C++ Program For Binary To Octal Conversion

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

Examples:

Input: 110001110
Output: 616

Input: 1111001010010100001.010110110011011
Output: 1712241.26633 

Simple Approach

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

1. Get the length of the 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 a min number of 0’s in the beginning to make the length of the left substring a multiple of 3.
3. If right_len is not a multiple of 3 add a min number of 0’s in the end to make the length of the 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.

Below is the C++ program to implement the above approach:

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 code
int main()
{
    string bin = "1111001010010100001.010110110011011";
    cout << "Octal number = " << convertBinToOct(bin);
    return 0;
}

Output

Octal number = 1712241.26633

The complexity of the above method

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

Auxiliary space: O(1).

Optimized Approach

The steps for this approach are as follows:

1. Convert the given binary number into groups of three digits starting from the rightmost side.
2. Convert each group of three digits to its corresponding octal digit.
3. Concatenate the octal digits obtained in step 2 to get the octal equivalent of the given binary number.

Below is the C++ program to implement the above approach:

C++

// C++ program to convert binary
// to octal
#include <cmath>
#include <iostream>
#include <string>
using namespace std;

// Function to convert binary to
// octal
string binaryToOctal(string binary)
{
    // Check if the binary number is valid
    if (binary.find_first_not_of("01.") != string::npos) {
        return "Invalid binary number";
    }

    // Convert the integer part of the binary number
    // to octal
    int decimal = stoi(binary.substr(0, binary.find('.')),
                       nullptr, 2);
    string octal = "";
    while (decimal > 0) {
        octal = to_string(decimal % 8) + octal;
        decimal /= 8;
    }

    // Convert the fractional part of the binary number
    // to octal
    if (binary.find('.') != string::npos) {
        double fractional = stod(
            "0." + binary.substr(binary.find('.') + 1));
        octal += ".";
        for (int i = 0; i < 5; i++) {
            fractional *= 8;
            octal += to_string((int)floor(fractional));
            fractional -= floor(fractional);
        }
    }

    return octal;
}

// Dtriver code
int main()
{
    string binary1 = "110001110";
    string octal1 = binaryToOctal(binary1);
    cout << "Octal equivalent of " << binary1 << " is "
         << octal1 << endl;

    string binary2 = "1111001010010100001";
    string octal2 = binaryToOctal(binary2);
    cout << "Octal equivalent of " << binary2 << " is "
         << octal2 << endl;

    string binary3 = "11011.10";
    string octal3 = binaryToOctal(binary3);
    cout << "Octal equivalent of " << binary3 << " is "
         << octal3 << endl;

    string binary4 = "1002";
    string octal4 = binaryToOctal(binary4);
    cout << "Octal equivalent of " << binary4 << " is "
         << octal4 << endl;

    return 0;
}

Output

Octal equivalent of 110001110 is 616
Octal equivalent of 1111001010010100001 is 1712241
Octal equivalent of 11011.10 is 33.06314
Octal equivalent of 1002 is Invalid binary number

Complexity of the above method

Time Complexity: O(n), where n is the number of digits in the given binary number. 
Auxiliary Space: O(n/3), since we need to store the octal digits obtained for each group of three binary digits.