# Convert a binary number to octal

• Difficulty Level : Medium
• Last Updated : 17 Dec, 2021

The problem is to convert the given binary number (represented as string) to its equivalent octal number. The input could be very large and may not fit even into unsigned long long int.

Examples:

```Input : 110001110
Output : 616

Input  : 1111001010010100001.010110110011011
Output : 1712241.26633 ```

The idea is to consider the binary input as a string of characters and then follow the steps:

1. Get length of substring to the left and right of the decimal point(‘.’) as left_len and right_len.
2. If left_len is not a multiple of 3 add min number of 0’s in the beginning to make length of left substring a multiple of 3.
3. If right_len is not a multiple of 3 add min number of 0’s in the end to make length of right substring a multiple of 3.
4. Now, from the left extract one by one substrings of length 3 and add its corresponding octal code to the result.
5. If in between a decimal(‘.’) is encountered then add it to the result.

## C++

 `// C++ implementation to convert a binary number``// to octal number``#include ``using` `namespace` `std;`` ` `// function to create map between binary``// number and its equivalent octal``void` `createMap(unordered_map *um)``{``    ``(*um)[``"000"``] = ``'0'``;``    ``(*um)[``"001"``] = ``'1'``;``    ``(*um)[``"010"``] = ``'2'``;``    ``(*um)[``"011"``] = ``'3'``;``    ``(*um)[``"100"``] = ``'4'``;``    ``(*um)[``"101"``] = ``'5'``;``    ``(*um)[``"110"``] = ``'6'``;``    ``(*um)[``"111"``] = ``'7'``;   ``}`` ` `// Function to find octal equivalent of binary``string convertBinToOct(string bin)``{``    ``int` `l = bin.size();``    ``int` `t = bin.find_first_of(``'.'``);``     ` `    ``// length of string before '.'``    ``int` `len_left = t != -1 ? t : l;``     ` `    ``// add min 0's in the beginning to make``    ``// left substring length divisible by 3``    ``for` `(``int` `i = 1; i <= (3 - len_left % 3) % 3; i++)``        ``bin = ``'0'` `+ bin;``     ` `    ``// if decimal point exists   ``    ``if` `(t != -1)   ``    ``{``        ``// length of string after '.'``        ``int` `len_right = l - len_left - 1;``         ` `        ``// add min 0's in the end to make right``        ``// substring length divisible by 3``        ``for` `(``int` `i = 1; i <= (3 - len_right % 3) % 3; i++)``            ``bin = bin + ``'0'``;``    ``}``     ` `    ``// create map between binary and its``    ``// equivalent octal code``    ``unordered_map bin_oct_map;``    ``createMap(&bin_oct_map);``     ` `    ``int` `i = 0;``    ``string octal = ``""``;``     ` `    ``while` `(1)``    ``{``        ``// one by one extract from left, substring``        ``// of size 3 and add its octal code``        ``octal += bin_oct_map[bin.substr(i, 3)];``        ``i += 3;``        ``if` `(i == bin.size())``            ``break``;``             ` `        ``// if '.' is encountered add it to result``        ``if` `(bin.at(i) == ``'.'``)   ``        ``{``            ``octal += ``'.'``;``            ``i++;``        ``}``    ``}``     ` `    ``// required octal number``    ``return` `octal;   ``}`` ` `// Driver program to test above``int` `main()``{``    ``string bin = ``"1111001010010100001.010110110011011"``;``    ``cout << ``"Octal number = "``         ``<< convertBinToOct(bin);``    ``return` `0;    ``} `

