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)); |
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> |
Yes
Time Complexity: O(N)
Auxiliary Space: O(N)
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