Given a decimal number n without floating-point. The problem is to convert the decimal number to octal number with minimum use of arithmetic operators.
Examples:
Input : n = 10 Output : 12 12 is octal equivalent of decimal 10. Input : n = 151 Output : 227
Approach: Following are the steps:
- Perform decimal to binary conversion without using arithmetic operators of the given number n. Refer this post. Let this number be bin.
- Convert the binary number bin to octal. Refer this post.
C++
// C++ implementation of decimal to octal conversion// with minimum use of arithmetic operators#include <bits/stdc++.h>using namespace std;// function for decimal to binary conversion// without using arithmetic operatorsstring decToBin(int n){ if (n == 0) return "0"; // to store the binary equivalent of decimal string bin = ""; while (n > 0) { // to get the last binary digit of the // number 'n' and accumulate it at the // beginning of 'bin' bin = ((n & 1) == 0 ? '0' : '1') + bin; // right shift 'n' by 1 n >>= 1; } // required binary number return bin;} // Function to find octal equivalent of binarystring convertBinToOct(string bin){ int l = bin.size(); // add min 0's in the beginning to make // string length divisible by 3 for (int i = 1; i <= (3 - l % 3) % 3; i++) bin = '0' + bin; // create map between binary and its // equivalent octal code unordered_map<string, char> 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'; 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; } // required octal number return octal; }// function to find octal equivalent of decimalstring decToOctal(int n){ // convert decimal to binary string bin = decToBin(n); // convert binary to octal // required octal equivalent of decimal return convertBinToOct(bin);}// Driver program to test aboveint main(){ int n = 151; cout << decToOctal(n); return 0;} |
Java
// Java implementation of decimal to octal // conversion with minimum use of // arithmetic operatorsimport java.util.*;class GFG {// function for decimal to binary conversion// without using arithmetic operatorsstatic String decToBin(int n){ if (n == 0) return "0"; // to store the binary equivalent of decimal String bin = ""; while (n > 0) { // to get the last binary digit of the // number 'n' and accumulate it at the // beginning of 'bin' bin = ((n & 1) == 0 ? '0' : '1') + bin; // right shift 'n' by 1 n >>= 1; } // required binary number return bin;}// Function to find octal equivalent of binarystatic String convertBinToOct(String bin){ int l = bin.length(); // add min 0's in the beginning to make // string length divisible by 3 for (int i = 1; i <= (3 - l % 3) % 3; i++) bin = '0' + bin; // create map between binary and its // equivalent octal code Map<String, Character> bin_oct_map = new HashMap<>(); bin_oct_map.put("000", '0'); bin_oct_map.put("001", '1'); bin_oct_map.put("010", '2'); bin_oct_map.put("011", '3'); bin_oct_map.put("100", '4'); bin_oct_map.put("101", '5'); bin_oct_map.put("110", '6'); bin_oct_map.put("111", '7'); 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; } // required octal number return octal; }// function to find octal equivalent of decimalstatic String decToOctal(int n){ // convert decimal to binary String bin = decToBin(n); // convert binary to octal // required octal equivalent of decimal return convertBinToOct(bin);}// Driver Codepublic static void main(String[] args) { int n = 151; System.out.println(decToOctal(n));}} // This code is contributed by Rajput-Ji |
Python3
# Python3 implementation of decimal to octal conversion# with minimum use of arithmetic operators# function for decimal to binary conversion# without using arithmetic operatorsdef decToBin(n): if (n == 0): return "0" # to store the binary equivalent of decimal bin = "" while (n > 0): # to get the last binary digit of the # number 'n' and accumulate it at the # beginning of 'bin' bin = ('0' if(n & 1) == 0 else '1') + bin # right shift 'n' by 1 n >>= 1 # required binary number return bin# Function to find octal equivalent of binarydef convertBinToOct(bin): l = len(bin) # add min 0's in the beginning to make # string length