Check if decimal representation of Binary String is divisible by 9 or not

Given a binary string S of length N, the task is to check if the decimal representation of the binary string is divisible by 9 or not.

Examples:

Input: S = 1010001
Output:Yes
Explanation: The decimal representation of the binary string S is 81, which is divisible by 9. Therefore, the required output is Yes.

Input: S = 1010011
Output: No
Explanation: The decimal representation of the binary string S is 83, which is not divisible by 9. Therefore, the required output is No.

Naive Approach: The simplest approach to solve this problem is to convert the binary number into a decimal number and check if the decimal number is divisible by 9 or not. If found to be true then print True. Otherwise, print False.

C++

#include<iostream>
#include<bitset>
using namespace std;

void is_binary_divisible_by_9(string s){
    bitset<32> decimal_num(s);
    if(decimal_num.to_ulong() % 9 == 0){
        cout << "Yes\n";
    } else {
        cout << "No\n";
    }
}

// Example Usage
int main() {
    string s = "1010001";
    is_binary_divisible_by_9(s);
    return 0;
}

Java

public class BinaryDivisibleByNine {

    // Function to check if binary string
    // is divisible by 9 or not
    public static String isBinaryDivisibleByNine(String s) {

        // Convert the binary string to its decimal representation
        int decimal_num = Integer.parseInt(s, 2);

        // Check if the decimal representation is divisible by 9
        if (decimal_num % 9 == 0) {
            return "Yes";
        } else {
            return "No";
        }
    }

    public static void main(String[] args) {

        String s = "1010001";

        System.out.println(isBinaryDivisibleByNine(s));
    }
}

Python3

def is_binary_divisible_by_9(s):
    decimal_num = int(s, 2)
    if decimal_num % 9 == 0:
        print("Yes")
    else:
        print("No")

# Example Usage
s = '1010001'
is_binary_divisible_by_9(s)

C#

using System;

public class BinaryDivisibleByNine {
    // Function to check if binary string is divisible by 9
    // or not
    public static string IsBinaryDivisibleByNine(string s)
    {
        // Convert the binary string to its decimal
        // representation
        int decimal_num = Convert.ToInt32(s, 2);

        // Check if the decimal representation is divisible
        // by 9
        if (decimal_num % 9 == 0) {
            return "Yes";
        }
        else {
            return "No";
        }
    }

    public static void Main(string[] args)
    {
        string s = "1010001";
        Console.WriteLine(IsBinaryDivisibleByNine(s));
    }
}

Javascript

function isBinaryDivisibleByNine(s) {

    // Convert the binary string to its decimal representation
    let decimal_num = parseInt(s, 2);

    // Check if the decimal representation is divisible by 9
    if (decimal_num % 9 === 0) {
        return "Yes";
    } else {
        return "No";
    }
}

let s = "1010001";

console.log(isBinaryDivisibleByNine(s));

Output

Yes

Time Complexity: O(N)
Auxiliary Space: O(1)

Efficient Approach: If the length of a binary string is greater than 64 then the decimal representation of the binary string will cause an overflow. Therefore, to reduce the overflow issue the idea is to convert the binary string into the octal representation and check if the octal representation of the binary string is divisible by 9 or not. Follow the steps below to solve the problem:

• Convert the binary string into octal representation.
• Initialize a variable, say Oct_9 to store the octal representation of 9.
• Find the sum of digits, say evenSum present at even positions in the octal representation of the binary string.
• Find the sum of digits, say oddSum present at odd positions in the octal representation of the binary string.
• Check if abs(oddSum – EvenSum) % Oct_9 == 0 or not. If found to be true, then print Yes.
• Otherwise, print No.

