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.

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 ``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

 ``

Javascript

 ``
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.

C++

 `// C++ program to check if the binary string is divisible``// by 3.``#include ``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

 ``
Output
`Yes`

Time Complexity:   where N is the number of bits.

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