divisible by 3 for i in range(1,((3 - l % 3) % 3) + 1): bin = '0' + bin # create map between binary and its # equivalent octal code bin_oct_map = dict() 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' 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 # required octal number return octal# function to find octal equivalent of decimaldef decToOctal(n): # convert decimal to binary bin = decToBin(n) # convert binary to octal # required octal equivalent of decimal return convertBinToOct(bin)# Driver program to test aboveif __name__=='__main__': n = 151 print(decToOctal(n)) # This code is contributed by pratham76 |
C#
// C# implementation of decimal to octal // conversion with minimum use of // arithmetic operatorsusing System;using System.Collections.Generic; class GFG{ // Function for decimal to binary// conversion without using // arithmetic operatorsstatic string decToBin(int n){ if (n == 0) return "0"; // To store the binary equivalent // of decimal string bin = ""; while (n > 0) { // To get the last binary digit of the // number 'n' and accumulate it at the // beginning of 'bin' bin = ((n & 1) == 0 ? '0' : '1') + bin; // Right shift 'n' by 1 n >>= 1; } // Required binary number return bin;} // Function to find octal equivalent of binarystatic string convertBinToOct(string bin){ int l = bin.Length; // Add min 0's in the beginning to make // string length divisible by 3 for(int j = 1; j <= (3 - l % 3) % 3; j++) bin = '0' + bin; // Create map between binary and its // equivalent octal code Dictionary<string, char> bin_oct_map = new Dictionary<string, char>(); bin_oct_map.Add("000", '0'); bin_oct_map.Add("001", '1'); bin_oct_map.Add("010", '2'); bin_oct_map.Add("011", '3'); bin_oct_map.Add("100", '4'); bin_oct_map.Add("101", '5'); bin_oct_map.Add("110", '6'); bin_oct_map.Add("111", '7'); 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[bin.Substring(i, 3)]; i += 3; if (i == bin.Length) break; } // Required octal number return octal; } // Function to find octal equivalent of decimalstatic string decToOctal(int n){ // Convert decimal to binary string bin = decToBin(n); // Convert binary to octal // required octal equivalent // of decimal return convertBinToOct(bin);} // Driver Codepublic static void Main(string[] args) { int n = 151; Console.Write(decToOctal(n));}} // This code is contributed by rutvik_56 |
Javascript
<script>// Javascript implementation of decimal to octal conversion// with minimum use of arithmetic operators// function for decimal to binary conversion// without using arithmetic operatorsfunction decToBin(n){ if (n == 0) return "0"; // to store the binary equivalent of decimal var bin = ""; while (n > 0) { // to get the last binary digit of the // number 'n' and accumulate it at the // beginning of 'bin' bin = ((n & 1) == 0 ? '0' : '1') + bin; // right shift 'n' by 1 n >>= 1; } // required binary number return bin;} // Function to find octal equivalent of binaryfunction convertBinToOct(bin){ var l = bin.length; // add min 0's in the beginning to make // string length divisible by 3 for (var i = 1; i <= (3 - l % 3) % 3; i++) bin = '0' + bin; // create map between binary and its // equivalent octal code var bin_oct_map = new Map(); bin_oct_map.set("000",'0'); bin_oct_map.set("001",'1'); bin_oct_map.set("010",'2'); bin_oct_map.set("011",'3'); bin_oct_map.set("100",'4'); bin_oct_map.set("101",'5'); bin_oct_map.set("110",'6'); bin_oct_map.set("111",'7'); var i = 0; var octal = ""; while (1) { // 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; } // required octal number return octal; }// function to find octal equivalent of decimalfunction decToOctal(n){ // convert decimal to binary var bin = decToBin(n); // convert binary to octal // required octal equivalent of decimal return convertBinToOct(bin);}// Driver program to test abovevar n = 151;document.write( decToOctal(n));ing// This code is contributed by itsok.</script> |
Output
227
Time Complexity: O(n), where n is the length of the binary string.
Auxiliary Space: O(n) because using extra space for string octal
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