Below is the implementation of the above approach:

C++

// C++ program to implement
// the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to convert the binary string
// into octal representation
string ConvertequivalentBase8(string S)
{
    // Stores binary representation of
    // the decimal value [0 - 7]
    map<string, char> mp;

    // Stores the decimal values
    // of binary strings [0 - 7]
    mp["000"] = '0';
    mp["001"] = '1';
    mp["010"] = '2';
    mp["011"] = '3';
    mp["100"] = '4';
    mp["101"] = '5';
    mp["110"] = '6';
    mp["111"] = '7';

    // Stores length of S
    int N = S.length();

    if (N % 3 == 2) {

        // Update S
        S = "0" + S;
    }
    else if (N % 3 == 1) {

        // Update S
        S = "00" + S;
    }

    // Update N
    N = S.length();

    // Stores octal representation
    // of the binary string
    string oct;

    // Traverse the binary string
    for (int i = 0; i < N; i += 3) {

        // Stores 3 consecutive characters
        // of the binary string
        string temp = S.substr(i, 3);

        // Append octal representation
        // of temp
        oct.push_back(mp[temp]);
    }

    return oct;
}

// Function to check if binary string
// is divisible by 9 or not
string binString_div_9(string S, int N)
{
    // Stores octal representation
    // of S
    string oct;

    oct = ConvertequivalentBase8(S);

    // Stores sum of elements present
    // at odd positions of oct
    int oddSum = 0;

    // Stores sum of elements present
    // at odd positions of oct
    int evenSum = 0;

    // Stores length of oct
    int M = oct.length();

    // Traverse the string oct
    for (int i = 0; i < M; i += 2) {
        // Update oddSum
        oddSum += int(oct[i] - '0');
    }

    // Traverse the string oct
    for (int i = 1; i < M; i += 2) {
        // Update evenSum
        evenSum += int(oct[i] - '0');
    }

    // Stores cotal representation
    // of 9
    int Oct_9 = 11;

    // If absolute value of (oddSum
    // - evenSum) is divisible by Oct_9
    if (abs(oddSum - evenSum) % Oct_9
        == 0) {
        return "Yes";
    }
    return "No";
}

// Driver Code
int main()
{
    string S = "1010001";
    int N = S.length();
    cout << binString_div_9(S, N);
}

Java

// Java program to implement
// the above approach
import java.util.*;
import java.lang.*;
import java.io.*;

class GFG{

// Function to convert the binary string
// into octal representation
static String ConvertequivalentBase8(String S)
{

    // Stores binary representation of
    // the decimal value [0 - 7]
    HashMap<String,
            Character> mp = new HashMap<String,
                                        Character>();

    // Stores the decimal values
    // of binary Strings [0 - 7]
    mp.put("000", '0');
    mp.put("001", '1');
    mp.put("010", '2');
    mp.put("011", '3');
    mp.put("100", '4');
    mp.put("101", '5');
    mp.put("110", '6');
    mp.put("111", '7');

    // Stores length of S
    int N = S.length();

    if (N % 3 == 2)
    {

        // Update S
        S = "0" + S;
    }
    else if (N % 3 == 1)
    {

        // Update S
        S = "00" + S;
    }

    // Update N
    N = S.length();

    // Stores octal representation
    // of the binary String
    String oct = "";

    // Traverse the binary String
    for(int i = 0; i < N; i += 3)
    {

        // Stores 3 consecutive characters
        // of the binary String
        String temp = S.substring(i, i + 3);

        // Append octal representation
        // of temp
        oct += mp.get(temp);
    }
    return oct;
}

// Function to check if binary String
// is divisible by 9 or not
static String binString_div_9(String S, int N)
{

    // Stores octal representation
    // of S
    String oct = "";

    oct = ConvertequivalentBase8(S);

    // Stores sum of elements present
    // at odd positions of oct
    int oddSum = 0;

    // Stores sum of elements present
    // at odd positions of oct
    int evenSum = 0;

    // Stores length of oct
    int M = oct.length();

    // Traverse the String oct
    for(int i = 0; i < M; i += 2)

        // Update oddSum
        oddSum += (oct.charAt(i) - '0');

