Skip to content
Related Articles

Related Articles

Improve Article

Convert a binary number to octal

  • Difficulty Level : Medium
  • Last Updated : 11 Jun, 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:  

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

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 <bits/stdc++.h>
using namespace std;
  
// function to create map between binary
// number and its equivalent octal
void createMap(unordered_map<string, char> *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<string, char> 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<String, Character> 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<String,
        Character> bin_oct_map = new HashMap<String,
                                             Character>();
    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

Javascript




<script>
 
// Javascript implementation to convert a
// binary number to octal number
 
// Function to create map between binary
// number and its equivalent hexadecimal
function createMap(um)
{
    um.set("000", '0');
    um.set("001", '1');
    um.set("010", '2');
    um.set("011", '3');
    um.set("100", '4');
    um.set("101", '5');
    um.set("110", '6');
    um.set("111", '7');
}
 
// Function to find octal equivalent of binary
function convertBinToOct(bin)
{
     
    let l = bin.length;
    let t = bin.indexOf('.');
 
    // Length of string before '.'
    let len_left = t != -1 ? t : l;
 
    // Add min 0's in the beginning to make
    // left substring length divisible by 3
    for(let i = 1;
            i <= (3 - len_left % 3) % 3;
            i++)
        bin = '0' + bin;
 
    // If decimal point exists
    if (t != -1)
    {
         
        // Length of string after '.'
        let len_right = l - len_left - 1;
 
        // add min 0's in the end to make right
        // substring length divisible by 3
        for(let i = 1;
                i <= (3 - len_right % 3) % 3;
                i++)
            bin = bin + '0';
    }
 
    // Create map between binary and its
    // equivalent octal code
    let bin_oct_map = new Map();
 
    createMap(bin_oct_map);
 
    let i = 0;
    let octal = "";
 
    while (true)
    {
         
        // One by one extract from left, substring
        // of size 3 and add its octal code
        octal += bin_oct_map.get(bin.substr(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
let bin = "1111001010010100001.010110110011011";
document.write("Octal number = " +
               convertBinToOct(bin));
 
// This code is contributed by gfgking
 
</script>

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
Recommended Articles
Page :