## Java

 `// Java implementation to convert a``// binary number to octal number``import` `java.io.*;``import` `java.util.*;` `class` `GFG{` `// Function to create map between binary``// number and its equivalent hexadecimal``static` `void` `createMap(Map um)``{``    ``um.put(``"000"``, ``'0'``);``    ``um.put(``"001"``, ``'1'``);``    ``um.put(``"010"``, ``'2'``);``    ``um.put(``"011"``, ``'3'``);``    ``um.put(``"100"``, ``'4'``);``    ``um.put(``"101"``, ``'5'``);``    ``um.put(``"110"``, ``'6'``);``    ``um.put(``"111"``, ``'7'``);``}` `// Function to find octal equivalent of binary``static` `String convertBinToOct(String bin)``{``    ``int` `l = bin.length();``    ``int` `t = bin.indexOf(``'.'``);` `    ``// Length of string before '.'``    ``int` `len_left = t != -``1` `? t : l;` `    ``// Add min 0's in the beginning to make``    ``// left substring length divisible by 3``    ``for``(``int` `i = ``1``;``            ``i <= (``3` `- len_left % ``3``) % ``3``;``            ``i++)``        ``bin = ``'0'` `+ bin;` `    ``// If decimal point exists``    ``if` `(t != -``1``)``    ``{``        ` `        ``// Length of string after '.'``        ``int` `len_right = l - len_left - ``1``;` `        ``// add min 0's in the end to make right``        ``// substring length divisible by 3``        ``for``(``int` `i = ``1``;``                ``i <= (``3` `- len_right % ``3``) % ``3``;``                ``i++)``            ``bin = bin + ``'0'``;``    ``}` `    ``// Create map between binary and its``    ``// equivalent octal code``    ``Map bin_oct_map = ``new` `HashMap();``    ``createMap(bin_oct_map);` `    ``int` `i = ``0``;``    ``String octal = ``""``;` `    ``while` `(``true``)``    ``{``        ` `        ``// One by one extract from left, substring``        ``// of size 3 and add its octal code``        ``octal += bin_oct_map.get(``            ``bin.substring(i, i + ``3``));``        ``i += ``3``;``        ` `        ``if` `(i == bin.length())``            ``break``;` `        ``// If '.' is encountered add it to result``        ``if` `(bin.charAt(i) == ``'.'``)``        ``{``            ``octal += ``'.'``;``            ``i++;``        ``}``    ``}` `    ``// Required octal number``    ``return` `octal;``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``String bin = ``"1111001010010100001.010110110011011"``;``    ``System.out.println(``"Octal number = "` `+``                        ``convertBinToOct(bin));``}``}` `// This code is contributed by jithin`

## Python3

 `# Python3 implementation to convert a binary number``# to octal number` `# function to create map between binary``# number and its equivalent octal``def` `createMap(bin_oct_map):``    ``bin_oct_map[``"000"``] ``=` `'0'``    ``bin_oct_map[``"001"``] ``=` `'1'``    ``bin_oct_map[``"010"``] ``=` `'2'``    ``bin_oct_map[``"011"``] ``=` `'3'``    ``bin_oct_map[``"100"``] ``=` `'4'``    ``bin_oct_map[``"101"``] ``=` `'5'``    ``bin_oct_map[``"110"``] ``=` `'6'``    ``bin_oct_map[``"111"``] ``=` `'7'` `# Function to find octal equivalent of binary``def` `convertBinToOct(``bin``):``    ``l ``=` `len``(``bin``)``    ` `    ``# length of string before '.'``    ``t ``=` `-``1``    ``if` `'.'` `in` `bin``:``        ``t ``=` `bin``.index(``'.'``)``        ``len_left ``=` `t``    ``else``:``        ``len_left ``=` `l``    ` `    ``# add min 0's in the beginning to make``    ``# left substring length divisible by 3``    ``for` `i ``in` `range``(``1``, (``3` `-` `len_left ``%` `3``) ``%` `3` `+` `1``):``        ``bin` `=` `'0'` `+` `bin``    ` `    ``# if decimal point exists``    ``if` `(t !``=` `-``1``):``        ` `        ``# length of string after '.'