    // Traverse the String oct
    for(int i = 1; i < M; i += 2)
    {

        // Update evenSum
        evenSum += (oct.charAt(i) - '0');
    }

    // Stores octal representation
    // of 9
    int Oct_9 = 11;

    // If absolute value of (oddSum
    // - evenSum) is divisible by Oct_9
    if (Math.abs(oddSum - evenSum) % Oct_9 == 0)
    {
        return "Yes";
    }
    return "No";
}

// Driver Code
public static void main(String[] args)
{
    String S = "1010001";
    int N = S.length();

    System.out.println(binString_div_9(S, N));
}
}

// This code is contributed by grand_master

Python3

# Python3 program to implement
# the above approach

# Function to convert the binary
# string into octal representation
def ConvertequivalentBase8(S):

    # Stores binary representation of
    # the decimal value [0 - 7]
    mp = {}

    # Stores the decimal values
    # of binary strings [0 - 7]
    mp["000"] = '0'
    mp["001"] = '1'
    mp["010"] = '2'
    mp["011"] = '3'
    mp["100"] = '4'
    mp["101"] = '5'
    mp["110"] = '6'
    mp["111"] = '7'

    # Stores length of S
    N = len(S)

    if (N % 3 == 2):

        # Update S
        S = "0" + S

    elif (N % 3 == 1):

        # Update S
        S = "00" + S

    # Update N
    N = len(S)

    # Stores octal representation
    # of the binary string
    octal = ""

    # Traverse the binary string
    for i in range(0, N, 3):

        # Stores 3 consecutive characters
        # of the binary string
        temp = S[i: i + 3]

        # Append octal representation
        # of temp
        if temp in mp:
           octal += (mp[temp])

    return octal

# Function to check if binary string
# is divisible by 9 or not
def binString_div_9(S, N):

    # Stores octal representation
    # of S
    octal = ConvertequivalentBase8(S)

    # Stores sum of elements present
    # at odd positions of oct
    oddSum = 0

    # Stores sum of elements present
    # at odd positions of oct
    evenSum = 0

    # Stores length of oct
    M = len(octal)

    # Traverse the string oct
    for i in range(0, M, 2):

        # Update oddSum
        oddSum += ord(octal[i]) - ord('0')

    # Traverse the string oct
    for i in range(1, M, 2):

        # Update evenSum
        evenSum += ord(octal[i]) - ord('0')

    # Stores cotal representation
    # of 9
    Oct_9 = 11

    # If absolute value of (oddSum
    # - evenSum) is divisible by Oct_9
    if (abs(oddSum - evenSum) % Oct_9 == 0):
        return "Yes"

    return "No"

# Driver Code
if __name__ == "__main__":

    S = "1010001"
    N = len(S)

    print(binString_div_9(S, N))

# This code is contributed by chitranayal

C#

// C# program to implement
// the above approach
using System;
using System.Collections.Generic;

class GFG{

// Function to convert the binary string
// into octal representation
static String ConvertequivalentBase8(String S)
{

    // Stores binary representation of
    // the decimal value [0 - 7]
    Dictionary<String,
               char> mp = new Dictionary<String,
                                         char>();

    // Stores the decimal values
    // of binary Strings [0 - 7]
    mp.Add("000", '0');
    mp.Add("001", '1');
    mp.Add("010", '2');
    mp.Add("011", '3');
    mp.Add("100", '4');
    mp.Add("101", '5');
    mp.Add("110", '6');
    mp.Add("111", '7');

    // Stores length of S
    int N = S.Length;

    if (N % 3 == 2)
    {

        // Update S
        S = "0" + S;
    }
    else if (N % 3 == 1)
    {

        // Update S
        S = "00" + S;
    }

    // Update N
    N = S.Length;

    // Stores octal representation
    // of the binary String
    String oct = "";

