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

• Difficulty Level : Hard
• Last Updated : 15 Jun, 2021

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.

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 ``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 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 mp = ``new` `HashMap();` `    ``// 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 mp = ``new` `Dictionary();` `    ``// 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

 ``

Output:

`Yes`

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

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