c, $size))
        echo "YES\n";
    else
        echo "NO\n";
 
//This code is contributed by mits
?>

Javascript




<script>
 
// JavaScript Program to illustrate
// DFA for multiple of 3
 
    // checks if binary characters
    // are multiple of 3
    function isMultiple3(c, size)
    {
        // initial state is 0th
        let state = '0';
      
        for (let i = 0; i < size; i++) {
      
            // storing binary digit
            let digit = c[i];
      
            switch (state) {
      
            // when state is 0
            case '0':
                if (digit == '1')
                    state = '1';
                break;
      
            // when state is 1
            case '1':
                if (digit == '0')
                    state = '2';
                else
                    state = '0';
                break;
      
            // when state is 2
            case '2':
                if (digit == '0')
                    state = '1';
                break;
            }
        }
      
        // if final state is 0th state
        if (state == '0')
            return true;
        return false;
    }
   
  
// Driver code
 
        // size of binary array
        let size = 5;
      
        // array of binary numbers
        // Here it is 21 in decimal
        let c = [ '1', '0', '1', '0', '1' ];
      
        // if binary numbers are a multiple of 3
        if (isMultiple3(c, size))
            document.write ("YES");
        else
            document.write ("NO");
 
</script>
Output
YES

Approach 2:We will check if the difference between the number of non-zero odd bit positions and non-zero even bit positions is divisible by 3 or not.

Mathematically -> |odds-evens| divisible by 3.

Code:

C++




// C++ program to check if the binary string is divisible
// by 3.
#include <bits/stdc++.h>
using namespace std;
// Function to check if the binary string is divisible by 3.
void CheckDivisibilty(string A)
{
    int oddbits = 0, evenbits = 0;
    for (int counter = 0; counter < A.length(); counter++) {
        // checking if the bit is nonzero
        if (A[counter] == '1') {
            // checking if the nonzero bit is at even
            // position
            if (counter % 2 == 0) {
                evenbits++;
            }
            else {
                oddbits++;
            }
        }
    }
    // Checking if the difference of non-zero oddbits and
    // evenbits is divisible by 3.
    if (abs(oddbits - evenbits) % 3 == 0) {
        cout << "Yes" << endl;
    }
    else {
        cout << "No" << endl;
    }
}
// Driver Program
int main()
{
    string A = "10101";
    CheckDivisibilty(A);
    return 0;
}

Java




// Java program to check if the binary String is divisible
// by 3.
import java.util.*;
 
class GFG
{
   
// Function to check if the binary String is divisible by 3.
static void CheckDivisibilty(String A)
{
    int oddbits = 0, evenbits = 0;
    for (int counter = 0; counter < A.length(); counter++)
    {
       
        // checking if the bit is nonzero
        if (A.charAt(counter) == '1')
        {
           
            // checking if the nonzero bit is at even
            // position
            if (counter % 2 == 0) {
                evenbits++;
            }
            else {
                oddbits++;
            }
        }
    }
   
    // Checking if the difference of non-zero oddbits and
    // evenbits is divisible by 3.
    if (Math.abs(oddbits - evenbits) % 3 == 0) {
        System.out.print("Yes" +"\n");
    }
    else {
        System.out.print("No" +"\n");
    }
}
   
// Driver Program
public static void main(String[] args)
{
    String A = "10101";
    CheckDivisibilty(A);
}
}
 
// This code is contributed by umadevi9616

Python3




# Python program to check if the binary String is divisible
# by 3.
 
# Function to check if the binary String is divisible by 3.
def CheckDivisibilty(A):
    oddbits = 0;
    evenbits = 0;
    for counter in range(len(A)):
 
        # checking if the bit is nonzero
        if (A[counter] == '1'):
 
            # checking if the nonzero bit is at even
            # position
            if (counter % 2 == 0):
                evenbits += 1;
            else:
                oddbits += 1;
         
    # Checking if the difference of non-zero oddbits and
    # evenbits is divisible by 3.
    if (abs(oddbits - evenbits) % 3 == 0):
        print("Yes" + "");
    else:
        print("No" + "");
     
# Driver Program
if __name__ == '__main__':
    A = "10101";
    CheckDivisibilty(A);
     
# This code is contributed by umadevi9616.

C#




// C# program to check if the binary String is divisible
// by 3.
using System;
 
public class GFG
{
   
// Function to check if the binary String is divisible by 3.
static void CheckDivisibilty(String A)
{
    int oddbits = 0, evenbits = 0;
    for (int counter = 0; counter < A.Length; counter++)
    {
       
        // checking if the bit is nonzero
        if (A[counter] == '1')
        {
           
            // checking if the nonzero bit is at even
            // position
            if (counter % 2 == 0) {
                evenbits++;
            }
            else {
                oddbits++;
            }
        }
    }
   
    // Checking if the difference of non-zero oddbits and
    // evenbits is divisible by 3.
    if (Math.Abs(oddbits - evenbits) % 3 == 0) {
        Console.Write("Yes" +"\n");
    }
    else {
        Console.Write("No" +"\n");
    }
}
   
// Driver Program
public static void Main(String[] args)
{
    String A = "10101";
    CheckDivisibilty(A);
}
}
 
// This code is contributed by umadevi9616

Javascript




<script>
// javascript program to check if the binary String is divisible
// by 3.
 
    // Function to check if the binary String is divisible by 3.
    function CheckDivisibilty( A) {
        var oddbits = 0, evenbits = 0;
        for (var counter = 0; counter < A.length; counter++) {
 
            // checking if the bit is nonzero
            if (A[counter] == '1') {
 
                // checking if the nonzero bit is at even
                // position
                if (counter % 2 == 0) {
                    evenbits++;
                } else {
                    oddbits++;
                }
            }
        }
 
        // Checking if the difference of non-zero oddbits and
        // evenbits is divisible by 3.
        if (Math.abs(oddbits - evenbits) % 3 == 0) {
            document.write("Yes" + "\n");
        } else {
            document.write("No" + "\n");
        }
    }
 
    // Driver Program
        var A = "10101";
        CheckDivisibilty(A);
 
// This code is contributed by umadevi9616
</script>
Output
Yes

Time Complexity:  O(N)            where N is the number of bits.




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