    // Traverse the binary String
    for(int i = 0; i < N; i += 3)
    {

        // Stores 3 consecutive characters
        // of the binary String
        String temp = S.Substring(0, N);

        // Append octal representation
        // of temp
        if (mp.ContainsKey(temp))
            oct += mp[temp];
    }
    return oct;
}

// Function to check if binary String
// is divisible by 9 or not
static String binString_div_9(String S, int N)
{

    // Stores octal representation
    // of S
    String oct = "";

    oct = ConvertequivalentBase8(S);

    // Stores sum of elements present
    // at odd positions of oct
    int oddSum = 0;

    // Stores sum of elements present
    // at odd positions of oct
    int evenSum = 0;

    // Stores length of oct
    int M = oct.Length;

    // Traverse the String oct
    for(int i = 0; i < M; i += 2)

        // Update oddSum
        oddSum += (oct[i] - '0');

    // Traverse the String oct
    for(int i = 1; i < M; i += 2)
    {

        // Update evenSum
        evenSum += (oct[i] - '0');
    }

    // Stores octal representation
    // of 9
    int Oct_9 = 11;

    // If absolute value of (oddSum
    // - evenSum) is divisible by Oct_9
    if (Math.Abs(oddSum - evenSum) % Oct_9 == 0)
    {
        return "Yes";
    }
    return "No";
}

// Driver Code
public static void Main(String[] args)
{
    String S = "1010001";
    int N = S.Length;

    Console.WriteLine(binString_div_9(S, N));
}
}

// This code is contributed by shikhasingrajput

Javascript

<script>

// Javascript program to implement
// the above approach

// Function to convert the binary string
// into octal representation
function ConvertequivalentBase8(S)
{

    // Stores binary representation of
    // the decimal value [0 - 7]
    let mp = new Map();

    // Stores the decimal values
    // of binary Strings [0 - 7]
    mp.set("000", '0');
    mp.set("001", '1');
    mp.set("010", '2');
    mp.set("011", '3');
    mp.set("100", '4');
    mp.set("101", '5');
    mp.set("110", '6');
    mp.set("111", '7');

    // Stores length of S
    let N = S.length;

    if (N % 3 == 2)
    {

        // Update S
        S = "0" + S;
    }
    else if (N % 3 == 1)
    {

        // Update S
        S = "00" + S;
    }

    // Update N
    N = S.length;

    // Stores octal representation
    // of the binary String
    let oct = "";

    // Traverse the binary String
    for(let i = 0; i < N; i += 3)
    {

        // Stores 3 consecutive characters
        // of the binary String
        let temp = S.substring(i, i + 3);

        // Append octal representation
        // of temp
        oct += mp.get(temp);
    }
    return oct;
}

// Function to check if binary String
// is divisible by 9 or not
function binString_div_9(S, N)
{

    // Stores octal representation
    // of S
    let oct = "";

    oct = ConvertequivalentBase8(S);

    // Stores sum of elements present
    // at odd positions of oct
    let oddSum = 0;

    // Stores sum of elements present
    // at odd positions of oct
    let evenSum = 0;

    // Stores length of oct
    let M = oct.length;

    // Traverse the String oct
    for(let i = 0; i < M; i += 2)

        // Update oddSum
        oddSum += (oct[i] - '0');

    // Traverse the String oct
    for(let i = 1; i < M; i += 2)
    {

        // Update evenSum
        evenSum += (oct[i] - '0');
    }

    // Stores octal representation
    // of 9
    let Oct_9 = 11;

    // If absolute value of (oddSum
    // - evenSum) is divisible by Oct_9
    if (Math.abs(oddSum - evenSum) % Oct_9 == 0)
    {
        return "Yes";
    }
    return "No";
}

// Driver Code
let S = "1010001";
let N = S.length;

document.write(binString_div_9(S, N));

// This code is contributed by avanitrachhadiya2155

</script>

Output:

Yes

Time Complexity: O(N)
Auxiliary Space: O(N)