``        ``len_right ``=` `l ``-` `len_left ``-` `1``        ` `        ``# add min 0's in the end to make right``        ``# substring length divisible by 3``        ``for` `i ``in` `range``(``1``, (``3` `-` `len_right ``%` `3``) ``%` `3` `+` `1``):``            ``bin` `=` `bin` `+` `'0'``    ` `    ``# create dictionary between binary and its``    ``# equivalent octal code``    ``bin_oct_map ``=` `{}``    ``createMap(bin_oct_map)``    ``i ``=` `0``    ``octal ``=` `""``    ` `    ``while` `(``True``) :``        ` `        ``# one by one extract from left, substring``        ``# of size 3 and add its octal code``        ``octal ``+``=` `bin_oct_map[``bin``[i:i ``+` `3``]]``        ``i ``+``=` `3``        ``if` `(i ``=``=` `len``(``bin``)):``            ``break``            ` `        ``# if '.' is encountered add it to result``        ``if` `(``bin``[i] ``=``=` `'.'``):``            ``octal ``+``=` `'.'``            ``i ``+``=` `1``            ` `    ``# required octal number``    ``return` `octal` `# Driver Code``bin` `=` `"1111001010010100001.010110110011011"``print``(``"Octal number = "``,``       ``convertBinToOct(``bin``))` `# This code is contributed``# by Atul_kumar_Shrivastava`

## C#

 `// C# implementation to convert a``// binary number to octal number` `using` `System;``using` `System.Collections.Generic;` `public` `class` `GFG{` `// Function to create map between binary``// number and its equivalent hexadecimal``static` `void` `createMap(Dictionary um)``{``    ``um.Add(``"000"``, ``'0'``);``    ``um.Add(``"001"``, ``'1'``);``    ``um.Add(``"010"``, ``'2'``);``    ``um.Add(``"011"``, ``'3'``);``    ``um.Add(``"100"``, ``'4'``);``    ``um.Add(``"101"``, ``'5'``);``    ``um.Add(``"110"``, ``'6'``);``    ``um.Add(``"111"``, ``'7'``);``}` `// Function to find octal equivalent of binary``static` `String convertBinToOct(String bin)``{``    ``int` `l = bin.Length;``    ``int` `t = bin.IndexOf(``'.'``);``    ``int` `i = 0;``    ``// Length of string before '.'``    ``int` `len_left = t != -1 ? t : l;` `    ``// Add min 0's in the beginning to make``    ``// left substring length divisible by 3``    ``for``(i = 1;``            ``i <= (3 - len_left % 3) % 3;``            ``i++)``        ``bin = ``'0'` `+ bin;` `    ``// If decimal point exists``    ``if` `(t != -1)``    ``{``        ` `        ``// Length of string after '.'``        ``int` `len_right = l - len_left - 1;` `        ``// add min 0's in the end to make right``        ``// substring length divisible by 3``        ``for``(i = 1;``                ``i <= (3 - len_right % 3) % 3;``                ``i++)``            ``bin = bin + ``'0'``;``    ``}` `    ``// Create map between binary and its``    ``// equivalent octal code``    ``Dictionary bin_oct_map = ``new` `Dictionary();``    ``createMap(bin_oct_map);` `    ``i = 0;``    ``String octal = ``""``;` `    ``while` `(``true``)``    ``{``        ` `        ``// One by one extract from left, substring``        ``// of size 3 and add its octal code``        ``octal += bin_oct_map[``            ``bin.Substring(i,  3)];``        ``i += 3;``        ` `        ``if` `(i == bin.Length)``            ``break``;` `        ``// If '.' is encountered add it to result``        ``if` `(bin[i] == ``'.'``)``        ``{``            ``octal += ``'.'``;``            ``i++;``        ``}``    ``}` `    ``// Required octal number``    ``return` `octal;``}` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``String bin = ``"1111001010010100001.010110110011011"``;``    ``Console.WriteLine(``"Octal number = "` `+``                        ``convertBinToOct(bin));``}``}` `// This code is contributed by 29AjayKumar`

## Javascript

 ``

Output:

`Octal number = 1712241.26633`

Time Complexity: O(n), where n is the length of string.

This article is contributed by Ayush Jauhari. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

My Personal Notes arrow_drop_up